University of Al-mustansiryah College of Science Department of Chemistry First Grade - First Term

Analytical Chemistry

Dr. Amer Saleh Mahdi Dr. Khitam Jaber Nabhan

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Analytical ChemistryFirst Grade-First Term (2018-2019)

References:-(1) “Analytical Chemistry” by Gary D. Christian, 6th Edition, 2004, John Wiley and Sons, Inc.(2) “Fundamental of Analytical Chemistry” by Doglas A.Skooge, Donald M.West and James Holler, 8th Edition, 2004.(3)-“An Introduction to Analytical Chemistry” by Doglas A. Skooge & Donald M.West, 4th Edition, 1986.

) 4 -( ’ الكمي التحليل الالعضوية التحليلية للكيمياء النظرية االساسيات . ’ واخرون، عوض كاظم هادي د تأليف ، الحجمي و .1986الوزني

Analytical Chemistry: The science seeks ever improved means of measuring the

chemical composition of natural and artificial materials by using techniques to

identify the substances which may be present in a material and to determine the

exact amounts of the identified substance. Analytical chemistry involves the

analysis of matter to determine its composition and the quantity of each kind of

matter that is present. Analytical chemists detect traces of toxic chemicals in water

and air. They also develop methods to analyze human body fluids for drugs,

poisons, and levels of medication.

Analytical chemistry consists of:(A) Qualitative analysis which deals with the identification of elements, ions, or compounds present in a sample (tells us what chemicals are present in a sample).

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(B) Quantitative analysis which is dealing with the determination of how much of one or more constituents is present (tells how much amounts of chemicals are present in a sample). This analysis can be divided into three branches:(1) Volumetric analysis (Titrimetric analysis): The analyte reacts with a measured volume of reagent of known concentration, in a process called titration. (1st grade) (2) Gravimetric analysis: usually involves the selective separation of the analyte by precipitation, followed by the very non-selective measurement of mass (of the precipitate). (2nd grade)

(3) Instrumental analysis: They are based on the measurement of a physical property of the sample, for example, an electrical property or the absorption of electromagnetic radiation. Examples are spectrophotometry (ultraviolet, visible, or infrared), fluorimetry, atomic spectroscopy (absorption, emission), mass spectrometry, nuclear magnetic resonance spectrometry (NMR), X-ray spectroscopy (absorption, fluorescence). (4th grade)

Solutions.Solution: Homogeneous mixture of two or more substance produce from dissolved (disappeared) solute particle (ions, atoms, molecules) (lesser amount) between solvent particle (larger amount).

Solute (lesser amount) + Solvent (larger amount) SolutionNaCl(s)+H 2O(l)→ Salt Solution

Concentrated Solution has a large amount of solute. Dilute Solution has a small amount of solute.

Solute

SolventGas Liquid Solid

Gas O2(g) in N2(g), Air CO2(g) in H2O(l), Soda H2(g) in Pd(s), H2 catalystLiqui

d Perfume Alcohol(l) in H2O(l), Martini

Hg(l) in Ag(s), Dental filling

Solid Dust air, Smoke industry

NaCl(s) in H2O(l), salt water, saline sol. Zn(s) in Cu(s), Brass alloy

Classification of solutions according to amount of solute:

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(1) Unsaturated solutions: if the amount of solute dissolved is less than the solubility limit, or if the amount of solute is less than capacity of solvent.

(2) Saturated solutions: is one in which no more solute can dissolve in a given amount of solvent at a given temperature, or if the amount of solute equal to capacity of solvent.(3) Super saturated solutions: solution that contains a dissolved amount of solute

that exceeds the normal solubility limit (saturated solution). Or a solution contains

a larger amount of solute than capacity of solvent at high temperature.

Classification of solution based on solute particle size:

(1) True solution: A homogeneous mixture of two or more substance in which substance (solute) has a particle size less than 1 nm dissolved in solvent. Particles of true solution cannot be filtered through filter paper and are not visible to naked eye (NaCl in water).(2) Suspension solution: heterogeneous mixtures which settles on standing and its components can be separated by filtrating (Amoxcycilline Antibiotics), particle of solute visible to naked eye.

(3) Colloidal solution: homogeneous mixture which does not settle nor are their components filterable, solute particle visible with electron microscope (milk).

Stoichiometric Calculations:-Gram atomic weight (g.Aw sometime A.wt): Is the weight of a specified number of atoms of that element (contains exactly the same number of atoms of that element as there are carbon atoms in exactly 12g of carbon 12 (this number is Avogadro’s number = 6.0221023 atoms).Gram molecular weight (gMw sometimes M.wt): Defined as the sum of the atomic weight of the atoms that make up a molecular compound.Gram formula weight (gFw sometime F.wt): The sum of the atomic weight of the atoms that make up an ionic formula. (is the more accurate description for substances that do not exist as molecules but exist as ionic compounds e.q strong electrolytes-acids, bases, salts). Sometimes use

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the term molar mass (Molecular weight, M.wt) in place of gram formula weight, gm Fw).Example (1):-Calculate the number of grams in one mole of CaSO4.7H2O (calculate gram molecular or formula weight).Solution: One mole is the formula weight expressed in grams. The formula weight is (Ca=40.08; S=32.06; O=16.00; H=1.01)

CaSO4 .7H 2O=40.08+32.06+(16.0 × 4 )+7 [ (2×1.01 )+16.00 ]=262.25 g /mol

Mole Concept:-Mole: which is Avogadro’s number (6.0221023) of atoms, molecules, ions or other species. Numerically: it is the atomic, molecular, or formula weight of a substance expressed in grams.

Mole=weight ( g )

formula weight( gmole )

,

mMole= weight (mg)

formulaweight ( mgmmole

)

Where formula weight represents the atomic or molecular weight of the substance.

Example (2):-Calculate the number of moles in 500 mg Na2WO4.

Solution:

mMole= wt (mg)

M . wt ( mgmmole

)= 500(mg)

293.8 ( mgmmole

)=1.706 mmole

Mole=mmole1000

=1.7061000

=0.00170 mol

Example (3): How many molecules are contained in 25.0 g H2. Solution:

moles H2=wt (g )Fwt

= 25.0 g

2.016 gmole

=12.40 mole

No. Molecules = No. moles × Avogadro number

=12.40×6.022X1023=7.74×1024 molecule

Example (4):-How many milligrams are in 0.250 mmole Fe2O3 (ferric oxide). Solution:

wt (mg )=mmole× M . wt ( mgmmol )

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¿0.250 mmole ×159.7 mgmmol

=39.9 mg

Problems 1

Q1: Find the number of Na+ ions in 2.92 g of Na3PO4?

Q2: Find the number of K+ ions in 3.41 mol. of K2HPO4?

Q3: Find the amount of the indicated element (in moles) in:

(a) 8.75 g of B2O3. (b) 167.2 mg of Na2B4O7 . 10H2O.

(c) 4.96 g of Mn3O4. (d) 333 mg of CaC2O4.

Q4: Find the amount in millimoles of the indicated species in:

(a) 850 mg of P2O5. (b) 40.0 g of CO2.

(c) 12.92 g of NaHCO3. (d) 57 mg of MgNH4PO4

The Molarity Concentration: - also called (Molarity, amount of concentration or substances concentration):- is measure of the concentration of a chemical species in a particular of a solute in a solution, in terms of amount of substance per unit volume of solution. In chemistry, the most commonly used unit for molarity is the number of moles per litre, having the unit symbol (mol. /L). A solution with a concentration of 1 mol. /L is said to be 1 Molar, commonly designated as 1M.

How do we express concentrations of solutions:-

1-A:-Molarity concentration for solutions prepared from dissolving solid solute in liquid solvent:-Molarity: - defined as a number of solute moles dissolved in solution volumes in litre.

Molarity ( M )= No .of mole solutesolutionvolume (L )

=moleL

Molarity ( M )= No . of mmole solutesolutionvolume (mL)

=mmolemL

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M= No of mole soluteVolume solution(L)

=

wt (g)

M . wt ( gmole

)

V (mL)

1000(mLL )

=wt (g)

M . wt ( gmole

)× 1000

V (mL)

Example (5):-A solution is prepared by dissolving 1.26 gm AgNO3 in a 250 mL volumetric flask and diluting to volume. Calculate the molarity of the silver nitrate solution. How many millimoles AgNO3 were dissolved?

Solution:

M=wt (g)

M . wt ( gmol

)× 1000

V (mL)

¿1.26(g)

169.9( gmol

)× 1000

250 (mL)=0.0297 mol /L

Millimoles=M (mmolmL )×V (mL)

¿0.0297( mmolmL )× 250 mL=7.42 mmole

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Example (6):-How many grams per millilitre of NaCl are contained in a 0.250 M solution. Solution:

M=wt (g)

M . wt ( gmol

)× 1000

V (mL)

0.250 M= wt (g)

58.4( gmol

)× 1000

1(mL) , wt ( g )=0.0146 gmL

Example (7):-How many grams Na2SO4 should be weight out to prepare 500 mL of a 0.100 M solution. Solution:

M=wt (g)

M . wt ( gmol

)×

1000( mLL

)

V (mL)

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0.10 mol /L=wt (g)

142( gmol

)×

1000( mLL

)

500(mL)

wt ( g )=0.10(mol

L )×142( gmol )

2( 1L )

=7.1 g

Example (8):- Calculate the concentration of potassium ion in grams per litter after mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2SO4. Solution:

mmol ¿¿

¿V (mL) × M ( mmolmL )+2[V (mL )× M ( mmol

mL )]¿100 (mL )× 0.250(mmol

mL )+2[200(mL)×0.1(mmolmL ) ]

¿25 mmol+2[20mmol ]¿25 mmol+40 mmol=65mmol∈300 mL

mmole=wt (mg )

M . wt ( mgmmol

)

wt=65 (mmol ) ×39.1( mgmmol )=2541.5 mg

¿2541.5(mg)

1000( mgg

)=2.541 gm∈300 mL

¿2.541 gm ×1000( mL

L)

300(mL)=8.47 ( gm

L)

Homework-(a) How many grams of K2SO4 are contained in 50 mL of 0.200 M, (b) How many millimoles of K2SO4 are present?

1-B:-Molarity concentration for solution prepared from dissolved liquid solute in liquid solvent.

M=% × Density ×1000M . wt

=% × Sp .Gr .×1000M . wt

=¿

%=(wtwt

%)= wt solute ( g )wt solution (g )

×100

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Density: is the weight per unit volume at the specified temperature, usually (gm/mL) or (gm/cm3) or (gm.cm-3) in 20C (is the ratio of the mass in (gm) and volume (mL).

Specific Gravity (sp. gr.): defined as the ratio of the mass of a body (e.g. a solution) usually at 20C to the mass of an equal volume of water at 4C (or sometimes 20C) or (is the ratio of the densities of the two substances).

Example (9):-Calculate the molarity of 28.0% NH3, specific gravity 0.898.Solution:

M . wt NH 3=14+(3 ×1 )=17

M=% × sp . gr .∨(density )× 1000

M . wt

M=

28100

×0.898 ×1000

17=16.470 mol

L=16.470 mmol

mL=16.470 M

Example (10):-How many millilitres of concentrated sulphuric acid, 94.0% (g/100g solution), density 1.831 g/cm3, are required to prepare 1 liter of a 0.100 M solution.Solution:

M=

94100

×1.831 ×1000

98.1=17.5( mmol

mL)

no . of mmol (conc . )=no . mmol(dilu .)(M 1× V 1)conc .=(M 2× V 2)dilu .

17.5 ×V 1=0.1×1000V 1=5.71 mL

Of concentrated H2SO4 must be diluted to 1L (1000mL) to prepare (become) 0.1M.Another solution:no .of mmol (conc . )=no .mmol(dilu .)

(M 1× V 1)conc .=(M 2× V 2)dilu .

%× sp . gr .∨(density )× 1000M . wt

× V 1=(M 2×V 2)dilu .

94100

×1.831× 1000

98.1×V 1=0.1× 1000

V 1=5.71 mL

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Diluting Solutions:- We often must prepare dilute solutions from more concentrated stock solutions. For example, we may prepare a dilute HCL solution from concentrated HCL to be used for titration .Or, we may have a stock standard solution from which we wish to prepare a series of more dilute standards. The millimoles of stock solution taken for dilution will be identical to the millimoles in the final diluted solution. Mstock × Vstock = Mdiluted × Vdiluted

Example (11):-You wish to prepare a calibration curve for the spectrophotometric determination of permanganate. You have a stock 0.100 M solution of KMnO4 and a series of 100 mL volumetric flasks. What volumes of the stock solution will you have to pipet into the flasks to prepare standards of 1.00, 2.00, 5.00, and 10.0x10 -3

M KMnO4 solutions?Solution-1:

1.0x10-3 M(M 1× V 1)conc .=(M 2× V 2)dilu .

0.1( mmolmL )×V 1=1.0× 10−3( mmol

mL )×100(mL)

V 1=1.0 mL stock solution (conc . ) ,Also to prepare 2.0, 5.0, 10.0x10-3 M

Example (12):-A solution contains 3.30 gm of Na2CO3.10H2O in each 15 mL.What

is its molarity? What is its normality? With how many millilitres of 3.1N Acetic

Acid, CH3COOH, will 25 ml of the carbonate react according to the equation?

Solution 2CH3COOH 2CH3COO + 2H+

Na2CO3 2Na+ + CO3=

2H+ + CO2= H2O + CO2?

With how many millilitres of 3.1 N H2SO4 will 25 ml of the carbonate react?

Solution:

M=wt (g)

M . wt ( gmol

)× 1000

V (mL)

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M= 3.30

286( gmol

)× 1000

15(mL)

M=0.77 M

N = a M = 2 × 0.77 = 1.53N

N1 V1= N2 V2 , 3.1 ×V1 =1.53 ×25

V1 =12.4 ml acetic acid.

V1 =12.4 ml H2SO4.

Example (13):-You wish to prepare 500 mL of 0.1 M K2Cr2O7 solution from a 0.250 M solution. What volume of the 0.250 M solution must be diluted to 500 mL.

(no . of mmol)conc .=(no . of mmol)dilu .

(M 1× V 1)conc .=(M 2× V 2)dilu .

0.250(mmolmL )×V 1(mL)=0.1(mmol

mL )×500 mL

V 1=200 mL

Example (14):-What is the molarity and normality of a 13.0% solution of H2SO4? To what volume should 100 ml of acid be diluted in order to prepare a 1.50 N solution?Solution: From specific gravity table in the appendix, the specific gravity of the acid is

1.090.

M=%×SP.gr×1000/M.wt =0.13×1.090×1000/98=1.45M.

N=a M = 2×1.45=2.9 N.

N1 V1 =N2 V2 2.9×100 = 1.50×V2

V2 = 193 mL.

Example (15):-Prepare 500mL of 0.010M solution of Na+ from Na2CO3.

Na2CO3→ 2 Na+¿+CO32−¿ ¿¿

0.012

0.01

M Na2 CO3= wt

M . wt× 1000

V (mL)

0.005= wt106

× 1000500

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2×wt=0.005 ×106

wt=0.005× 1062

=0.265 g Na2CO3

Example (16):- How many mL of permanganate KMnO4 (0.100 M) should be used to prepare 100 mL of 1.0×10-3 M solution.Solution:

(M 1× V 1)conc .=(M 2× V 2)dilu .

