Ch 8: Aqueous Solutions: Chemistry of the Hydrosphere · 4 Concentration of Solutions - Molarity...

1

Ch 8: Aqueous Solutions:

Chemistry of the Hydrosphere

H2S + Cu2+ → CuS(s) + 2H+

(Fe, Ni, Mn also)

HS- + 2 O2 HSO4- + energy

(supports life)

3

Figure taken from “Principles of Biochemistry”, 2nd Ed. By Lehninger, Nelson, and Cox

Chapter Outline

8.1 Solutions and Their Concentrations

8.2 Dilutions

8.3 Electrolytes and Nonelectrolytes

8.4 Acids, Bases, and Neutralization Reactions

8.5 Precipitation Reactions

8.6 Oxidation-Reduction Reactions

8.7 Titrations

8.8 Ion Exchange

4

Concentration of Solutions - Molarity

The concentration of a solution is the amount of solute

present in a given quantity of solvent or solution.

Molarity (M) = liters of solution

moles of solute

Mass-to-mass ratios:

Mass-to-volume ratios:

Parts per million:

mg/kg solvent

mg/L solvent

1 g 106 g

= mg 103 g

= mg L

dH2O = 1.00 g/mL @ 25 oC

Sample Exercise 8.1: Converting Mass-per-

Volume Concentrations into Molarity

Vinyl chloride (C2H3Cl, MW = 62.49) is one of the most widely used industrial chemicals. It is also one of the most toxic, and it’s a known carcinogen. The maximum concentration of vinyl chloride allowed in drinking water in the U.S. is 0.002 mg/L. What is that concentration in moles per liter?

Plan: mg/L g/L mol/L = Molarity

0.002 mg

L x

1 g

1000 mg x

1 mol

62.49 g = 3.2 x 10-8

mol/L

5

Sample Exercise 8.2: Converting Mass-per-Mass

Concentrations into Molarity

A water sample from the Great Salt Lake in Utah contains 83.6 mg Na+ per gram of lake water (soln). What is the molar concentration of Na+ ions if the density of the lake water is 1.160 g/mL? AW Na = 22.99

Plan: mg Na+ g mol

1 g soln mL L

dsoln = 1.160 g/mL

mol

L = M

83.6 mg Na+ x 1 g

1000 mg x 1 mol Na+

22.99 g = 3.636 x 10-3 mol Na+

1 g soln x 1 mL soln

1.160 g x 1 L

1000 mL = 8.6221 x 10-4 L

mol

L M =

8.6221 x 10-4 L

3.636 x 10-3 mol Na+

= = 4.22 M

Sample Exercise 8.3: Calculating the Quantity of

Solute Needed to Prepare a Solution

An aqueous solution called phosphate-buffered saline (PBS) is used in biology research to wash and store living cells. It contains ionic solutes including 10.0 mM Na2HPO42H2O (MW = 178.0). How many grams of this solute would you need to prepare 10.0 L of PBS?

Plan: mM M mol g

use the volume

MM

Useful equation: M x V = moles

x 1 mol

1000 mmol x

178.0 g

mol = 17.8 g 10.0 mmol L

x 10.0 L

M x V = moles

6

Chapter Outline


8.2 Dilutions





8.7 Titrations

8.8 Ion Exchange

Dilution is the procedure for preparing a less concentrated

solution from a more concentrated solution.

Dilution

Add Solvent

Moles of solute

before dilution (i)

Moles of solute

after dilution (f) = MiVi MfVf =

7

Dilution Equation: Mi Vi = Mf Vf

The solution used in hospitals for intravenous infusion—called physiological or normal saline - is 0.155 M NaCl. It may be prepared by diluting a commercially available standard solution that is 1.76 M NaCl. What volume of standard solution (Vi) is required to prepare 10.0 L of physiological saline?