0.1( mmolmL )×V 1=1.0× 10−3( mmol

mL )×100(mL)

V 1=1.0 mL stock solution (conc . ) ,Also to prepare 2.0, 5.0, 10.0x10-3 M

Example (17):- What volume of the 0.250 M of K2Cr2O7 solution must be diluted

to prepare 500 mL of 0.1 M solution.

Solution: (M 1× V 1)conc .=(M 2× V 2)dilu .

0.250(mmolmL )×V 1(mL)=0.1(mmol

mL )×500 mL

V 1=200 mL

Example (18):- What volume of 0.40 M Ba (OH)2 must be added to 50 mL of 0.30 M NaOH to give a solution 0.50 M in OH.Solution: Let x = mL Ba (OH)2 and the final volume is (50+x) mL

mmole OH−¿=mmole NaOH +2 ×mmol Ba(OH )2¿

0.5 M × (50+x )mL=0.30 M ×50mL+2 ×(0.40 M ×(x )mL)

x=33 mL Ba(OH )2

Example (19):- Prepare 500mL of 0.010M solution of Na+ from solid Na2CO3.

Solution:

Na2CO3→ 2 Na+¿+CO32−¿ ¿¿

0.012

0.01

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M Na 2 CO3= wt

M . wt× 1000

V (mL), 0.005= wt

106× 1000

500

2 ×wt=0.005 ×106 , wt=0.01× 1062

=0.265 g Na2 CO3

Problems 2

Q1: Calculate the formula weights of the following substance :(a) BaCl2.2H2O, (b) Ag2Cr2O7 ,(c)Ca3(PO4)2.

Q2: A solution containing 10 mmol. CaCl2 is diluted to 1 L. Calculate the number of grams of CaCl2. 2H2O per milliliter of the final solution?

Q3: Calculate the molarity of each of the following solutions:

(a) 10 gm H2SO4 in 250 ml of solution,(b)6 gm NaOH in 500 ml of solution,(c)25 gm AgNO3 in 1L of solution.

Q4:Calculate the number of grams in 500 ml of each of the following solutions:(a) 0.1 M Na2SO4 ,(b) 0.250 M Fe(NH4)2(SO4)2.6H2O ,(c) 0.667 M Ca(C9H6ON)2 .

Q5: Calculate the grams of each substance required to prepare the following solutions :- (a) 250 mL of 0.1 M KOH.

(b) 1L of 0.0275 M K2Cr2O7 ,(c)500 ml of 0.05 M CuSO4 .

Q6: How many milliliters of concentration hydrochloric acid, 38 %( wt/wt), specific gravity 1.19, are required to prepare 1L of a 0.1 M solution?

Q7: Calculate the molarity of each of the following commercial acid or base solutions :( a) 70% HClO4, sp. grv. 1.668, (b) 69% HNO3, sp. grv. 1.409, (c) 85% H3PO4, sp. grv. 1.689, (d) 99.5% HC2H3O2 (acetic acid) , sp. grv. 1.051, (e) 28% NH3, sp. grv. 0.898.

Dilution calculation

Q8/A 12.5-mL portion of a solution is diluted to 500 mL, and its molarity is determined to be 0.125. What is the molarity of the original solution?

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Q9/ What volume of 0.50 M H2SO4 must be added to 65 mL of 0.20 M H2SO4 to give a final solution of 0.35 M? Assume volumes are additive.

Q10/You are required to prepare working standard solutions of 1.00 × 10−5, 2.00 × 10−5, 5.00 × 10−5, and 1.00 × 10−4 M glucose from a 0.100 M stock solution. You have available (100 mL) volumetric flasks and pipets of 1.00-, 2.00-, 5.00-, and 10.00-mL volume. Outline a procedure for preparing the working standards.

Formal concentration (Formality) F: - Chemists sometime use the term formality for solutions of ionic salts that do not exist as molecules in the solid or in solution. The concentration is given as formal (F). Formality is numerically the same as molarity.

F=wt (g)

F . wt ( gmol

)× 1000

V (mL)

The term use for solutions of ionic salts that do not exist as molecules in the solid or in solution. Operationally, formality is identical to molarity.

Example (20): Exactly 4.57 g of BaCl2.2H2O are dissolved in sufficient water to give 250mL of solution. Calculate the formal concentration of BaCl2 and Cl- in this solution. Solution:

FBaCl2= wt

Fwt× 1000

V mL=4.57

244× 1000

250=0.0749 F Ba Cl2 .2H 2O

BaCl2→ Ba2+¿+2Cl−¿¿ ¿

0.0749 F 0.0749 F ×2=0.149 F Cl−¿¿

Example (21):- Calculate the formal concentration of: (a) an aqueous solution that contains 1.80g of ethanol in 750mL. (b) An aqueous solution that contains 365mg of iodic acid HIO3 in 20.0mL (the acid is 71.0% ionized in this solution).Solution: (a)

FC2 H 5 OH= wtFwt

× 1000V mL

=1.8046.1

× 1000750

=0.0521 F C2 H 5OH

The only solute species present in significant amount in an aqueous solution of

ethanol is C2H5OH, therefore;M=F=0.0521

(b)

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F= wtFwt

× 1000V mL

=

3651000176

× 100020

=0.104 F

Here, only 29.0% (100% −¿71.0%) of the solute exists as undissociated HIO3.

Thus, the molar concentration of this species will be:29.0100

× 0.104 F=0.0302 M HI O3

Normality (N): Normal concentration: Number of equivalent solute in solution volume in litre.

N= No . of equivalentSolutionVolumn(L)

=wt (gm)

Eq . wt ¿¿¿

N= wtEq . wt

× 1000V (mL)

N=( Eq .L

)=(meq .mL

)

Equivalent weight (Eq.wt):- is the formula weight divided by the number of reacting units (H+ for acid-base and electron for oxidation-reduction reaction).

( Eq . wt ) for acid−basereaction= formula weight (F .wt )No .of H+¿∨OH−¿¿ ¿

( Eq . wt ) for oxidation−reductionreaction= formula weight (F . wt )No . of electron

Number of equivalent (Eq)=wt (gm)

Eq. wt ( gmEq

)

Number of equivalent ( Eq )=N ( EqL )×Volume (L)

Number of milliequivalent (meq )= wt (mg)

Eq . wt ( mgmeq

)

Number of milliequivalent (meq )=N (meq .mL )×Volume (mL )

Example (21):- Calculate the equivalent weight of the following substances: (a) NH3, (b) H2C2O4 (in reaction with NaOH), and (c) KMnO4 [Mn7+ is reduced to Mn2+].

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Solution: (a)

Eq wt= Mwt

No . of H+¿∨OH−¿= 17.03

1=17.03 gm

Eq .¿¿

(b)Eq wt=90.04

2=45.02 gm /Eq

(c)MnO4

−¿+8 H +¿+5 e=Mn+2+4 H 2O ¿¿

Eq wt= M . wtNo . of electron

=158.045

=31.608 gm / Eq

Example (22):-Calculate the normality of the solutions containing the following: (a) 5.300gm/L Na2CO3 (when the CO3

-2 reacts with two protons), (b) 5.267 gm/L K2Cr2O7 (the Cr6+ is reduced to Cr3+).Solution :( a)

N= wtEq . wt

× 1000V (mL)

= 5.3105.99

2

× 10001000

=0.10 Eq / L

(b)Cr2O7

2−¿+14 H+¿+6 e−¿→ 2Cr3+¿+7H 2O ¿

¿¿¿

¿ 5.267294.19

6

× 10001000

=0.1074 Eq /L

Example (23):-How many millilitres of a 0.25M solution of H2SO4 will react with 10mL of a 0.25M solution of NaOH.Solution:N=nM ¿

N H 2 SO4=2×0.25

¿0.5( EqL )∨( meq

mL )∨N

N NaOH=1 ×0.25=0.25 N

(N ×V )H 2 SO 4=(N × V )NaOH

(0.5 ×V ) H2 SO4=(0.25 ×10)NaOHV H 2 SO4

=5.0 mL

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Example (24):-A solution of sodium carbonate is prepared by dissolving 0.212 gm Na2CO3 and diluting to 100mL. Calculate the normality of the solution (a) if it is used as a monoacidic base, and (b) if it is used as a di acidic base.Solution: (a)

N= wtEq . wt

× 1000V (mL)

= 0.212106.0

1

× 1000100

=0.020 meq /mL

(b)

N= 0.212106.0

2

× 1000100

=0.040 meq /mL

Example (25):-Iodine (I2) is an oxidizing agent that in reactions with reducing agent is reduced to iodide (I-). How many grams I2 would you weigh out to prepare 100mL of a 0.10N I2 solution?Solution:I 2+2e →2 I−¿¿

N= wtEq . wt

× 1000V (mL )

0.1= wt254

2

× 1000100

wt=1.27 g

Example (26):-Calculate the normality of a solution of 0.25 g/L H2C2O4, both as an acid and as a reducing agent.Solution:

H 2C2 O4→ 2 CO2+H2 O

N= wtEq .wt

× 1000V (mL)

= 0.25990.04

2

× 10001000

¿0.00555 meq ./mL

Example (27):-How many milliequivalents are involved in 43.50mL of 0.1379 N K2Cr2O7?Solution:no .of milliequivalents=N × V

¿0.1379 meq.mL

×43.50 mL

¿5.9987 meq .

Problems 3

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Q1: To what volume must 750 ml of 2.4 N solution be diluted in order to make it 1.7N?Q2: How much 0.6 N base must be added to 750 ml of a 0.2 N base in order the solution to be 0.3 N?Q3: How many volume of 0.1421 N KOH solution are required to neutralize 13.72

mL of 0.0686 M H2SO4?

(5) Molality m (Molal concentration):-The solution concentration produce from dissolved solute (mole) in solvent (kg), molality does not change with temperature and used for physicochemical measurements.

Example (28): -Calculate the molal concentration for solution preparing from mixing 4 g NaOH with 500 gm water.Solution:

Molality (m)= wtM . wt

× 1000wt (g )

= 4 g40g /mol

× 1000500

=0.2m

Mole fraction concentration (X):- The ratio between number of mole for solute or solvent to solution, the terms used in physical chemistry (phases equilibrium for example).

Example (29):-One litter of acetic acid solution contain 80.8 g of acetic acid, the solution density 1.00978 g/cm3 or g/mL. (cm3 = mL)Solution:

Mole fraction for solute (X1)=no . mole solute (n1)

no . mole solute ( n1 )+no . mole solvent (n2)

¿( wt (g)

M . wt ( gmol

))CH3 COOH

( wt (g)

M . wt ( gmol

))CH 3COOH

+(density of solution( g

mL)× volumeof solution (mL )−wt of solute(g))(Mwt of Solvent ( g

mol))

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¿

80.8(g)

60( gmol

)

80.8(g)

60( gmol

)+( 1.009791( g

mL )×1000 mL−80.8 g

18( gmol

) )=0.025

Mole fraction for solvent ( X2)=no . mole solvent

no .mole solute ( n1 )+no . mole solvent (n2)

¿

(1.009791( gmL )× 1000 mL−80.8 g

18( gmol

) )80.8(g)

60( gmol

)+( 1.009791( g

mL )×1000 mL−80.8 g

18( gmol

) )=0.975

X1+X 2=1unit=0.025+0.975=1.00unit

Example (30):- Calculate the mole fraction(x) of water in a mixture consisting of 9.0 gm water, 120 gm Acetic Acid, and 115 gm ethyl alcohol CH3CH2OH?

Solution:-

The molecular weight of water, acetic acid, and ethyl alcohol are 18, 60 and 46 gm/mol, respectively.

Mol. water =wt / M.wt 9.0 / 18 = 0.5 mol.

Mol. acetic acid =wt / M.wt 120 / 60 = 2.0 mol.

Mol. ethyl alcohol =wt / M.wt 115 / 46 = 2.5 mol.

Sum of mole = 0.5 + 2.0 + 2.5 = 5.0 mole

Mole fraction for water X=no . mole water ( n1 )

∑ of mole

Mole fraction for water X=0.5mole5.0mole

¿0.10

20

Equilibrium Molarity:

The equilibrium molarity expresses the molar concentration of a particular species in a solution at equilibrium. To state the species molarity, it is necessary to know how the solute behaves when it is dissolved in a solvent. For example, the species molarity of H2SO4 in a solution with an analytical concentration of 1.0 M is 0.0 M because the sulphuric acid is entirely dissociated into a mixture of H+, HSO4

-, and SO42- ions; essentially no H2SO4 molecules as such are present in this

solution. The equilibrium concentrations and thus the species molarity of these three ions are 1.01, 0.99, and 0.01 M, respectively.

Example (32):- Calculate the analytical and equilibrium molar concentration of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid, Cl3CCOOH (F.wt =163.4) in 10.0mL (the acid is 73% ionized in water). Employing HA as the symbol for Cl3CCOOH, we substitute into equation (law) to obtain the analytical or total concentration of the acid.

Solution:-

CHA=wt

M . wt× 1000

V mL=

2851000163.4

× 100010

=0.174 mmol HA /mL=0.174 M HA

Because all but 27% of the acid is undissociated into H3O+ and A-, the species

concentration of HA is:

[ HA ]=CHA × 27.0100

=0.047 mmol /mL=0.047 M

The molarity of H3O+ as well as that of A- equal to the analytical concentration of

the acid minus the species concentration of undissociated acid¿

Note: the analytical concentration of HA is the sum of the species concentration of

HA and A-CHA=[ HA ]+¿

21

Concentration by percent:-

( wtwt

%)= wt solute ( g )wt solution∨sample (g )

×100

¿ wt solute (mg )wt solution∨sample (mg )

× 100

(wtV

%)= wt solute ( g )V solution∨sample (mL )

×100

¿ wt solute (mg )V solution∨sample (μL )

×100

(VV

%)= V solute ( mL)V solution∨sample (mL)

× 100

¿V solute (µ L )

V solution∨sample (μL )×100

Example (33):-Calculate the weight percentage of solution prepare by mixing 5.0g AgNO3 with 100mL water (density 1g/cm3).Solution:

(wtwt

%)= wt solute (g )wt solution ( g )

× 100

( wtwt

%)= wt solute( AgNO3) (g )wt solute+wt solvent (H 2O)( g )

×100

( wtwt

%)= 5 g

5 g+(100 mL× 1 gmL

)×100= 5 g

105 g×100=4.76 %

Example (34):-Calculate number of grams in 500mL silane solution (wt/v % = 0.859%).Solution:

( wtV

%)= wt solute ( g )V solution (mL)

× 100

0.859 %=wt NaCl (g )

500×100 , wt NaCl=0.859 ×500

100=4.25 g NaCl

Example (35):-Calculate the weight of glucose in litter solution(wt/v % = 5 %).Solution:

( wtV

%)= wt solute ( g )V solution (mL)

× 100= wt glucose ( g )V solution ( mL)

× 100

22

5 %=wt glucose (g )1000 (mL)

×100 ,wt glucose=5×1000100

=50 g

Example (36):-Calculate the volume percentage of solution preparing by mixing 50mL methyl alcohol with 200mL water.Solution:

(VV

%)= V solute ( mL)V solution∨sample (mL)

× 100

¿ V methylalcohol (mL)V methyl alcohol+V water (mL)

× 100= 50 mL(50+200 ) mL

×100=20 %

The relationship between molarity and normality with percentage concentration:-

Example (37):-Calculate the molar concentration for 0.85% (w/v %) sodium chloride solution.Solution:-

M=wt ( g )M .wt

× 1000V mL

=wt ( g )M . wt

× 1000100

=wtV

%× 1000M . wt

¿ 0.85100

× 100058.5

=0.145 M

Concentration in parts per thousand or million or billion:-

part per thousand ( ppt )( wtwt )= wt solute (g)

wt solution(sample)(g)×103=

wt (mg)wt (g)

=wt (g)wt (kg)

part per million ( ppm )(wtwt )= wt solute(g)

wt solution(sample)(g)×106=

wt (μg)wt (g)

=wt (mg )wt (kg )

23

part per billion ( ppb )( wtwt )= wt solute(g)

wt solution(sample)(g)× 109=wt (ng)

wt (g)=wt (μ g)

wt (kg)

Example (38):-A 2.6 g sample of plant tissue was analyzed and found to contain 3.6 g zinc, what is the concentration of zinc in the plant in ppm? In ppb?Solution:-

ppm=wt (μ g)wt (g)

=3.6 μ g2.6 g

=1.4 μ gg

=1.4 ppm

ppb=wt (ng )wt ( g )

=3.6 × 103 ng2.6 g

=1.4 ×103 ngg

=1400 ppb

part per thousand ( ppt)( wtV

)= wt solute (g)V solution ( sample )(mL)

×103=wt (mg )V (mL )

=wt ( g )V ( L )

part per million( ppm)( wtV

)=wt solute (g)

V solution ( sample )(mL)×106=

wt ( μ g )V (mL )

=wt (mg )V ( L )

part per billion(ppb)( wtV )= wt solute ( g )

V solution (sample ) (mL )×109=

wt (ng )V (mL)

=wt ( μ g )V ( L )

part per trillion( ppt)(wtV )= wt solute (g )

V solution ( sample ) (mL )×1012=wt ( pg )

V (mL)=wt (ng )

V ( L )

Example (39):- A 25.0 L serum sample was analyzed for glucose content and found to contain 26.7 g. Calculate the concentration of glucose in ppm and in mg/dL.

24

Solution: Note: 1dL = 100 mL

ppm=wt (μ g)V (mL)

= 26.7(μ g)25.0 (μ L)

1000(μ LmL )

= 26.7(μ g)0.025(mL)

¿1.07 ×103( μ gmL

)=1.07 ×103 ppm

wt (mg )V (dL )

=26.7μ g ×10−3 mg

μg

25 μ L ×10−5 dLμ L

=107 mg /dL

The relationship between molarity, normality and part per million:-

M= ppmM .wt × 1000

N= ppmEq . wt ×1000

Example (40):-(a) Calculate the molar conc. of 1.0 ppm solutions each of Li+ and Pb+2.(b) What weight of Pb(NO3)2 will have to be dissolved in 1 liter of water to prepare a 100 ppm Pb+2 solution.Solution:

M= ppmM . wt × 1000

(a)M

Li+¿= 1.06.94×1000

=1.44×10−4 mole /L¿

M Pb+2=1.0

207 ×1000=4.83× 10−6 mole /L

(b)

M= 100207 ×1000

=4.83× 10−4 mole / L

M= wtM . wt

× 1000V (mL)

4.83 × 10−4= wt331

× 10001000

wt=0.1598 gm Pb(NO3)2

Example (41):-The concentration of Zinc ion in blood serum is about (1ppm). Express this as meq/L.Solution:-

25

N= ppmEq .wt ×1000

= ppmA .wt

2×1000

= 165.4

2× 1000

=3.06 ×10−5 Eq / L

¿3.06 ×10−5× 1000=3.06 ×10−2meq /L

Problems 4

Q1: A solution contains 6 µ mole Na2SO4 in 250 mL. How many ppm sodium does it contain? of sulfate ?

Q2: Calculate the molar concentration of 1 ppm solutions of each of the following?

a) AgNO3 b) Al (SO4)3 c) CO2 d) HClO4

Q3: Calculate the ppm conc. Of 2.5 10 -4 M solutions of each of the following?

a) Ca+2 b) CaCl2 c) HNO3 d) KCN

Q4: You want to prepare 1L of a solution containing 1ppm Fe+2. How many grams ferrous ammonium sulfate, Fe SO4 (NH4)2 SO4.6H2O, must be dissolved and diluted in 1L? What would be the molarity of this solution?

Q5: How many grams NaCl should be weighed out to prepare 1L of a 100 ppm solution of (a) Na+ and (b) Cl-

Q6: One liter of a 500 ppm solution of KClO3 contains how many grams of K+?

p-Functions

Scientists frequently express the concentration of a species in terms of itsp-function, or p-value. The p-value is the negative logarithm (to the base 10) ofthe molar concentration of that species. Thus, for the species X,

pX = -log [X]

26

As shown by the following examples, p-values offer the advantage of allowing

concentrations that vary over ten or more orders of magnitude to be expressed interms of small positive numbers.

Example (42): Calculate the p-value for each ion in a solution that is 2.00 X 10-3 M in NaCl and 5.4 X 10-4 M in HCl.

Solution pH = -log [H+] = -log (5.4 × 10-4) = 3.27

pNa = -log (2.00 × 10-3) = -log 2.00 × 10-3 = 2.699

[Cl-] = 2.00 × 10-3 M + 5.4 × 10-4 M

= 2.00 × 10-3 M + 0.54 × 10-3 M = 2.54 × 10-3 M

pCl = -log 2.54 × 10-3 M = 2.595

Example (43): Calculate the molar concentration of Ag+ in a solution that has a pAg of 6.372.

Solution: pAg = -log [Ag+] = 6.372,

log [Ag+] = -6.372,[Ag +] = 4.246 × 10-7 = 4.25 ×10-7

Volumetric analysis (titration analysis) Are the most useful and accurate analytical techniques, especially for

millimole amounts of analyte. They are rapid and can be automated, and they can

be applied to smaller amounts of analyte when combined with a sensitive

instrumental technique for detecting the completion of the titration reaction, for 27

example, pH measurement. In a titration the test substance (analyte) in a flash react

with a reagent added from a buret as a solution of known concentration. This is

referred to as a standard solution and is called the titrant. The volume of titrant

required to just completely react with the analyte is measured. Since we know the

concentration as well as the reaction between the analyte and the reagent, we can

calculate the amount of analyte.

The requirements of a titration are as follows:-(1) The reaction must be stoichiometric: That is, there must be a well defined and

known reaction between the analyte and the titrant.

CH 3COOH +NaOH →CH 3COONa+H 2O

(2) The reaction should be rapid. Most ionic reactions.

(3) There should be no side reaction, and the reaction should be specific.

(4) There should be a marked change in some property of the solution when the

reaction is complete. This may be a change in color of the solution or in some

electrical or other physical property of the solution (by used indicator or pH

meter).

(5) The point at which an equivalent or stoichiometric amount of titrant is added is

called the equivalence point. The point at which the reaction is observed to be

complete is called the end point, that is, when a change in some property of the

solution.

(6) The reaction should be quantitative. That is, the equilibrium of the reaction

should be far to the right so that a sufficiently sharp change will occur at the end

point to obtain the desired accuracy. The equivalence point is the theoretical end of

the titration where the number of moles of titrant = number of moles of analyte.

The end point is the observed end of the titration.

28

Standard solution: A solution is prepared by dissolving an accurately weighed

quantity of a highly pure material called a primary standard and diluting to an

accurately known volume in a volumetric flask.

A primary standard should fulfil these requirements:-(1) It should be 100.00% pure, although 0.01 to 0.02% impurity is tolerable if it is

accurately known.

(2) It should be stable to drying temperature, and it should be stable indefinitely at

room temperature. The primary standard is always dried before weighing.

(3) It should be readily available and fairly inexpensive.

(4) Althogh not necessary, it should have a high formula weight.

(5) If it is to be used in titration, it should possess the properties required for a

titration listed above. In particular, the equilibrium of the reaction should be far to

the right so that a very sharp end point will be obtained.

A solution standardized by titrating a primary standard is itself a secondary

standard. It will be less accurate than a primary standard solution due to the errors

of titrations. A high formula weight means a larger weight must be taken for a

given number of moles. This reduces the error in weighing.

Classification of volumetric or titration methods:- (1) Neutralization (acid-base) titrations: Many compounds ,both inorganic and

organic ,are either acids or bases can be titrated with a standard solution of a strong

base or a strong acid. The end point of these titrations are easy to detect ,either by

means of indicator or by following the change in pH with a pH meter.29

(2) Precipitation titrations: In the case of precipitation, the titrant forms an

insoluble product with the analyte. An example is the titration of chloride ion with

silver nitrate solution to form silver chloride precipitate.

(3) Complexometric titrations :In complexometric titrations ,the titrant is a reagent

that forms a water-soluble complex with the analyte , a metal ion .The titrant is

often a chelating.

(4) Reduction-Oxidation titrations :These (redox) titrations involve the titration of

an oxidizing agent with a reducing agent ,or vice versa .An oxidizing agent gains

electrons and a reducing agent loses electrons in a reaction between them.

Example (44):-A 0.471 g sample containing sodium bicarbonate was dissolved

and titrated with standard 0.1067 M hydrochloric acid solution, requiring 40.72

mL. The reaction is:HC O3−¿+H +¿→ H2 O+CO2¿ ¿

Calculate the percent sodium bicarbonate in the sample.

Solution:millimoles NaH CO3=millimoles HCl

(wt (g)

M . wt ( gmol

))

NaHCO3

X 1000=(M × V )HCl

(wt (g )

84.01( gmol )

)X 1000=0.1067 × 40.72

wt=0.365 g

% NaHCO3=wt NaHCO3

wt sample×100= 0.365 g

0.4671 g× 100=78.14 %

Example (45):-A 0.2638 g soda ash sample is analyzed by titrating the sodium

carbonate with the standard 0.1288 M hydrochloric solution, requiring 38.27 mL.

The reaction is:C O32−¿+2H +¿→ H2O +CO2¿ ¿

Calculate the percent sodium carbonate in the sample.

Solution:

30

millimoles Na2 CO3=12

millimoles HCl

( wtM .wt

)Na2 CO3

X 1000=12(M ×V )HCl

( wt105.99 )X 1000=1

2(0.1288 ×38.27 ) ,wt=0.2612 g

% Na2 CO3=wt Na2 CO3

wt sample× 100=0.2612

0.2638× 100=99.02 % Na2CO3

Example (46):-How many millilitres of 0.25 M solution of H2SO4 will react with

10 mL of a 0.25 M solution of NaOH.H 2 SO4+2 NaOH → Na2 SO4+2 H2 O

Solution:

One half as many millimoles of H2SO4 will react one millimole NaOH

(M×V) H2SO4 = 12 (M×V) NaOH

0.25×V = 12 (0.25×10)

V = 5.0 mL H2SO4

Example (47): An approximate 0.1M hydrochloric acid solution is prepared by

120-fold dilution of concentration hydrochloric acid. It is standardized by titrating

0.1876 g of dried primary standard sodium carbonate: CO3 2− +2H+ → H2O+CO2

The titration required 35.86 mL acid. Calculate the molar concentration of the

hydrochloric acid.

Solution:

millimoles HCl equal to twice the millimoles of Na2CO3

(𝐌×𝐕)𝐇𝐂i= 21( .𝐰𝐭 )𝐍𝐚𝟐𝐂𝐎𝟑

𝐌 ×𝟑𝟓.𝟖𝟔 = 𝟐 ( 𝟏𝟖𝟕.𝟔 𝐦𝐠 𝟏𝟎𝟓.𝟗𝟗 𝐦𝐠/𝐦𝐦𝐨𝐥 )

𝐌𝐇𝐂𝐥 = 𝟎.𝟎𝟗𝟖𝟕 𝐦𝐦𝐨𝐥/𝐦𝐋

31

Example (48): The iron (II) in an acidified solution is titrated with a 0.0206 M

solution of potassium permanganate: 5Fe2+ +MnO4 − +8H+ → 5Fe3+ +Mn2+ + 4H2O

If the titration required 40.2 mL, how many milligrams iron are in the solution.

Solution:

( wtM . wt )Fe=5

1( M × V ) KMnO4

wt55.8= 𝟓 (𝟎.𝟎𝟐𝟎𝟔 × 𝟒𝟎.𝟐)

𝐰𝐭 = 𝟐𝟑𝟏.𝟎 𝐦𝐠

Example (49): Aluminium is determined by titrating with EDTA:

Al3+ +H2Y2− → AlY− + 2H+

A 1.00 g sample requires 20.5 mL EDTA for titration. The EDTA was

standardized by titrating 25.0 mL a 0.100M CaCl2 solution, requiring 30.0 mL

EDTA. Calculate the percent Al2O3 in the sample.

Solution: 𝟏 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐂𝐚𝟐+ = 𝟏 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐄𝐃𝐓𝐀

(𝐌×𝐕)𝐄𝐃𝐓𝐀 = (𝐌×𝐕)𝐂𝐚𝐂𝐥𝟐

𝐌×𝟑𝟎.𝟎 = 𝟎.𝟏𝟎×𝟐𝟓.𝟎 𝐌 = 𝟎.𝟎𝟖𝟑𝟑 𝐦𝐦𝐨𝐥/𝐦𝐋 𝟏 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐀𝐥𝟐𝐎𝟑 =

12 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬

𝐄𝐃𝐓𝐀

( wtM .wt 𝐗𝟏𝟎𝟎𝟎)𝐀𝐥𝟐𝐎𝟑 =

12 ×(𝐌×𝐕)𝐄𝐃𝐓𝐀

( wt1101.96 𝐗𝟏𝟎𝟎𝟎)𝐀𝐥𝟐𝐎𝟑 = 1

2 × (𝟎.𝟎𝟖𝟑𝟑×𝟐𝟎.𝟓)𝐄𝐃𝐓𝐀

𝐰𝐭 = 𝟎.𝟎𝟖𝟕𝟏 𝐠m

% 𝐀𝐥𝟐𝐎𝟑 = wt component

wt sample ×𝟏𝟎𝟎

= 0.08711 ×𝟏𝟎𝟎 = 𝟖.𝟕𝟏% 𝐀𝐥𝟐𝐎𝟑

32

Example (50): - In acid solution, potassium permanganate reacts with H2O2 to

form Mn2+: 5H2O2 + 2MnO4− + 6H+ → 5O2 + 2Mn2+ + 8H2O

In neutral solution, it reacts with MnSO4 to form MnO2:

3Mn2+ + 2MnO4− + 4OH− → 5MnO2 +2H2O

Calculate the number of millilitres of 0.100M KMnO4 that will react with 50.0mL

of 0.200M H2O2 and with 50.0mL of 0.200 M MnSO4.