Sample Exercise 8.4: Diluting Solutions

Collect data:

Mi = 1.76 M Mf = 0.155 M

Vi = ? Vf = 10.0 L

Mi Vi = Mf Vf

Vi = Mf Vf

Mi

Vi = (0.155 M )(10.0 L)

(1.76 M)

= 0.881 L = 881 mL

add

H2O

Add 881 mL of

1.76 M NaCl,

and then “dilute

to the mark”

(10.0 L)

8

Chapter Outline


8.2 Dilutions





8.7 Titrations

8.8 Ion Exchange

9

Strong Electrolytes:

• Nearly 100% dissociated

into ions

• Conduct current efficiently

Examples: ionic compounds such as

NaCl, strong acids like HCl

• NaCl(s) → Na+(aq) + Cl−(aq)

• HCl(aq) + H2O(l) H3O+ + Cl-(aq)

Figure from John J. Fortman, J. Chem. Ed. Vol. 71, No. 1, 1994, p. 27-28

10

Weak Electrolytes:

• Only partially dissociate into

ions

• Slightly conductive

Example: weak acids such as

acetic acid

Hydronium Ion: H3O+

+

The higher the concentration of H3O+ in solution, the stronger the acid.

11

Non-electrolytes

Substances in which no ionization

occurs. There is no flow of electrical

current through the solution

Examples: molecular or covalent

compounds such as ethanol

12

Chapter Outline


8.2 Dilutions





8.7 Titrations

8.8 Ion Exchange

13

Acids

Have a sour taste. Vinegar owes its taste to

acetic acid. Citrus fruits contain citric acid.

React with certain metals to produce hydrogen gas.

React with carbonates and bicarbonates to produce carbon

dioxide gas

Cause color changes in plant dyes (litmus paper).

2HCl (aq) + Mg (s) MgCl2 (aq) + H2 (g)

2HCl (aq) + CaCO3 (s) CaCl2 (aq) + CO2 (g) + H2O (l)

Aqueous acid solutions conduct electricity.

Limestone

dissolving in acid =

cave

Most metals

dissolve in acids to

produce H2

14

Strong Acids and Bases

Strong Acids/Bases:

• Completely ionized

in aqueous solution

(i.e., strong electrolytes)

• Memorize the 6 strong

acids

• All other acids are weak

KNOW!

15

Have a bitter taste.

Feel slippery. Many soaps contain bases.

Bases

Cause color changes in plant dyes (litmus paper).

Aqueous base solutions conduct electricity.

A Brønsted acid is a proton donor

A Brønsted base is a proton acceptor

acid base conjugate

acid

conjugate

base

For every acid there is a conjugate base, and for every

base there is a conjugate acid (conj. acid-base pairs).

16

Water: Acid or Base? Water as Base:

HCl(aq) + H2O(l) → H3O+(aq) + Cl−(aq)

Water as Acid:

NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH−(aq)

Amphiprotic: Acts as an Acid or Base.

Preface: Double-Displacement Reactions Many of the reactions in this chapter follow this pattern:

AB + CD → AD + CB (+) (-) (+) (-) (+) (-) (+) (-)

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Acid-Base (Neutralization) Reactions

acid + base salt + water

HCl(aq) + NaOH(aq)

H2SO4(aq) + 2 KOH(aq)

HOH(aq) + NaCl(aq)

2 HOH(aq) + K2SO4(aq)

Net Ionic Equations

1. Write the balanced molecular equation.

2. Write the ionic equation showing the strong

electrolytes completely dissociated into cations and

anions.

3. Cancel the spectator ions on both sides of the ionic

equation

Tips:

• Do not break apart solids, liquids (such as H2O), and gases (e.g. CO2)

• Do not break apart weak electrolytes (such as weak acids, e.g.

HC2H3O2)

• Carbonic acid, H2CO3, decomposes into CO2 and H2O

• An acid such as H2SO4 breaks apart into 2H+(aq) + SO42-, not H2

2+ +

SO42-

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H2SO4(aq) + 2 KOH(aq) 2 H2O(aq) + K2SO4(aq)


2. Write the ionic equation showing the strong electrolytes

completely dissociated into cations and anions.