Solution: 𝟏 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐌𝐧𝐎𝟒− = 25 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐇𝟐𝐎𝟐

(𝐌×𝐕)𝐌𝐧𝐎𝟒− = 25 (𝐌×𝐕)𝐇𝟐𝐎𝟐

(𝟎.𝟏𝟎×𝐕)𝐌𝐧𝐎𝟒− = 25(𝟎.𝟐𝟎×𝟓𝟎.𝟎)𝐇𝟐𝐎𝟐 𝐕𝐊𝐌𝐧𝐎𝟒 = 𝟒𝟎.𝟎 𝐦𝐋 𝟏 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐌𝐧𝐎𝟒− =

23 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐌𝐧𝟐+

(𝐌×𝐕)𝐌𝐧𝐎𝟒− = 23 (𝐌×𝐕)𝐌𝐧𝟐+

(𝟎.𝟏𝟎×𝐕)𝐌𝐧𝐎𝟒− = 23 (𝟎.𝟐𝟎×𝟓𝟎.𝟎)𝐌𝐧𝟐+ 𝐕𝐊𝐌𝐧𝐎𝟒 = 𝟔𝟔.𝟕 𝐦𝐋

Example(51):- Oxalic acid, H2C2O4, is a reducing agent that reacts with KMnO4 as follows:

5H2C2O4 +2MnO4− + 6H+ → 10CO2 + 2Mn2+ + 8H2O

Its two protons are also titratable with a base. How many millilitres of 0.100M NaOH 0.100 M KMnO4 will react with 500 mg H2C2O4?

Solution: millimoles NaOH = 2 millimoles H2C2O4

(𝟎.𝟏𝟎 mmol

mL × 𝒙 𝐦𝐋)𝐍𝐚𝐎𝐇 = 𝟐 ( 500 mg

90.0 mgmmol

)

𝒙 = 𝟏𝟏𝟏 𝐦𝐋 𝐍𝐚𝐎𝐇

𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐊𝐌𝐧𝐎𝟒 = 25 𝐦𝐢𝐥𝐥𝐢𝐦𝐨𝐥𝐞𝐬 𝐇𝟐𝐂𝟐𝐎𝟒

33

(𝟎.𝟏𝟎×𝒙 𝐦𝐋)𝐊𝐌𝐧𝐎𝟒 = 25 ( 500 mg

90.0 mgmmol

)

𝒙 = 𝟐𝟐.𝟐 𝐦𝐋 𝐊𝐌𝐧𝐎𝟒

Example(52):- Pure Na2C2O4 plus KHC2O4.H2C2O4 (three replaceable protons, KH3A2) are mixed in such a proportion that each gram of the mixture will react with equal volumes of 0.100 M KMnO4 and 0.100 M NaOH. What is the proportion? Assume 10.0 mL titrant, so there is 1.00 mmol NaOH or KMnO4. The acidity is due to KHC2O4.H2C2O4 (KH3A2):

Solution:

𝐦𝐦𝐨𝐥 𝐊𝐇𝟑𝐀𝟐 = 𝐦𝐦𝐨𝐥 𝐍𝐚𝐎𝐇× 13 ( mmol K H 3 A2

mmolOH−¿¿ )

= (1.0 mmol× 13 )NaOH = 0.333 mmol KH3A2

From example 5.26, each mmol Na2C2O4 (Na2A) reacts with 2/5 mmol KMnO4:-

mmol KMnO4 =mmol Na2A× 25 ( mmol Mn

O 4−¿

mmol N a2 A ¿ ) + mmol KH3A2 × 45 (

mmol MnO4

−¿

mmol K H 3 A2¿ )

1.00 mmol KMnO4 = mmol Na2A× 25 + 0.333 mmol KH3A2 × 4

5 mmol Na2A

mmol Na2A = 1.83 mmol

The ratio is:1.83 mmol N a2 A

0.333 mmol K H 3 A2

= 5.50 mmol N a2 A

mmol K H 3 A2

the weight ratio is = 5.50 mmol N a2 A ×134 mg

mmol

218kgK H 3 A2

mmol

= 3.38 g Na2A/gm KH3A2

Example (53):- A 0.4671 g sample containing sodium bicarbonate (a monoacidic base) is dissolved and titrated with a standard solution of hydrochloric acid, requiring 40.72 mL. The hydrochloric acid was standardized by titrating 0.1876

34

gm sodium carbonate, which required 37.86 mL acid (see example 5.18 for reaction). Calculate the percent sodium bicarbonate in the sample.

Solution:

(𝐍×𝐕)𝐇𝐂𝐥 = (wt

Eq.wt¿Na2CO3

N ×37.86 mL = (0.1876 ×1000

105.992

)Na2CO3

𝐍 = 𝟎.𝟎𝟗𝟑𝟓 ( 𝐦𝐞𝐪 𝐦𝐋)𝐇𝐂𝐥(𝐍×𝐕)𝐇𝐂𝐥 = ( wt

Eq. wt )𝐍𝐚𝐇𝐂𝐎𝟑

(𝟎.𝟎𝟗𝟑𝟓×𝟒𝟎.𝟕𝟐) = wt

84.011𝐰𝐭 = 𝟎.𝟑𝟏𝟗 𝐠𝐦

%𝐍𝐚𝐇𝐂𝐎𝟑 = wtwt sample ×𝟏𝟎𝟎

=0.3190.467×𝟏𝟎𝟎 = 𝟔𝟕.𝟏𝟖% 𝐍𝐚𝐇𝐂𝐎𝟑 𝐢𝐧 𝐬𝐚𝐦𝐩𝐥𝐞

Example(54): The organic matter in a 3.776 g sample of amercuric ointment was decomposed with HNO3. After dilution, the Hg2+ was titrated with 21.30 mL of a 0.1144 M solution of NH4SCN. Calculate the percent Hg in ointment. Solution:

Hg2+ +2SCN− → Hg(SCN)2 (aq.)

mmol Hg2+ = mmol NH4SCN× 1 mmol. Hg+22mmol . NH 4 SCN

( wtA . wt

×1000)𝐇𝐠𝟐+ = (𝐌×𝐕×12)𝐍𝐇𝟒𝐒𝐂𝐍

( wt200.59×𝟏𝟎𝟎𝟎)𝐇𝐠𝟐+ = 𝟎.𝟏𝟏𝟒𝟒×𝟐𝟏.𝟑𝟎× 1

2

wt ×1000200.59 = 𝟏.𝟐𝟏𝟖 , 𝐰𝐭𝐇𝐠𝟐+ = 𝟎.𝟐𝟒𝟒𝟑 𝐠m

𝐇𝐠𝟐+% = wt Hg+2wt sample ×𝟏𝟎𝟎 = 0.2443

3.776 ×𝟏𝟎𝟎 = 𝟔.𝟒𝟕%

35

Example (55): Calculate the normality of an I2 solution if 37.34 mL were needed to titrate 0.2040 g of primary standard As2O3.

Reaction: I2 +H2AsO3 − +H2O → 2I− + H2AsO4 − +2H+

Solution:

At equivalence point: meq I2 = meq As2O3

(𝐍×𝐕)𝐈𝟐 = wt As 2O 3

Eq. wt As 2O 3×𝟏𝟎𝟎𝟎Arsenic is oxidized from the +3 to +5 state in this reaction. Since two atoms of As are contained in each As2O3:- As2O3 ≡ 2H2AsO3 − ≡ 2I2 ≡ 4e-

(𝐍×𝟑𝟕.𝟑𝟒) 𝐈𝟐 = (0.204197.8

4 ×𝟏𝟎𝟎𝟎)𝐀𝐬𝟐𝐎𝟑

𝐍 = 𝟎.𝟏𝟏𝟎𝟓 EqL = 𝟎.𝟏𝟏𝟎𝟓

meqmL

Back titration

Sometimes a reaction is slow to go to completion, and a sharp end point cannot be obtained. A back titration will often yield useful results. In this technique, a measured amount of the reagent, which would normally be the tritrant, is added to the sample so that there is a slight excess. After the reaction with the analyte is allowed to go to completion, the amount of excess (unreacted) reagent is determined by titration with another standard solution.

In back-titration, a known number of millimoles of reaction it is taken, in excess of the analyte. The unreacted portion is titrated.

mmol reagent reacted = mmol taken – mmol back titrted

mg analyte = mmol reagent reacted x factor (mmol analyte/mmol reagent)

Example(56):- A 0.50 g sample containing Na2CO3 plus inert matter in analyzed by adding 50.0mL of 0.1M HCl, a slight excess, boiling to remove CO2, and then

36

back-titrating the excess acid with 0.1M NaOH. If 5.6mL of NaOH is required for the back-titration, what is the percent Na2CO3 in the sample.

Solution:

𝐦𝐦𝐨𝐥 𝐍𝐚𝟐𝐂𝐎𝟑 = (𝐦𝐦𝐨𝐥 𝐇𝐂𝐥 ×1mmol . Na2 CO3

2mmol .HCl ) − 𝒎𝒎𝒐𝒍 𝑵𝒂𝑶𝑯( wt

M . wt×𝟏𝟎𝟎𝟎)𝐍𝐚𝟐𝐂𝐎𝟑 = (𝐌 × 𝐕 ×1

2) 𝐇𝐂𝐥 − (𝐌×𝐕) 𝐍𝐚𝐎𝐇( wt

106×𝟏𝟎𝟎𝟎) = (𝟎.𝟏×𝟓𝟎×12) − (𝟎.𝟏×𝟓.𝟔)

𝐰𝐭𝐍𝐚𝟐𝐂𝐎𝟑 = 𝟎.𝟐𝟎𝟓 𝐠m

%𝐍𝐚𝟐𝐂𝐎𝟑 = wt Na2CO3

wt sample ×𝟏𝟎𝟎= 0.205

0.500×𝟏𝟎𝟎 = 𝟒𝟏.𝟏𝟐𝟖%

Example (57):- Chromium (III) is slow to react with EDTA (H4Y) and is therefore determined by back-titration. A pharmaceutical preparation containing chromium (III) is analyzed by treating a 2.63 gm sample with 5.00 mL of 0.0103M EDTA. Following reaction, the unreacted EDTA is back-titrated with 1.32 mL of 0.0122M Zinc solution. What is the percent chromium chloride in the pharmaceutical preparation?

Solution: 𝐦𝐦𝐨𝐥 𝐂𝐫𝐂𝐥𝟑 = 𝐦𝐦𝐨𝐥 𝐄𝐃𝐓𝐀−𝐦𝐦𝐨𝐥 𝐙𝐧𝟐+

( wtM . wt

×𝟏𝟎𝟎𝟎) 𝐂𝐫𝐂𝐥𝟑 = (𝐌×𝐕)𝐄𝐃𝐓𝐀 −(𝐌×𝐕)𝐙𝐧𝟐+

( wt158.4

×𝟏𝟎𝟎𝟎) = (𝟎.𝟎𝟏𝟎𝟑×𝟓)−(𝟎.𝟎𝟏𝟐𝟐×𝟏.𝟑𝟐)

wt158.4

×𝟏𝟎𝟎𝟎 = 𝟎.𝟎𝟓𝟏𝟓−𝟎.𝟎𝟏𝟔wt

158.4×𝟏𝟎𝟎𝟎 = 𝟎.𝟎𝟑𝟓𝟓

𝐰𝐭 = 𝟎.𝟎𝟎𝟓𝟖 𝐠m 𝐂𝐫𝐂𝐥𝟑

% 𝐂𝐫𝐂𝐥𝟑 =0.0058

2.63 ×𝟏𝟎𝟎

37

= 𝟎.𝟐𝟐𝟏% 𝐂𝐫𝐂𝐥𝟑

Example (58):- A 0.200 gm sample of pyrolusite is analyzed for manganese content as follows. Add 50.0 mL of a 0.100M solution of ferrous ammonium sulphate to reduce the MnO2 to Mn2+. After reduction is complete, the excess ferrous ion is titrated in acid solution with 0.0200M KMnO4, requiring 15.0mL.

Calculate the percent manganese in the sample as Mn3O4 (only part or none of the manganese may exist in this form, but we can make the calculations on the assumption that it does).

Solution:-

m𝐦𝐨𝐥 𝐅𝐞𝟐+ 𝐫𝐞𝐚𝐜𝐭𝐞𝐝 = 𝐦𝐦𝐨𝐥 𝐅𝐞𝟐+ −𝟓×𝐦𝐦𝐨𝐥 𝐊𝐌𝐧𝐎𝟒𝐦𝐦𝐨𝐥 𝐅𝐞𝟐+ 𝐫𝐞𝐚𝐜𝐭𝐞𝐝 = (𝐌×𝐕)𝐅𝐞𝟐+ −𝟓×(𝐌×𝐕)𝐊𝐌𝐧𝐎𝟒

= (𝟓𝟎×𝟎.𝟏𝟎)−𝟓×(𝟎.𝟎𝟐×𝟏𝟓) = 𝟓−𝟏.𝟓 = 𝟑.𝟓

𝐦𝐦𝐨𝐥 𝐌𝐧𝐎𝟐 = 𝟑.𝟓 𝐦𝐦𝐨𝐥 𝐅𝐞𝟐+ × 12(𝐦𝐦𝐨𝐥 𝐌𝐧𝐎𝟐/𝐦𝐦𝐨𝐥 𝐅𝐞𝟐+)

= 𝟏.𝟕𝟓 𝐦𝐦𝐨𝐥 𝐦𝐦𝐨𝐥 𝐌𝐧𝟑𝐎𝟒 = 𝟏.𝟕𝟓 𝐦𝐦𝐨𝐥 𝐌𝐧𝐎𝟐 × 13 (𝐦𝐦𝐨𝐥 𝐌𝐧𝟑𝐎𝟒/𝐦𝐦𝐨𝐥 𝐌𝐧𝐎𝟐)

= 𝟎.𝟓𝟖𝟑 𝐦𝐦𝐨𝐥 𝐰𝐭𝐌𝐧𝟑𝐎𝟒 = 𝟎.𝟓𝟖𝟑 𝐦𝐦𝐨𝐥×𝟐𝟐𝟖.𝟖 𝐦𝐠/𝐦𝐦𝐨𝐥 = 𝟏𝟑𝟑.𝟑𝟗𝟎 𝐦𝐠

% 𝐌𝐧𝟑𝐎𝟒 = 133.390

200 ×𝟏𝟎𝟎 = 𝟔𝟔.𝟕% 𝐌𝐧𝟑𝐎𝟒Acid-Base Equilibria

Acid-base theories:-

1) Arrhenius Theory (H+ and OH-):-

Acid:-any substance that ionizes (partially or completely) in water to give hydrogen ion (which associate with the solvent to give hydronium ion H3O+):

HA+H2 O↔ H3O+¿+A−¿¿ ¿

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Base:-any substance that ionizes in water to give hydroxyl ions. Weak (partially ionized) to generally ionize as follows:-

B+H 2O ↔BH +¿−OH−¿ ¿¿

While strong bases such as metal hydroxides (e.g. NaOH) dissociate as

M (OH )n ↔ M n+¿+nOH−¿¿¿

This theory is obviously restricted to water as the solvent.

2) Bronsted-Lowry Theory (taking and giving protons, H+):-

Acid:-any substance that can donate a proton.

Base:-any substance that can accept a proton. Thus, we can write a half reaction:

Acid = H+ + Base

3) Lewis Theory (taking and giving electrons):-

Acid:-a substance that can accept an electron pair.

Base:-a substance that can donate an electron pair.

H2O+H+¿↔ H 2 O : H +¿ ¿¿¿

Acid-Base Equilibria in water:-when an acid or base is dissolved in water, it will dissociate, or ionize, the amount of ionization being dependent on the strength of the acid or base. A strong electrolyte is completely dissociated, while a weak electrolyte is partially dissociated.

HCl+H2O → H3O+¿+Cl−¿(strongacid ,completely ionized)¿ ¿

HOAC +H2O H3O+ +OAC-

(weak acid , partially ionized , a few percent )

Thermodynamic acidity constant K

K °=a H3O+¿ ×C

aH 2 O× aHOAC¿

39

Cl3Al:O RRRRAlCl3 + :O

In dilute solutions, the activity of water remains essentially constant, and is taken as unity at standard state:

Ka° =a H 3O

+¿ × aOAC−¿

aHOAC¿¿

Pure water ionizes slightly, or undergoes autoprotolysis (self-ionization of solvent to give a cation and anion):-

2 H 2O ↔ H3 O+¿+OH−¿¿ ¿

The equilibrium constant for this is:

Kw° =a H3O

+¿ × aOH−¿

aH 2 O2¿

¿

Again, the activity of water is constant in dilute solution (its concentration is essentially constant at ~ 55.3M), so:-

Kw° =a H 3O+¿ ×aOH−¿¿ ¿

Kw° , thermodynamicoutoprotolysisorself −ionization , constant ¿

We will use H+ in place of H3O+ for simplification, also, molar concentration will generally be used instead of activities and represented by square brackets [ ] around the species).

HCl ↔ H+¿+Cl−¿¿ ¿

HOAC ↔H+¿+OAC−¿,K a=¿¿ ¿¿

H 2O ↔ H+¿+OH−¿,K w=¿¿ ¿

1.0 ×10−14=¿

Therefore ¿

Example (59):-A 1.010-3 M solution of HCl prepared. What is the hydroxyl ion concentration [OH-]?

Solution: Kw=¿

1.0 ×10−3× ¿

¿

The pH scale: pH=−log¿¿40

Example (60):- Calculate the pH of a 210-3M HCl?

Solution: ¿

¿−log(2.0 × 10−3)=3−log 2.0=2.70

pOH=−log ¿¿

Example (61):-Calculate the pOH and pH of a 510-2 M NaOH?

Solution: ¿

¿−log(5 ×10−2)=2−log5=2−0.70=1.30

pH +pOH=14 pH =14−pOH=14−1.30=12.70

Example (62):-Calculate the pH of a solution prepared by mixing 2.0 mL of a

strong acid solution (keep track of millimoles) of Ph=3.0 and 3.0 mL of a strong

base of pH= 10.0?

Solution:- ¿

¿1.0×10−3×2.0=2×10−3 mmol

pOH=14−pH =14−10=4.0¿

mmol O H−¿=M× V=1 .0× 10−4 ×3.0mL=3.0×10−4 mmol ¿

There is an excess of acid:-

mmol H+¿=2.0× 10−3−3.0×10−4=1 . 7 ×10−3 mmol ¿

¿

pH=−log3.4 × 10− 4=4−0.53=3.47

Example (63):-The pH of a solution is 9.67. Calculate the hydrogen ion concentration in the solution?

41

Solution:-

pH=−log¿¿

¿

Note: - if the concentration of an acid or base is much less than 10-7 M, then its contribution to the acidity or basicity will be negligible compared with the contribution from water. The pH of a 10-8 M sodium hydroxide solution would therefore not differ significantly from 7. If the concentration of the acid or base is around 10-7 M, then its contribution is not negligible and neither is that from water, hence the sum of the two contributions must be taken.

Example (64):- Calculate the pH and pOH of a 1.0 × 10-7 solution HCl?

Solution: 𝐇𝐂𝐥 → 𝐇+ + 𝐂𝐥−

𝐇𝟐𝐎 ↔ 𝐇+ + 𝐎𝐇−

[𝐇+][𝐎𝐇−] = 𝟏.𝟎×𝟏𝟎−𝟏𝟒

[𝐇+]𝐇𝟐𝐎 𝐝𝐢𝐬𝐬. =[𝐎𝐇−]𝐇𝟐𝐎𝐝𝐢𝐬𝐬. = 𝐗

Since the H+ contributed from the ionization of water are not negligible compared to the HCl added.

[𝐇+] = 𝐂𝐇𝐂𝐥 + [𝐇+]𝐇𝟐𝐎 𝐝𝐢𝐬𝐬.

[𝐇+][𝐎𝐇−] = 𝟏.𝟎×𝟏𝟎−𝟏𝟒

(𝟏.𝟎×𝟏𝟎−𝟕 +𝐗)(𝐗) = 𝟏.𝟎×𝟏𝟎−𝟏𝟒𝐗𝟐 +𝟏.𝟎×𝟏𝟎−𝟕 +𝟏.𝟎×𝟏𝟎−𝟏𝟒 = 𝟎

Using the quadratic equation to solve:-

𝐗 = −𝐁 ± √ B 2−4 AC

2 A

= −1.0 ×10−7± √ (1. 0 ×10−7)2−4 (1 .0 × 10−14)2

42

= 𝟔.𝟐×𝟏𝟎−𝟖 ,𝐭𝐡𝐞𝐫𝐞𝐟𝐨𝐫𝐞 𝐭𝐡𝐞 𝐭𝐨𝐭𝐚𝐥 𝐜𝐨𝐧𝐜𝐞𝐧𝐭𝐫𝐚𝐭𝐢𝐨𝐧 𝐇+ = 𝟏.𝟎×𝟏𝟎−𝟕 + 𝟔.𝟐×𝟏𝟎−𝟖 = 𝟏.𝟔𝟐×𝟏𝟎−𝟕 𝐌 𝐩𝐇 = −𝐥𝐨𝐠𝟏.𝟔𝟐×𝟏𝟎−𝟕 = 𝟕−𝟎.𝟐𝟏 = 𝟔.𝟕𝟗𝐩𝐎𝐇 = 𝟏𝟒−𝟔.𝟕𝟗 = 𝟕.𝟐𝟏Example (65):- Calculate the pH and pOH of a 1.0×10-3 M solution acetic acid, Ka = 1.75×10-5?

Solution: 𝐇𝐎𝐀C ↔ 𝐇+ + 𝐎𝐀C−

𝐈𝐧𝐢𝐭𝐢𝐚𝐥 𝟏.𝟎×𝟏𝟎−𝟑 𝟎 + 𝟎 𝐄𝐪𝐮𝐢𝐥𝐢𝐛𝐫𝐢𝐮𝐦 𝟏.𝟎×𝟏𝟎−𝟑 − 𝐱 𝐱 + 𝐱𝐊𝐚 = [𝐇+][𝐎𝐀C−] [𝐇𝐎𝐀C]𝟏.𝟕𝟓×𝟏𝟎−𝟓 =𝐱 . 𝐱 𝟏.𝟎×𝟏𝟎−𝟑 −¿ x

If CHA > 100 Ka, x can be neglected compared to CHA

𝟏.𝟕𝟓×𝟏𝟎−𝟓 = X 21.0× 10−3

x = 1.32×10−4M = [H+]

pH = −log[H+] = −log 1.32×10−4 = 4−0.12 = 3.88𝐩𝐇+𝐩𝐎𝐇 = 𝟏𝟒𝐩𝐎𝐇 = 𝟏𝟒−𝐩𝐇 = 𝟏𝟒−𝟑.𝟖𝟖 = 𝟏𝟎.𝟏𝟐

General law to calculate [H+] for weak acids: [𝐇+] = √𝐊𝐚 𝐂𝐚𝐜𝐢𝐝

Example (66):- The basicity constant kb for ammonia is 1.75×10-5 at 25°C, (H is only coincidental that this is equal to Ka for acetic acid). Calculate the pH and pOH for a 1.0×10-3 M solution of ammonia?

Solution:

NH3 + H2O ↔ NH4 + + OH−

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Equilibrium 1.0×10−3 −x x x𝐊𝐛 =[𝐍𝐇𝟒 +][𝐎𝐇−] [𝐍𝐇𝟑]𝟏.𝟕𝟓×𝟏𝟎−𝟓 = 𝐱.𝐱 𝟏.𝟎×𝟏𝟎−𝟑𝐱 = √𝟏.𝟕×𝟏𝟎−𝟓 ×𝟏.𝟎×𝟏𝟎−𝟑 = 𝟏.𝟑𝟐×𝟏𝟎−𝟒𝐌 = [𝐎𝐇−] 𝐩𝐎𝐇 = −𝐥𝐨𝐠 [𝐎𝐇−] = −𝐥𝐨𝐠𝟏.𝟑𝟐×𝟏𝟎−𝟒 = 𝟑.𝟖𝟖𝐩𝐇 = 𝟏𝟒−𝐩𝐎𝐇 = 𝟏𝟒−𝟑.𝟖𝟖 = 𝟏𝟎.𝟏𝟐

General law to calculate [OH-] for weak bases:- [OH−] = √KbCbase

Salts:any compound that produce from reaction of acid with base, and can be classified to:

Salts of weak acids and bases:-

Example (67):- Calculate the pH of a 0.1M solution of CH3COONa,

ka=1.75 ×10-5?

Solution:

(𝐢𝐨𝐧𝐢𝐳𝐚𝐭𝐢𝐨𝐧) 𝐍𝐚𝐎𝐀C → 𝐍𝐚+ +𝐎𝐀C−

(𝐡𝐲𝐝𝐫𝐨𝐥𝐲𝐬𝐢𝐬) 𝐎𝐀C− + 𝐇𝟐𝐎 ↔ 𝐇𝐎𝐀C + 𝐎𝐇−

𝐊𝐰 = 𝐊𝐚𝐊𝐛 , 𝐊𝐛 = [HOAc]¿ ¿

Kb= K wKa =

X 20.1𝐂𝐀− > 𝟏𝟎𝟎 𝐊𝐛 → 𝐱 𝐜𝐚𝐧 𝐛𝐞 𝐧𝐞𝐠𝐥𝐞𝐜𝐭𝐞𝐝

1 ×10−141.75× 10−5 = X 2

0.1

𝐱 = [𝐎𝐇−]= 𝟕.𝟔×𝟏𝟎−𝟔[𝐇+] =

1× 10−147.6× 10−6 = 1.3×10−9 M

𝐩𝐇 = −𝐥𝐨𝐠[𝐇+] = −𝐥𝐨𝐠 𝟏.𝟑×𝟏𝟎−𝟗 = 𝟖.𝟖𝟗For salt of weak acids:-

44

[𝐎𝐇−] = √ KwKa .𝐂𝐀− = √𝐊𝐛𝐂𝐀−

Example (68):- Calculate the pH of a 0.25M solution of ammonium chloride NH4Cl?

Solution:

(𝐢𝐨𝐧𝐢𝐳𝐚𝐭𝐢𝐨𝐧) 𝐍𝐇𝟒𝐂𝐥 → 𝐍𝐇𝟒 + + 𝐂𝐥−

(𝐡𝐲𝐝𝐫𝐨𝐥𝐲𝐬𝐢𝐬) 𝐍𝐇𝟒 + +𝐇𝟐𝐎 ↔ 𝐍𝐇𝟒𝐎𝐇 + 𝐇+ 𝐍𝐇𝟒 + +𝐇𝟐𝐎 ↔ 𝐍𝐇𝟑 + 𝐇𝟑𝐎+

𝐊𝐚= ¿ [NH 4OH ]¿ ¿=KwKb =

X 20.25 =

1×10−141.75× 10−5 =

X 20.25𝐱 = [𝐇+] = 𝟏.𝟐×𝟏𝟎−𝟓 𝐌 𝐩𝐇 = −𝐥𝐨𝐠𝟏.𝟐×𝟏𝟎−𝟓 = 𝟒.𝟗𝟐

For salt of weak bases:- [𝐇+] = √ KwKb .𝐂𝐁𝐇+

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Buffer solution

A buffer is defined as a solution that resists change in pH when a small amount of an acid or base is added or when the solution is diluted.

Example (69):-Calculate the pH of a buffer prepared by adding 10mL of 0.1M acetic acid to 20mL of 0.1M sodium acetate?

Solution:

Henderson-Hasselbalch equation:-

pH=pK a+ log [salt ][acid ]

pH=−log1.75 ×10−5+log

0.1×2030

0.1×1030

=4.76+ log 2.0=5.06

We can use millimoles of acids and salt in place of molarity.

Example (70):-Calculate the pH of a solution prepared by adding 25mL of 0.1M sodium hydroxide to 30mL of 0.2M acetic acid (this would actually be a step in a typical titration?

Solution:

mmol HOAc=0.2 M ×30mL=6.0 mmol

mmol NaOH=0.1 M × 25 mL=2.5 mmol

These react as follows:-

HOAc+NaOH ↔ NaOAc+H2 O

mmol NaO Ac=2.5mmolmmol HOAc=6.0−2.5=3.5 mmol ]after reaction

pH=pK a+ log [salt ][acid ]

=4.76+ log 2.53.5

=4.61

Example (71):-Calculate the volume of concentrated ammonia and the weight of ammonium chloride you would have taken to prepare 100mL of a buffer at pH 10.0 if the final concentration of salt is to be 0.2M?

46

Solution:

pOH=pK b+ log [ salt ][ base ]

pH+POH=14 ,10+POH=14

POH=14-10 = 44=4.76+ log 0.2[ NH 3]

−0.76=log 0.2[NH 3 ]

10−0.76= 0.2[ NH3]

[ NH 3 ]= 0.210−0.76 =

0.20.173

=1.15 M

M 1V 1=M 2 V 2(0.28× 0.9 ×1000

17)×V 1=1.15 × 100

14.8 ×V 1=1.15 ×100 , V 1=7.8 mL

M NH 4 Cl=wt

M . wt× 1000

V mL

0.2= wt53.5

× 1000100 wt=1.07 gm NH 4 Cl

Example (72):-How many grams ammonium chloride and how many millimilters 3.0M sodium hydroxide should be added to 200mL water and diluted to 500mL to prepare a buffer of pH9.50 with salt concentrated of 0.10M?

Solution:

pOH=pK b+ log [ salt ][ base ]

4.5=4.76+ log ¿¿

−0.26=log ¿¿

0.549= 0.1[ NH 3 ] [ NH 3 ]=0.18 M

mmol NH 4+¿∈ final solution=0.1M × 500mL=50 mmol ¿

mmol NH3∈final solution=0.18 M ×500 mL=90 mmol

The NH3 formed from an equal number of millimoles of NH4Cl. Therefore, a total of 50 + 90 = 140 mmol NH4Cl must be taken.

mmol NH 4Cl= wt (mg)M . wt (mg /mL)

47

140= wt (mg)53.5

wt=7.49 gm

The millimoles of NaOH needed are equal to the millimoles of NH3:

mmol NH3=M ×V

90mmol=3.0 M ×V V=30mL NaOH

The amount of acid or base that can be added without causing a large change in pH is governed by the buffering capacity (B), the buffering capacity increases with the concentration of the buffering species. The buffering capacity is maximum at pH=pKa.

Example (73):- A buffer solution is 0.2M of acetic acid and sodium acetate. Calculate the change in pH upon adding 1.0mL of 0.1M hydrochloric acid to 10mL of this solution?