3. Cancel the spectator ions on both sides of the ionic equation

strong strong strong Not a strong

electrolyte!!

2 H+ SO42- 2 K+ 2OH- 2 H2O 2 K+ SO4

2-

2 H+ + 2 OH- 2 H2O

H2O H+ + OH- Net ionic equation:

Write the net ionic equation describing the reaction that takes place when acid rain containing sulfuric acid reacts with a marble statue (calcium carbonate) -

Sample Exercise 8.5: Writing the Net Ionic

Equation for a Neutralization Reaction


2. Write the ionic equation showing the strong electrolytes completely dissociated

into cations and anions.

*generally speaking, CaSO4 is insoluble in water, but not in strong acid. More on

this later when we study solubility rules.

H2SO4(aq) + CaCO3(s) H2O(l) + CO2(g) + CaSO4(aq)

strong solid strong* Not strong

electrolytes!!

2H+(aq) + SO42-(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca2+(aq) + SO4

2-(aq)

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Sample Exercise 8.5: Writing the Net Ionic

Equation for a Neutralization Reaction

3. Cancel the spectator ions on both sides of the ionic equation

2H+(aq) + SO42-(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca2+(aq) + SO4

2-(aq)

2H+(aq) + CaCO3(s) H2O(l) + CO2(g) + Ca2+(aq)

Net ionic equation:

Chapter Outline


8.2 Dilutions





8.7 Titrations

8.8 Ion Exchange

20

Double-Displacement Reactions Again

Precipitation reactions in this section follow the same

pattern as acid-base reactions:

AB + CD → AD + CB (+) (-) (+) (-) (+) (-) (+) (-)

Types of Solutions

Saturated solution:

• Contains the maximum amount of solute that can

dissolve in a given volume at a given temperature.

• Solubility = g solute/100 mL solvent.

Supersaturated solution:

• Contains more dissolved solute than predicted at a

given temperature.

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A Saturated Solution Example

Supersaturated Solution

Sodium acetate precipitates from a supersaturated solution.

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Precipitate:

• Solid product formed from reactants in

solution.

• AgNO3(aq) + NaCl(aq) → NaNO3(aq) + AgCl(s)

Can predict formation of precipitates

based on solubility “rules.”

We learned in Ch 4, p. 134, that the electrostatic energy (strength) of an

ionic bond depended on the product of the charges, divided by the

internuclear distance. The solubility of ionic compounds generally

follows the trend that the greater the charge, and the shorter the

internuclear distance, the more insoluble the compound is.

𝐸 α 𝑄+ 𝑥 𝑄−

𝑑

cations compounds anions compounds solubility

+1 alkali metals NH4

+

-1 NO3- CH3COO-

Always soluble

+2 -2 SO42- Soluble with exceptions

-2 S2-, O2- Insoluble with exceptions

+3 Al3+ -3 PO43- Insoluble with exceptions

α (alpha) means “is proportional to”

Q+ = charge on the cation

Q- = charge on the anion

d = internuclear distance

+ -

d

24

Always

soluble

Soluble

w/exceptions

Insoluble

w/exceptions

and Ag+ Add to table

Table 8.3 – Always Soluble Compounds

• Any compound containing Group I cations

• Any compound containing NH4+

• Any compound with NO3

− or CH3COO− (C2H3O2-)

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• Halides (Group 17)

» Exceptions: Cu+, Hg22+, Ag+, Pb2+

• Sulfates (SO42−)

» Exceptions: Hg22+, Ag+, Pb2+, Ca2+, Ba2+, Sr2+

Add to table

Table 8.3 - Soluble Compounds With Exceptions

Table 8.3 - Insoluble Compounds

All hydroxides (OH−)

All sulfides (S2−)

All carbonates (CO32−)

All phosphates (PO43−)

All chromates (CrO42-)

Don’t Forget:

except IA, NH4+

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Mnemonic device to help remember the

exceptions to the soluble compounds -

Cute HAPpy halides

Cl-, Br-, I-

Hg22+

Ag+ Pb2+

HAPpy sulfates "2"

Hg22+

Ag+

Pb2+ Cu+

Ca2+

Sr2+

Ba2+

from

Group 2

Mnemonic device to help remember the insoluble

compounds -

Shop for a chrome car ulfides

S2-

ydroxides

OH-

hosphates

PO43-

hromates

CrO42-

arbonates

CO32-

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Does a precipitate form when sodium chloride is mixed with silver nitrate?