Solution:

𝐩𝐇 = 𝐩𝐊𝐚 +𝐥𝐨𝐠 [salt ]−[adding acid ][ acid ]−[addingacid ]

(𝐩𝐇)𝟏 = 𝟒.𝟕𝟔+𝐥𝐨𝐠 [0 .2×10 ]−[0 .1×1]

[ 0.2×10 ]−[0.1×1]=¿𝟒.𝟕𝟏

(𝐩𝐇)𝟐 = 𝟒.𝟕𝟔+𝐥𝐨𝐠 [0 . 2][0.2] = 𝟒.𝟕𝟔

∆𝐩𝐇 = (𝐩𝐇)𝟏 −(𝐩𝐇)𝟐 = 𝟒.𝟕𝟏−𝟒.𝟕𝟔 = −𝟎.𝟎𝟓The change in pH is (-0.05). This is rather small especially if we consider that had the HCl been added to unbuffered neutral solution the final concentration would have been approximately 10-2M, and the pH would be 2.0.

Physiological Buffers (they keep you alive):-

The pH of the blood in healthy individual remains remarkably constant at (7.35-7.45). this is because the blood contains a number of buffers that protect

48

against pH change due to the presence of acidic or basic metabolites. From a physiological viewpoint, a change of ± 0.3 pH unit is extreme. An important diagnostic analysis is the CO2/HCO3

- balance in blood. This ratio is related to the pH of the blood by the Henderson-Hassellatch equation (pH = 6.10 + log [HCO3

-]/[H2CO3] where H2CO3 can be considered equal to the concentration of dissolved CO2 in blood, 6.10 is pKa1 of carbonic acid in blood at body temperature (37C).

Example (74):- The total carbon dioxide content (HCO3- + CO2) in a blood sample

is determined by acidifying the sample and measuring the volume of CO2 evolved with a VanSlyke manomertric apparatus. The total concentration was determined to be 28.5mmol/L. The blood pH at 37C was determined to be 7.48. What are the concentration of HCO3

- and CO2 in the blood?

Solution:

pH=pK a1+log ¿¿

7.48=6.10+ log¿¿

101.38=log ¿¿

¿

But ¿

24 [CO2 ]+[CO2]=28.5

25[CO2]=28.5❑⇒

[CO2 ]=28.525

=1.14 mmol /L

¿

Buffers for Biological and Clinical Measurements:-

Many biological reactions of interest occur in the pH range (6-8). A number, particularly specific enzyme reactions that might be used for analyses, may occur in the pH range (4-10). One useful series of buffers are phosphate buffers (H3PO4/H2PO4

-) or (H2PO4-/HPO4

-2) or (HPO4-2/PO4-3).

49

Example (75):- What weights of NaH2PO4 and Na2HPO4 would be required to prepare (1L) of a buffer solution of pH7.45 that has an ionic strength of 0.10?

Solution:

Let x= [Na2HPO4] and y= [NaH2PO4]. There are two unknowns, and two equations are needed (remember there must be the same number as unknowns to solve) our first equation is the ionic strength equation:-

μ=12∑C i(Z i)

2 μ=ionic strength, C i=molar conc . , Zi=equavelent

0.1=12¿

0.1=12[(2 x+ y) (1 )2+x (2 )2+ y (1 )2]

0.1=3x+ y ………(1)

pH=pK a2+log[ HPO4

−2]¿ ¿

7.4 5=7.12+log xy

100.33= xy

2.14 y=x subsitute∈(1)

0.1=3 (2.14 ) y+ y

0.0135 M= y=[ Na H2 P O4 ] Substitute in (2)

x=(2.14 ) ( 0.0135 )=0.0289 M=[Na2 HP O4 ]

g Na H 2 PO4=0.0135 mol /L ×120 g/mol=1.62 g / L

g Na2 H PO4=0.0289 mol /L ×142 g/mol=4.10 g/ L

A buffer that is widely used in the clinical laboratory and in biochemical studies in the physiological pH range is that prepared from tris (hydroxymethyl) aminomethane [(HOCH2)3CNH2, Tris, or THAM] and its conjugate acid (the amino group is protonated).

50

Polyprotic acids and their salts:-

H 3 PO4 ↔ H+¿+H 2 PO 4−¿K a1=1.1 ×10−2=¿ ¿¿¿

H 2 PO4−¿ ↔ H+¿+H PO4

2−¿K a2=7.5 × 10−8=¿¿¿¿ ¿

H PO42−¿↔ H+¿+PO4

3−¿K a3=4.8× 10−13=¿¿¿¿ ¿

H3 PO4 ↔ 3 H+¿+PO43−¿K a=Ka1× K a2× Ka3=4.0 ×10−22=¿¿¿ ¿

The stepwise Ka values of polyprotic acids get progressively smaller as the increased negative charge makes dissociation of the next proton more difficult. We can titrate the first two protons of H3PO4 separately. The third is too weak to titrate.

Example (76):- The pH of blood is 7.40. What is the ratio of [HPO42-]/[H2PO4

-] in the blood (assume 25C)?

Solution:

pH=pK a+ log [ proton accepter ][ proton donor ]

=pK a+log[ HPO4

−2 ]¿¿

7. 40=−log7.5 ×108−¿+log[ HPO4

−2 ]¿ ¿ ¿

7. 4 0=7.12+log[HPO4

−2]¿ ¿

7. 40−7.12=log[HPO4

−2]¿ ¿

100.28=[HPO4

−2]¿¿

Example (77):- Calculate the pH of a 0.1M H3PO4 solution? Note (treat H3PO4 as a monoprotic acid)?

Solution:

H 3 PO4 ↔ H+¿+H 2 PO 4−¿¿ ¿

Equilibrium 0.1−x x x

K a1=x . x

0.1−x=1.1 ×102−¿ ¿

51

In order to neglect x, C should be ≥ 100 Ka, here, it is only 10 times as large. Therefore, use the quadratic equation to solve:-x2+0.011 x−1.1× 103−¿=0¿

X=−B ±√B2−4 AC2 A

=−0.011±√(0.011)2−4¿¿¿

pH=−log¿¿

Example (78):- Calculate the equilibrium concentration of the different species in a 0.10M phosphoric acid solution at pH=3.0?

Solution:

¿

∝°=1¿¿

¿¿¿¿

∝°=1.0 × 109−¿

1.2 ×108−¿=8.3 ×102−¿ ¿¿¿

[ H 3 PO4 ]=CH 3 PO4∝°=0.1× 8.3× 102−¿=8.3× 103−¿ M ¿¿

∝1=Ka 1 ¿¿¿

¿

∝2=Ka 1 Ka 2¿¿

¿

∝3=Ka 1 K a2 K a3

¿¿¿

¿

We see that at pH=3.0, the majority (91%) of the phosphoric acid exists as H2PO4-

and 8.3% exists as H3PO4, only 3.31012-% exists as PO43-.

Salts of polyprotic (acids or bases):-

1. Amphoteric salts:-H2PO4- possesses both acidic and basic properties. That is, it

is amphoteric. It ionizes as a weak acid and it also is a Bronsted base that hydrolyzes:-

52

H 2 PO4−¿ ↔ H+¿+H PO4

2−¿K a2=7.5 × 10−8=¿¿¿¿ ¿

H 2 PO4−¿+H 2 O↔ H 3 PO4+OH

−¿K b=Kw

K a1=[H 3PO4]¿ ¿¿

¿

2. Unprotonated salt in a fairly strong Bronsted base in solution and ionizes as follows:-PO4

3−¿+H 2 O↔ H PO42−¿+OH

−¿K b=K wKa3

¿

¿ ¿

Example:- Calculate the pH of 0.1M Na3PO4 ?

PO43−¿+H 2 O↔ H PO4

2−¿+OH−¿¿¿¿

Equilibrium 0. 1−x x x

K b=K w

K a3=1.0 ×10−14

4.8 × 10−13 =0.02=¿¿

x2+0.02 x−2.0 × 103−¿=0¿

X=−0.2 ±√(0.02)2−4¿¿¿

pOH=1.44 , pH =12.56

Example (79):- EDTA is a polyprotic acid with four protons (H4Y). Calculate the hydrogen ion concentration of a 0.01M solution Na2EDTA (Na2H2Y)?

Solution:

H 2Y 2−¿+H 2 O↔ H 3 Y−¿+OH−¿ Kb=

K wK a2

=1.0 × 10−14

2.2 ×10−3 ¿

¿ ¿

H 2Y 2−¿ ↔ H ¿¿ ¿

Note (H2Y2- is the equivalent of HA- and H3Y- is the equivalent of H2A. The equilibrium constants involved are Ka2 and Ka3, the former conjugate acid H3Y- of the hydrolyzed salt).

Thus:-¿

53

Acid-Base TitrationsAn Acid-Base titration involves a neutralization reaction in which an acid is reacted with an equivalent amount of base at equivalence point or endpoint.

The titration is always a strong acid or strong base

A) Titration of strong acid versus strong base.

H+¿Cl−¿+Na+¿OH−¿→ Na+ ¿Cl+¿+H2O ¿

¿ ¿¿ ¿¿

HCl+NaOH → NaCl+H2 O

The equivalence point is where the reaction is theoretically complete while the endpoint where the colour of indicator were changed.

Detection of the endpoint and equivalence point (selection indicators):

Example (80):- Calculate the pH at 0, 10, 90, 100, 110% titration of 50.0mL of 0.10M HCl with 0.10M NaOH?

Solution:

(1) At 0% titration: before addition of 0.1M NaOH

pH=−log¿¿

(2) At 10% titration: before equivalence point

50 mL× 10100

=5 mL NaOH added

mmol NaOH added=M × V=0.1 ×5=0.5

mmol HCl=M ×V=0.1×50=50

mmol HCl remaining=mmol HCl total−mmol NaOH added

¿5.0−0.5=4.5

M HCl=mmol

V= 4.5

50+5=0.0818 M

pH=−log 0.0818=1.09

(3) At 90% titration: before equivalence point54

50 mL× 90100

=45 mL NaOH added

mmol NaOH added=M × V=0.1 × 45=4.5

mmol HCl remaining (unreacted )=5.0−4.5=0.5

M HCl=mmol

V= 0.5

50+45=0.00526 M

pH=−log 0.00526=2.88

(4) At 100% titration: equivalence point

50 mL× 100100

=50 mL NaOH added

mmol NaOH added=M × V=0.1 ×50=5

mmol NaOH added (5.0 )=mmol HCl (5.0)

2 H 2O ↔ H3 O+¿+OH−¿¿ ¿

Kw=1 ×10−14=¿

√1× 10−14=¿

pH=−log10−7=7 (neutrilization step)

(5) At 110% titration: after equivalence point

50 mL× 110100

=55 mL NaOH added

mmol NaOH added=M × V=0.1 ×55=5.5

mmol NaOH remaining (excess )=5.0−4.5=0.5

M NaOH =mmolV

= 0.550+55

=0.00476 M

pOH=−log 0.00476=2.32

pH=14−2.32=11.68

Construction (plot) titration curve of strong acid versus strong base:-

55

The relationship between pH calculated for HCl remaining or unreacted (excess) or NaOH on Y axis and the volume of titrant (0.1M NaOH) added on X axis, this curve called titration curve. This curve used for estimation the equivalence point (theoretically) and selection of the indicator for detecting the endpoint reaction by the colour change of the indicator.

Note: The selection of the indicator become more critical as the solution

become more dilute and the sharpness endpoint decrease as the concentration.

The point at which the reaction is observed to be complete at the indicator colour where changed is called endpoint.

Acid-Base indicators (also known as pH indicators):- are substances which change colour with change pH. They are usually weak acids or bases, which when dissolved in water dissociate slightly and form ions.

Phenolphthalein is a colourless, weak acid which dissociates in water forming pink anions. Under acidic conditions, the equilibrium is to the left, and the concentration of the anions is too low for the pink colour to be observed. However, under alkaline conditions, the equilibrium is to the right, and the concentration of the anion becomes sufficient for the pink colour to be observed.

56

1=(0.1M)versus (0.1M)

2=(0.1M)versus (0.1M)

3=(0.1M)versus (0.1M)

Acid–base titration is performed with a phenolphthalein indicator, when it is a strong acid – strong base titration, a bromthymol blue indicator in weak acid – strong base reactions, and a methyl orange indicator for strong acid – weak base reactions. If the base is off the scale, i.e. a pH of >13.5, and the acid has a pH >5.5, then an Alizarine yellow indicator may be used. On the other hand, if the acid is off the scale, i.e. a pH of <0.5, and the base has a pH <8.5, then a Thymol Blue indicator may be used.

Common Acid – Base Indicators

Indicator Approximate pH Range for Color Change

Color Change

Methyl Orange 3.2-4.4 Red to yellowBromthymol blue 6.0-7.6 Yellow to bluePhenolphthalein 8.2-10 Colorless to pink

Litmus 5.5-8.2 Red to blueBromcresol green 3.8-5.4 Yellow to blue

Thymol blue 8.0-9.6 Yellow to blue

Assume the indicator is a weak acid designated HIn, and assume that the nonionized form is red while the ionized form is blue and designated In-.

HIn ↔ H+¿+¿−¿,Ka=¿¿¿ ¿

Acidic (nonionized form), red color Basic (ionized form), blue color

57

We can write a Henderson–Hasselbalch equation for this just like other weak acids:

pH=pKa+log ¿¿

Your eyes can generally discern only one color if it is 10 times as intense as the other.

When only the color of the nonionized form is seen:

pH=pKa+log 110

=pKa−1

When only the color of the ionized form is observed:

pH=pKa+log 101

=pKa+1

So, the pH is going from one color to the other has changed from pKa-1 to pKa+1. This is a pH change of 2 units, and most indicators require a transition range of about two pH units. Choose an indicator with a pKa near the equivalence point.

Two drops (0.1mL) of 0.01m indicator (0.1% solution with Fwt=100) is equal to 0.01mL of 0.1M titrant.

B) Titration of weak acid versus strong base:-Acetic acid with sodium hydroxide

HOAc+Na+¿ OH−¿↔Na+¿OAc−¿+H 2O¿

¿¿¿

Example (81):- Calculate the pH at 0, 10, 25, 50, and 60mL titrant in the titration of 50mL of 0.1M acetic acid (Ka=1.75×10-5) with 0.10M NaOH?

Solution:

(1) At 0mL titrant (0.1M NaOH): HOAc solution only

¿

58

pH=−log¿¿

(2)At 10mL titrant (0.1M NaOH):before equivalence point (buffer formation region)

HOAc+Na+¿ OH−¿↔Na+¿OAc−¿+H 2O¿

¿¿¿

mmol NaOH added=M × V=0.1 ×10=1.0=mmolof NaOAc ( salt ) formed

mmol of HOAc ( total )=M × V=0.1 ×50=5.0

mmol of HOAc remaining (unreacted )=5.0−1.0=4.0

pH=pKa+log [Salt ][ Acid ]

=4.76+ log 1.04.0

=4.16

(3) At 25mL titrant (0.1M NaOH): before equivalence point (buffer formation region)

mmol NaOH added=M × V=0.1 ×25=2.5 mmol=mmol of NaOAc ( salt ) formed

mmol of HOAc remaining (unreacted )=5.0−2.5=2.5 mmol

pH=pKa+log [ Salt ][ Acid ]

=4.76+ log 2.52.5

=4.76

pH=pKa(mid−point )

(4) At 50mL titrant (0.1M NaOH): equivalence point

mmol NaOH added=M × V=0.1 ×50=5.0=mmolof NaOAc ( salt ) formed

All NaOH reacted with all HOAc and converted it to its salt sodium acetate.

OAc−¿+H 2O ↔ HOAc+OH −¿¿¿

¿

pOH=−log ¿¿

pH=14−5.27=8.73

(5) At 60mL titrant (0.1M NaOH): after equivalence point (NaOH solution alone)

mmol NaOH added=M × V=0.1 ×60=6.0 mmol

mmol of NaOH remaining (excess )=6.0−5.0=1.0 mmol

59

pOH=−log 1.0 mmol50 mL+60 mL

=2.04

pH=14−2.04=11.96

Construction (plotting) titration curve of weak acid versus strong base:-

The sharpness endpoint decreases as the concentration decreases.

Titration curves for 50mL 0.1M weak acids of different Ka value versus 0.1M NaOH.

The sharpness of the endpoint decreases as Ka decreases.

60

C) Titration of weak base versus strong acid:-Titration of ammonia solution versus hydrochloric acid.

Example:- Calculate the pH at 0, 10, 25, 50, and 60mL of titrant of 50mL of 0.1M NH3 (Kb=1.75×10-5) with 0.1M HCl?

(1) At 0mL titrant (0.1M HCl): NH3 solution alone

¿

pOH=−log ¿¿

pH=14−2.88=11.12

(2) At 10mL titrant (0.1M HCl): before equivalence point (buffer formation region)

mmol HCl added=M × V=0.1 ×10=1.0=mmolof NH 4 Cl (salt ) formed

mmol NH3 total=M × V=0.1×50=5.0

mmol NH3 remaining (unreacted )=5.0−1.0=4.0

pOH=pK b+ log [Salt ][Base ]

=4.75+ log 1.04.0

=4.15

pH=14−4.15=9.85

(3) At 25mL titrant (0.1M HCl): before equivalence point (buffer formation region)

mmol HCl added=M × V=0.1 ×25=2.5 mmol=2.5 mmol of NH 4Cl (salt ) formed

mmol NH3 remaining (unreacted )=5.0−2.5=2.5

pOH=pK b+ log [Salt ][Base ]

=4.75+ log 2.52.5

=4.75 pOH=pK b mid−point

(4) At 50mL titrant (0.1M HCl): equivalence point region

mmol HCl added=M × V=0.1 ×50=5.0 mmol

The all HCl added converted all NH3 to its salt NH4Cl

NH 4+¿+H 2 O↔ NH4 OH +H +¿¿¿

¿

61

pH=−log5.35 ×10−6=5.27

(5) At 60mL titrant (0.1M HCl): after equivalence point

mmol HCl added=M × V=0.1 ×60=6.0 mmol

mmol HCl remaining (excess )=6.0−5.0=1.0

M HCl=1.0 mmol

50 mL+60 mL=0.00909 M

pH=−log 0.00909=2.05

Construction (plotting) titration curve:-

Titration curve for 50mL 0.1M weak base of different Kb values versus 0.1M

HCl. The sharpness of the endpoint decreases as Kb decreases of weak bases.

Titration of sodium hydroxide versus mixture of acids:

Mixtures of acids (or bases) can be titrated stepwise if its difference in Ka

values of at least 104, unless perhaps a pH meter is used to construct the titration curve.

62

Note: one acid should be at least 104 weaker than the other to titrate separately.

Example:- A mixture of HCl and H3PO4 is titrated with 0.1M NaOH. The first endpoint (methyl red) occurs at 35.0mL, and the second endpoint (bromothymol blue) occurs at total of 50.0mL (15.0mL after the first point). Calculate the millimoles HCl and H3PO4 present in the solution?

The second endpoint corresponds to that the titration of one proton of H3PO4 (H2PO4

-→HPO42-).

mmol H 3 PO4=(M ×V )NaOH=0.1× 15=1.5 mmol

The HCl and the first proton of H3PO4 titrate. A 15mL portion of base was used to titrate the first proton of H3PO4 (same as for the second proton), leaving 20mL used to titrate the HCl.

mmol HCl=(M ×V )NaOH=0.1 × (35−15 )=0.1 ×20=2.0 mmol

The difference in Kb values must be at least 104 to titrate separately.

Example (82):- A 0.527g sample of mixture containing Na2CO3 and NaHCO3

and inert impurities is titrated with 0.109M HCl, requiring 15.7mL to reach the phenolphthalein endpoint and a total of 43.8mL to reach the modified methyl orange endpoint, what is the percent each of Na2CO3 and NaHCO3 in mixture?

Solution:

The phenolphthalein endpoint volume HCl equivalent for half amount of Na2CO3

Na2CO3+2 HCl ↔ H 2CO3+2 NaCl

mmol Na2CO3=mmol HCl× 12

( wtM . wt

×1000)Na2 CO3

= (M ×V )HCl ×12

63

( wt106

× 1000)Na2 CO 3

=(0.109 ×15.7 × 2 )HCl ×12

wt Na2 CO3=1.711× 106

1000=0.181 g Na2 CO3

Na2CO3 %=wtNa 2CO3

wtsample×100=0.181

0.527× 100=34.34 %

The volume HCl at second endpoint (modified methyl orange) equal for second half amount of Na2CO3 plus the amount of NaHCO3.

mmol NaH CO3=mmol HCl × 11

( wtM . wt

×1000)NaH CO3

=¿¿

( wt84

×1000)NaH CO3

=( 0.109×(43.8−2 ×15.7))HCl

wt NaHCO3=0.113 g

NaH CO3 %=wt NaH CO3

wt sample×100=0.113

0.527×100=21.5 %

Example (83):- Sodium hydroxide and sodium carbonate will titrate together to a phenolphthalein endpoint (OH-→H2O, CO3

2-→HCO3-). A mixture of

NaOH and Na2CO3 is titrated with 0.250M HCl, requiring 26.2mL for phenolphthalein endpoint and an additional 15.2mL to reach the modified methyl orange endpoint, how many milligrams NaOH and Na2CO3 are in the mixture.

Solution:

The volume HCl at first endpoint equivalent all hydroxide and half amount of Na2CO3= 26.2mL.mmol NaOH=mmol HCl × 1

1

( wtM . wt )=M ×(V 1−V 2)

( wt40 )=0.25 ×(26.2−15.2)

wt NaOH=110 mg

64

The volume HCl at second endpoint equivalent the second half of Na2CO3=15.2

mmol Na2CO3=mmol HCl× 12

( wtM . wt

×1000)Na 2 CO3

=( M ×2 V 2 )HCl ×12

wt106

=0.250 ×2×15.2 × 12

wt Na2 CO3=7.6 ×106 × 1

2=402.8 mg

65

External questions

Q1: Calculate the number of mole of ammonia, NH3, required to produce 2.5 mol of Cu (NH3)4SO4 according to the equation :

CuSO4 + 4NH3 Cu (NH3)4SO4

Solution:

No. of mol. of NH3 = No. of mol. of Cu (NH3)4SO4 × 4

= 2.5 × 4 = 10 mol. of NH3

Q2: Calculate the number of mole of Ca (HCO3)2 required to prepare 1.5 mol. of CO2 according to the equation:

Ca (HCO3)2 + 2HCl CaCl2 + 2CO2 + 2H2O

Solution:

No. of mol of Ca(HCO3)2 = No. of mol of CO2× 0.5

= 1.5 × 0.5 = 0.75 mol of Ca(HCO3)2

Q3: Calculate the mass of BaCO3 produced when excess CO2 is bubbled through a solution containing 0.205 mol of Ba(OH)2 .

Ba (OH)2 + CO2 BaCO3 + H2O

Solution:

No. of mol. of BaCO3 = No. of mol. of Ba (OH)2 = 0.205 mol

Mol= wtM . wt wt. of BaCO3 = mol. × M. wt

Wt. of Ba CO3 = 0.205 × 197.4= 40.5 gm

66

Q4: Caustic soda, NaOH, can be prepared commercially by the reaction of Na2CO3 with slaked lime, Ca(OH)2 . How many gram of NaOH can be obtained by treating 1.000 kg of Na2CO3 with Ca(OH)2 ?

Na2CO3 + Ca(OH)2 2NaOH + CaCO3

Solution:

mol= wtM . wt mol of Na2CO3 = 1000 gm / 106 gm . mol-1

mol of Na2CO3= 9.433 mol of Na2CO3

No. of mol of NaOH = 2 × No. of mol of Na2CO3 = 2× 9.433=18.87 mol

wt of NaOH = mol × M.wt = 18.87 × 40 = 755 gm NaOH

Q5: What mass of KI is needed to produce 69.6 gm of K2SO4 by the reaction :

8KI + 5H2SO4 4K2SO4 + 4I2 + H2S + 4H2O

Solution:

No. of mol of K2SO4= wt / M.wt 69.6 gm / 174 gm . mol-1= 0.4 mol

No. of mol of KI = 2× 0.4= 0.8 mol

wt of KI = mol × M.wt = 0.8 × 166 = 133gm K

Q6: What volume of 1.71 M NaCl solution contains 0.2 mol of NaCl ?

Solution:

Molarity = mol / volume volume=mol / molarity

V = 0.2 / 1.71 = 0.117 L = 117 ml

Q7: What volume of 0.3 M NaOH ( M.wt= 40 g mol-1) can be prepared with 84 gm NaOH ?

Solution:

M=wt (g)

M . wt ( gmol

)× 1000

V (mL)

67

3.0=84(g)

40( gmol

)× 1000

V (mL)=700 ml

Q8: How many of water must be added to 200 ml of 0.65 M HCl to dilute the solution to 0.2 M ?

Solution:

M1 V1 = M2 V2 200 × 0.65 = V2 × 0.2

V2 = 650 ml approximately 650 – 200 = 450 ml must be added

Q9: How much 1.00 M HCl should be mixed with what volume of 0.25 M HCl in order to prepare 1.00 L of 0.500 M HCl ?

Solution:

Let x= volume of 0.25 M HCl ; then 1.00- x = volume of 1.00 M HCl

No. of mol of S1 + No. of S2 = No. of S3

(1.00 – x) 1.00 + x (0.25) = 1.00(0.500)

Thus x= 0.667 L = 667 ml of 0.25 M HCl and 1.00 – x=

1.00 – 0.667=0.333 L = 333 ml

Q10: What is the molar concentration of a solution containing 16 gm CH3OH in 200 ml solution ?

Solution:

M= wt (g)

M .wt ( gmol

)× 1000

V (mL)= 16 (g)

32( gmol

)× 1000

200(mL )=2.5 M

Normality

Q11: What is the difference between the definition of an equivalent in an acid- base reaction and an equivalent in an oxidation – reduction reaction?

Solution:

68

An equivalent in an acid – base reaction is that amount of a substance which reacts with or liberates 1 mol of hydrogen ions ; an equivalent in a redox reaction is that amount of substance which reacts with or liberates 1 mol of electrons .

Q12: What volume of a 0.232 N solution contains (a) 3.17 meq of solute (b) 6.5 eq of solute ?

Solution:

(a ) N= No . of milliequivalentSolutionVolumn(ml)

SolutionVolumn(ml)= No .of milliequivalentN

= 3.170.232

=13.7 ml

(b ) SolutionVolumn(L)=No .of equivalentN

= 6.50.232

=28.0 L

Q13: How many (a) equivalent and (b) milliequivalent of solute are present in 60 ml of 4.0 N solution?

Solution:

(a) No. of equivalent = No. of liters × normality = 0.060 × 4.0 = 0.24 eq .

(b) 0.24 eq × 1000 = 240 meq .

Q14: How many equivalent of solute are contained in (a) 1 L of 2 N solution, (b) 1L of 0.5 N solution , (c) 0.5 L of 0.2 N solution ?

(a) No. of eq = 2 eq , (b) No. of eq = 0.5 eq , (c) No. of eq = 0.1 eq

Q15:What is normality of 0.3 M H3PO3 when it undergoes the following reaction ? H3PO3 + 2OH- HPO3

-2 + 2H2O

Solution:

Normality = a × molarity = 2 × 0.3 = 0.6 N .

Q16: How many cm3 of concentrated sulfuric acid, of density 1.84

g/ cm3 and containing 98.0 % H2SO4 by weight , should be taken to make (a) 1 L of 1 N solution (b) 1 L of 3 N solution (c) 200 cm3 of 0.5 N solution ?

69

(a) 27.2 cm3 ,(b) 81.7 cm3 , (c) 2.72 cm3.

Mole fraction and Molality

Q17: Calculate the mole fraction(x) of water in a mixture consisting of 9.0 gm water , 120 gm acetic acid , and 115 gm ethyl alcohol ?

Solution:

The molecular weight of water , acetic acid , and ethyl alcohol are 18 , 60 and 46 gm/mol , respectively .

Molwater =wt / M.wt 9.0 / 18 = 0.5 mol

Molacetic acid =wt / M.wt 120 / 60 = 2.0 mol

Molethyl alcohol =wt / M.wt 115 / 46 = 2.5 mol

Sum of mole = 0.5 + 2.0 + 2.5 = 5.0 mole

Mole fraction for waterX=no .mole water (n1 )

∑ of mole

Mole fraction for waterX=0.5 mole5.0 mole

=0.10

Q18: What is the mole fraction of H2 in a gaseous mixture containing 1.0 gm H2 , 8.0 gm O2 , and 16 gm CH4 ?

Solution:

Mol H2 =wt / M.wt 1.0 / 2.0 = 0.5 mol

Mol O2 =wt / M.wt 8.0 / 32.0 = 0.25mol

Mol CH4 =wt / M.wt 16 / 16 = 1.0 mol

70

Sum of mole = 0.5 + 0.25 + 1.0 = 1.75 mole

Mole fraction for H2( X)=no .mole H 2 (n1 )∑ of mole

Mole fraction for H2( X)= 0.5 mole1.75 mole

=0.29

Q19: A solution contains 116 gm acetone ( CH3COCH3 ) , 138 gm ethyl alcohol ( C2H5OH ) , and 126 gm water . Determine the mole fraction of each ?

Solution:

Mol acetone =wt / M.wt 116 / 58.0 = 2.0 mol

Mol C2H5OH =wt / M.wt 138 / 46.0 = 3.0mol

Mol water =wt / M.wt 126 / 18 = 7.0 mol

Sum of mole = 2.0 + 3.0 + 7.0 = 12.0 mole

Mole fraction for (X )=no .mole solute∑ of mole

Mole fraction for acetone(X )= 2.0 mole12.0 mole

=0.167

Mole fraction for C 2H 5OH ( X)= 3.0 mole12.0 mole

=0.250

Mole fraction for water (X )= 7.0 mole12.0 mole

=0.583

Q20: Determine the mole fraction of both substances in a solution containing 36.0 gm water and 46.0 gm glycerin , C3H5(OH)3 ?