If so, write the net ionic equation for the formation of the precipitate:

NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)

Applying the Solubility Rules – Will a Precipitate Form?

Na+(aq) + Ag+(aq) + Cl-(aq) + NO3-(aq) AgCl(s) + Na+(aq) + NO3

-(aq)

Ag+(aq) + Cl-(aq) AgCl(s)

Cute HAPpy halides

Cl-, Br-, I-

Hg22+ Ag+ Pb2+ Cu+

NaI(aq) + Pb(NO3)2(aq) →

Formation of an Insoluble Product

K+(aq) I

-(aq)

Pb2+(aq) NO3

-(aq)

The soluble compounds break up

into ions and recombine -

28

A precipitate forms when aqueous solutions of ammonium sulfate and

barium chloride are mixed. Write the net ionic equation for the reaction.

Sample Exercise 8.6

(NH4)2SO4(aq) + BaCl2(aq) 2 NH4Cl(aq) + BaSO4(s)

2NH4+(aq) + SO4

2-(aq) + Ba2+(aq) + 2Cl-(aq) BaSO4(s) + 2NH4+(aq) + 2Cl-(aq)

Ba2+(aq) + SO42-(aq) BaSO4(s)

Sample Exercise 8.7 – Prediction the Mass of a Precipitate

Barium sulfate is used to enhance Xray imaging of the upper and lower GI tract. In

upper GI imaging, patients drink a suspension of solid barium sulfate in H2O. The

compound is not toxic because of its limited solubility. To make pure barium sulfate,

a precipitation reaction is employed: aqueous solutions of soluble barium nitrate

and sodium sulfate are mixed together, and solid barium sulfate is separated by

filtration. How many grams of barium sulfate (MW = 233.40) will be produced if

exactly 1.00 L of 1.55 M barium nitrate is reacted with excess sodium sulfate?

Ba(NO3)2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaNO3(aq)

V = 1.00 L

1.55 M

? g

MW = 233.40

1.55 mol

L x 1.00 L

Plan: M x V = mol Ba(NO3)2 mol BaSO4 g BaSO4

ratio

x 1 mol BaSO4

1 mol Ba(NO3)2

233.40 g BaSO4

1 mol BaSO4

x g BaSO4 = = 362 g BaSO4

Excess

no LR

29

Sample Exercise 8.8 – Calculating Solute Concentration

from Mass of Precipitate

To determine the concentration of chloride ion in a 100.0 mL sample of

groundwater, a chemist adds a large enough volume of a solution of AgNO3

to precipitate all the Cl- ions as silver chloride. The mass of the resulting

precipitate is 71.7 mg. What is the Cl- concentration in the sample in mg/L?

Plan: mg AgCl g AgCl mol AgCl mol Cl- g Cl- ratio

AgNO3(aq) + Cl-(aq) AgCl(s) + NO3-(aq)

excess

no LR V = 100.0 mL

= 0.100 L

? mg/L

71.7 mg

= 0.0717 g

Then mg/L Cl- = g Cl- mg Cl- mg/L

Divide by liters of sample

Chapter Outline


8.2 Dilutions





8.7 Titrations

8.8 Ion Exchange

30

Section 8.6 - Oxidation-Reduction Reactions (electron transfer reactions)

Oxidation-Reduction Reactions (Redox):

Characterized by gain or loss of electrons by atoms involved in the reaction.

Oxidation:

Historical definition = gain of oxygen (sometimes with a simultaneous loss of hydrogen)

e.g. 2 C2H6 + 7 O2 4 CO2 + 6 H2O

Modern definition = loss of electrons

Reduction:

Historical definition = loss of oxygen (sometimes with a simultaneous gain in hydrogen)

e.g. Fe2O3 + 3 CO 2 Fe + 3 CO2

Modern definition = gain of electrons

“oil rig”

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Table 8.4 - Oxidation Number Rules

The charge the atom would have in a molecule (or an ionic

compound) if electrons were completely transferred, i.e.

applies to both ionic and covalent compounds.

1. The O.N. for pure elements = 0 (including diatomics).

2. O.N = the charge on monovalent ions

3. O.N. of fluorine = -1 for all of its compounds

H H Mg

Mg

Cl

O O

Mg2+ + 2 e + e

Cl

4. O.N. of oxygen = -2 in nearly all of its compounds

5. O.N. of hydrogen = +1 in nearly all of its compounds

6. O.N.‘s of the atoms in a neutral molecule sum up to

zero

7. O.N.’s of the atoms in a polyatomic ion sum up to the

charge on the ion

O O + 4 e 2

O

2- Exceptions are peroxides

(O.N. = -1) and superoxides

(O.N. = -1/2), e.g. H2O2, KO2

H H 2 H+ + 2 e Exception = hydrides, e.g. CaH2

CaCl2

PO4

3-

+2 + 2(Cl) = 0 so Cl = -1

P + 4(-2) = -3 so P = +5

32

What are the oxidation states for sulfur in (a) S8, (b) SO2, (c)

Na2S, and (d) CaSO4 ?

Sample Exercise 8.9

Electron Transfer in Redox Reactions

oxidation lose n electrons O.N. = +n

reduction gain n electrons O.N. = -n

What follows is a method that keeps track of the flow of electrons

within the atoms and molecules taking part in a chemical reaction.

Pb0 Pb2+ + 2 e

O.N. = final O.N. - initial O.N.

Cu2+ + 2 e Cu0

O.N. = (2 – 0) = +2

O.N. = (0 – 2) = -2

33

What is the net ionic equation for the reaction of zinc metal plus cupric sulfate?

34

Oxidizing Agent =

Reducing Agent =

oxidizes something else while being reduced (Cu2+ Cu0)

reduces something else while being oxidized (Zn Zn2+)

Sample Exercise 8.10: Identifying Oxidizing and Reducing Agents and Determining

Number of Electrons Transferred

Energy released by the reaction of hydrazine and

dinitrogen tetroxide is used to orient and maneuver

spacecraft and to propel rockets into space. Identify

the elements that are oxidized and reduced, the

oxidizing agent, and the reducing agent, and

determine the number of electrons transferred in

the balanced chemical equation.

2N + 4(-2) = 0

So N = +4

2N + 4(+1) = 0

So N = -2

(-2) (+1) (+4) (-2)

Element in its

natural state

(0)

O.N. rules

H = +1

O = -2

(+1) (-2)

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(-2) (0) (+4)

O.N. = 0 – (-2) = +2

lost electrons oxidation

O.N. = (0 – 4) = -4

gained electrons reduction

Net number of electrons transferred during the reaction:

= (#molecules) x (#atoms in molecule) x O.N.

2 x 2 x (+2) = +8 (oxidation)

O.N. from

previous slide

1 x 2 x (-4) = -8 (reduction)

O.N. from

previous slide

So the net number of electrons flowing is 8

36

Balancing Redox Reactions in Acid and Base (We’re going to use a simpler method than the book

called the Method of Half Reactions)

Examples of Half Reactions:

copper wire immersed in a silver nitrate solution:

Cu(s) + 2 AgNO3(aq) Cu(NO3)2(aq) + 2 Ag(s)

(+1) (+2) (+5) (+5)

(0) (0)

(-2) (-2)