Solution:

The molecular weight of C3H5(OH)3 is 92.0 ; of water , 18.0

Mol glycerin =wt / M.wt 46.0 / 92.0 = 0.5 mol

Mol water =wt / M.wt 36.0 / 18.0 = 2.0 mol

71

Sum of mole = 0.5+ 2.0 = 2.5 mole

Mole fraction for glycerin (X )=0.5 mole2.5 mole

=0.20

Mole fraction for water (X )=2.0 mole2.5 mole

=0.80

Check : Sum of mole fractions = 0.20 + 0.80 = 1.0

Q21: The density of a 2.0 M solution of acetic acid (M.wt = 60 ) in water is 1.02 kg/L . Calculate the mole fraction of acetic acid ?

Solution:

Per liter of solution :

M=wt (g)

M . wt ( gmol

)× 1000

V (mL)

0.2= wt (g)

60( gmol

)× 1000

1000(mL)=120 gm acetic

Weight = volume × density = 1.000L × 1.02 kg/ L = 1.02kg = 1020 gm of solution .

Weight og water = 1020 – 120 = 900 gm water .

Mol water =wt / M.wt 900 / 18.0 = 50.0 mol

Sum of mole = 50.0+ 2.0 = 52.0 mole

Mole fraction for aceticacid (X )= 2.0 mole52.0 mole

=0.038

Q22: A solution contains 10.0 gm acetic acid, CH3COOH, in 125 gm water . What is the concentration of solution expressed as (a) mole fractions of CH3COOH and water (b) molality?Solution:

(a) Mol CH3COOH =wt / M.wt 10.0 / 60.0 = 0.167 mol

Mol water =wt / M.wt 125 / 18.0 = 6.94 mol72

Sum of mole = 0.167+ 6.94 = 7.107 mole

Mole fraction for C H 3COOH (X )=0.167 mole7.107 mole

=0.0235

Mole fraction for water (X )= 6.94 mole7.107 mole

=0.965

(b) molality = 0.167 mol C H 3 COOH / 0.125 kg water = 1.34 m

Q23: Calculate the molalities and mole fractions of acetic acid in two solutions prepared by dissolving 120 gm acetic acid (a) in 100 gm water (b) in 100 gm ethanol?

Solution:

(a) in water m=wt (g)

M . wt ( gmol

)× 1000

wt (gm)

m=120(g)

60( gmol

)× 1000

100 gm=2.00 m acetic acid

Mole fraction for CH 3 COOH ( X)= 2.00 mole2.00+5.55 mole

=0.265

(b) in ethanol

m=wt (g)

M . wt ( gmol

)× 1000

wt (gm)

m= 120(g)

60( gmol

)× 1000

100 gm=2.00acetic acid

Mole fraction for C H 3COOH (X )= 2.00 mole2.00+2.17 mole

=0.480

Q24: What is the molality of a solution which contains 20.0 gm cane suger , C12H22O11 , dissolved in 125 gm water ?

73

Solution:

The molecular weight of C12H22O11 is 342 .

m=wt (g)

M . wt ( gmol

)× 1000

wt (gm)

m=20.0(g)

342( gmol

)× 1000

125 gm=0.468 m

Q25: The molality of a solution of ethanol , C2H5OH , in water is 1.54 m. How many gm of ethanol is dissolved in 2.50 kg water ?

Solution:

Molality= mole / wt(kg) mole = 2.50 × 1.54 = 3.85 mol

and mass of ethanol = mole × M.wt 3.85 × 46.1 = 177 gm

Q26: Calculate the molality of a solution containing (a) 0.65 mol glucose , C6H12O6 , in 250 gm water (b) 45 gm glucose in 1.00 kg water (c) 18 gm glucose in 200 gm water ?

Solution:

(a) Molality= mole / wt(kg) molality = 0.65/ 0.250 = 2.6 m

(b)

m=wt (g)

M . wt ( gmol

)× 1000

wt (gm)

m=45 (g)

180( gmol

)× 1000

1000 gm=0.25 m

(c) m=wt (g)

M . wt ( gmol

)× 1000

wt (gm)

74

m=18(g)

180( gmol

)× 1000

200 gm=0.50 m

Q27: How many gram CaCl2 should be added to 300 ml water to make up a 2.46 m solution?

Solution:

Assuming that water has a density of 1.00 gm/ mL.

Weight = density × volume wt = 300 × 1.00 = 300 gm

m= wt (g)

M . wt ( gmol

)× 1000

wt (gm)

2.46=wt (g)

111( gmol

)× 1000

300 gm=81.9 gm

Acids and bases

Q28: Distinguish between acid strength and acid concentration?

Solution:

The concentration of acid in a solution is determined by how many mol. of acid is dissolved per L of solution; its strength is determined by how completely it ionizes. Both of these factors affect the hydronium ion concentration.

Q29: Explain why a solution containing a strong base and its salts does not act as a buffer solution?

Solution:

Addition of OH- does not shift an equilibrium toward un-ionized base, as it would with a weak base and it’s conjugate.

Q30 : Explain the difference between a strong electrolyte and a weak electrolyte. Is an “insoluble” salt a weak or a strong electrolyte?

75

Solution:

A strong electrolyte is completely ionized in solution, while a weak electrolyte is only partially ionized in solution.

Q31: What is the Brønsted acid–base theory? What is the Lewis acid–base theory?

Solution:

The bronsted acid- base theory assumes that an acid is a proton donor, and a base is a proton accepter. In the Lewis theory, an acid is an electron accepter, while a base is an electron donor.

Q32: Calculate the pH of a solution which has a hydronium ion concentration

of 6 × 10-8 M?

Solution:

PH = -log[H+] PH = - log 6 – log 10-8 = -0.78 + 8= 7.22

Q33: Calculate the hydronium ion concentration of a solution which

has a pH of 11.73 ?

Solution:

[ H+] = 10-PH [H+] = 10-11.73 = 100.27× 10-12

From the logarithm table , 100.27 = 1.9 [H+] = 1.9×10-12

Q34: Calculate the hydrogen ion concentration and the hydroxide ion concentration in pure water at 25 0C ?

Solution:

2H2O H3O+ + OH-

76

Kw = [H+] [OH-] = 1.0 ×10-14

Let x = [H+] = [OH-] , hence x2 = 1.0 × 10-14 and x = 1.0 ×10-7M =[H+]= [OH-]

Q35: Calculate the hydronium ion concentration of a 0.100 M NaOH solution ?

Solution:

NaOH Na+ + OH-

0.100M 0.100M

Kw = [H+] [OH-] = 1.0 ×10-14

In this solution, 1.0 × 10-14 = [H+] (0.100) , thus [H+] = 1.0 × 10-13 M.

Q36: Calculate the PH values , assuming complete ionization , of (a) 4.9× 10-4 M monoprotic acid (b) 0.0016 M monoprotic base ?

Solution:

(a) [H+] = 4.9× 10-4 PH = -log [H+] = -log(4.9×10-4) =

-log 4.9 + 4 = 3.31

(b) Kw = [H+] [OH-] = 1.0 ×10-14 [H+] = 10-14 / [OH-]

[H+] = 10-14 / 1.6×10-3

PH = -log 10-14 / 1.6×10-3 = -(- 14- log 1.6 + 3) = 14 + 0.20 – 3 = 11.20

Q37: Calculate the PH of 1.0 × 10-3 M solutions of each the following: (a) HCl (b) NaOH (c) Ba(OH)2 (d) NaCl?

Solution:

(a) PH = 3.00 (b) PH = 11.00 (c) PH = 11.30 (d) PH = 7

Q38: Calculate the pH and pOH of the following strong acid solutions: (a) 0.020 M HClO4, (b) 1.3 × 10−4 M HNO3, (c) 1.2 M HCl?

Solution:

(a) PH = - log [ H+ ] = - log 2× 10-2 = 2 – 0.3 = 1.777

POH = 14 – 1.7 = 12.3

(b) PH = - log [ H+ ] = - log 1.3× 10-4 = 4 – 0.11 = 3.89 POH = 14 – 3.89 = 10.11

(c) PH = - log [ H+ ] = - log 1.2 = - 0.08

POH = 14 – (-0.08) = 14.08

Q39:Calculate the pH and pOH of the following strong base solutions: (a) 0.050 M NaOH, (b) 2.4 M NaOH,(c) 3.7 × 10−3 M KOH.?

Solution:

(a) POH = - log [ OH- ] = -log 5× 10-2 = 2- 0.7 = 1.3

PH = 14 – 1.3 = 12.7

(b) POH = - log [ OH- ] = -log 2.4 = - 0.38

PH = 14 – (-0.38) = 14.38

(c) POH = - log [ OH- ] = -log 3.7×10-3 = 3 – 0.57 = 2.43

PH = 14 – 2.43 = 11.57

Q40: Calculate the hydrogen ion concentration of the solutions with the following pH values: (a) 3.47, (b) 0.20, (c) 8.60?

Solution:

(a) [ H+ ] = 10-PH = 10-3.47 = 10-4 × 100.53 = 3.4 × 10-4 M .

(b) [ H+ ] = 10-PH = 10-0.02 = 10-1 × 100.80 = 6.3× 10-1 M .

(c) [ H+ ] = 10-PH = 10-8.6 = 10-9 × 100.40 = 2.5 × 10-9 M .

Q41: Calculate the pH and pOH of a solution obtained by mixing equal volumes of 0.10 M H2SO4 and 0.30 M NaOH ?Solution:

Assume the volume = 1 ml 78

Excess of NaOH = mmole of NaOH – mmole of H2SO4

= 0.3× 1 ml – 0.1× 1 ml× 2 = 0.1 mmole of NaOH

M = mmole / V(ml) = 0.1 / 2 ml = 0.5 M .

POH = -log [ OH- ] = -log 5× 10-2 = 2 – 0.7 = 1.3

PH = 14 – 1.3 = 12.70

Q42: Calculate the pH of a solution obtained by mixing equal volumes of a strong Solution:

Assume the volume = 1 ml [ H+ ] = 10-PH

[ H+ ] of acid solution = 1.0 ×10-3 M .

[ H+ ] of base solution = 1.0× 10-12 M .

[ H+ ] [ OH- ] = 1.0× 10-14

1.0× 10-12 [ OH- ] = 1.0 × 10-14 ; [ OH- ] = 1.0×10-14 / 1.0× 10-12 = 1.0×10-2 M .

Excess of base = mmole of NaOH – mmole of acid

= 1.0×10-2 * 1 ml – 1.0×10-3 * 1 ml = 9 / 2 × 10-3 = 4.5×10-3 M .

POH = -log [ OH- ] = -log 4.5 × 10-3 = 3 – 0.65 = 2.35

PH = 14 – 2.35 = 11.65

Problems 5

1 -How many moles and milIimoles of benzoic acid (Mw = 122.1 g/mol) are

contained in 2.00 g of the pure acid?

2- How many grams of Na+ (22.99 g/mol) are contained in 25.0 g of Na2S04

(142.0 g/mol).

79

3- Describe the preparation of 500 mL of 0.0740 M Cl- solution from solid

BaCl2.2H2O (244.3 g/mol).

4 -Find the number of millimoles of the indicated species in (a) 57 mg of P2O5 (b)

12.92 g of CO2 (c) 40.0 g of NaHCO3 (d) 850 mg of MgNH4PO4

5 -What is the mass in grams of

a- 7.1 mol of KBr b- 20.1 mmol of PbO c- 3.76 mol of MgSO4

6- Calculate the p-value for each of the indicated ions in the following:

a- Ba2+, Mn2+, and Cl- in a solution that is 7.65 × 10-3 M in BaC12 and 1.54 M in

MnC12.

b- Cu2+, Zn2+, and NO3- in a solution that is 4.78 × 10-2 M in Cu(NO3)2 and 0.104

M in Zn(NO3)2

c- H+, Ba2+, and CIO4 in a solution that is 3.35 × 10-4 M in Ba(ClO4)2 and 6.75 × 10-

4 M in HC1O4.

7- Sea water contains an average of 1.08 × l03 ppm of Na+ and 270 ppm of SO4-.

Calculate(a) the molar concentrations of Na+ and SO4- given that the average

density of sea water is1.02 g/mL. (b) the pNa and pSO4 for sea water.

8- A solution was prepared by dissolving 1210 mg of K3Fe(CN)6 (329.2 g/mol) in

sufficient water to give 775 mL. Calculate

(a) the molar analytical concentration of K3Fe(CN)6. (b) the molar concentration

of K+. (c) the molar concentration of Fe(CN)63- . (d) the weight/volume percentage

of K3Fe(CN)6. (e) the number of millimoles of K+ in 50.0 mL of this solution. (f)

ppm Fe(CN)63- . (g) pK+ for the solution. (h) pFe(CN)6

3- for the solution.

9- A 12.5% (w/w) NiCl2 (129.61 g/mol) solution has a density of 1.149 g/mL.

Calculate (a) the molar concentration of NiCl2 in this solution. (b) The molar Cl-

Concentration of the solution. (c) The mass in grams of NiCl2 contained in each

liter of this solution.

10- Describe the preparation of (a) 2.50 L of 21.0% (w/v) aqueous glycerol

(C3H8O3 (92.1 g/mol). (b) 2.50 kg of 21.0% (w/w) aqueous glycerol. (c) 2.50 L of

21.0% (v/v) aqueous glycerol.80

11- Describe the preparation of 900 mL of 3.00 M HNO3 from the commercial

reagent that is 70.5% HNO3 (w/w) and has a specific gravity of 1.42.

12- A solution containing 10.0 mmol CaCl2 is diluted to 1 L. Calculate the number

of grams of CaCl2. 2H2O per milliliter of the final solution.

13- Calculate the grams of each substance required to prepare the following

solutions: (a) 250mL of 0.100 M KOH, (b) 1.00 L of 0.0275 M K2Cr2O7, (c)

500mL of 0.0500 M CuSO4.

14- How many milliliters of concentrated hydrochloric acid, 38.0% (wt/wt),

specific gravity 1.19, are required to prepare 1 L of a 0.100 M solution?

15- Calculate the molar concentrations of 1.00-mg/L solutions of each of the

following. (a) AgNO3, (b) Al2(SO4)3, (c) CO2, (d) (NH4)4Ce(SO4)4. 2H2O, (e) HCl,

(f) HClO4.

16- How many grams NaCl should be weighed out to prepare 1 L of a 100-mg/L

solution of (a) Na+ and (b) Cl−?

17- What volume of 0.50 M H2SO4 must be added to 65mL of 0.20 M H2SO4 to

give a final solution of 0.35 M?

18- What is the molar concentration of NO3– in a solution prepared by mixing 50.0

mL of 0.050 M KNO3 with 40.0 mL of 0.075 M NaNO3? What is pNO3 for the

mixture?

19- For each of the following, explain how you would prepare 1.0 L of a solution

that is 0.10 M in K+. Repeat for concentrations of 1.0×102 ppm K+ and 1.0% w/v

K+. a. KCl b. K2SO4 c. K3Fe(CN)6

20- A solution was prepared by dissolving 5.76 g of KC1· MgC12· 6H20 (277.85

g/mol) in sufficient water to give 2.000 L. Calculate: (a) the molar analytical

concentration of KCl.MgCl2 in this solution. (b) The molar concentration of

Mg2+. (c) The molar concentration of Cl-. (d) The weight/volume percentage of

KCl.MgCl2 .6H20. (e) The number of millimoles of Cl- in 25.0 mL of this solution.

(f) ppm K+. (g) pMg for the solution.

81

82