O.N. Cu = +2 – (0) = +2 oxidation

O.N. Ag = 0 – (1) = -1 reduction

The half-reactions are:

Cu(s)

AgNO3(aq)

(clear soln)

Ag(s)

Cu(NO3)2(aq)

(blue soln)

ox: Cu(s) Cu2+(aq) + 2 e

red: Ag+(aq) + e Ag(s)

Examples of Half Reactions (cont’d):

copper wire immersed in a silver nitrate solution:


red: Ag+(aq) + e Ag(s)

The total reaction is obtained by adding the half reactions together, after

balancing electrons -

x 2


red: 2 Ag+(aq) + 2 e 2 Ag(s)

Net: Cu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s)

The next topic is how to balance a redox reaction that occurs in acid or base.

37

1. Determine the oxidation numbers for the reactants and compare to the products.

2. Write the oxidation and reduction half-reactions without electrons (yet)

3. Balance everything but oxygen and hydrogen

4. Balance oxygen by adding water

5. Balance hydrogen by adding (a) H+ in acidic solutions, (b) in basic solutions,

continue as if in acidic solution, but at the end each H+ ion will be neutralized by

adding OH- ions

6. Balance charge by adding electrons; for the oxidation half-reaction, the electrons

will be on the right, for the reduction half-reaction, the electrons will appear on the

left

7. Make sure the number of electrons in each half-reaction are the same. Then add

the half reactions together

8. Make sure that the equation is balanced for mass and for charge

NOTE: sometimes you have to cancel H2O’s that are on each side, as well

as H+ or OH-

Procedure for balancing a redox reaction in acid or base.

Sample Exercise 8.11

Balancing Redox Equations I: Acidic Solutions

Microorganisms called denitrifying bacteria that grow in waterlogged soil convert

NO3- ions into N2O gas as they feed on dead plant tissue (empirical formula =

CH2O), converting it into CO2 and H2O. Write a balanced net ionic equation

describing this conversion of dissolved nitrates to N2O gas. Assume the reaction

occurs in slightly acidic water.

NO3-(aq) + CH2O(s) N2O(g) + CO2(g) + H2O(l) Given:


(+5) (0) (+1) (-2) (-2) (+1) (+4) (+1) (-2) (-2) (-2)

O.N. C = +4 – (0) = +4 oxidation

O.N. N = 1 – (5) = -4 reduction

38


ox: CH2O(s) CO2(g)

red: NO3-(aq) N2O(g)



ox: H2O + CH2O(s) CO2(g)

red: 2 NO3-(aq) N2O(g) + 5 H2O

ox: CH2O(s) CO2(g)

red: 2 NO3-(aq) N2O(g)


continue as if in acidic solution, but at the end each H+ ion will be neutralized by

adding OH- ions

ox: H2O + CH2O(s) CO2(g) + 4 H+(aq)

red: 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O



left

ox: H2O + CH2O(s) CO2(g) + 4 H+(aq) + 4 e

red: 8 e + 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O

39

x 2



ox: H2O + CH2O(s) CO2(g) + 4 H+(aq) + 4 e

red: 8 e + 10 H+(aq) + 2 NO3-(aq) N2O(g) + 5 H2O

Net: 2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 5H2O


as H+ or OH-

2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 3H2O

2H2O + 2CH2O(s) + 10H+(aq) + 2NO3-(aq) 2CO2(g) + 8H+(aq) + N2O(g) + 5H2O

2CH2O(s) + 2H+(aq) + 2NO3-(aq) 2CO2(g) + N2O(g) + 3H2O


2CH2O(s) + 2H+(aq) + 2NO3-(aq) 2CO2(g) + N2O(g) + 3H2O

Mass: C = 2 C = 2

H = 6 H = 6

O = 8 O = 8

N = 2 N = 2

Charge: 2(+1) + 2(-1) = 0 = 0

40

Homework Problem 98(c)

Balancing Redox Equations II: Basic Solutions

Permanganate ion is used in water purification to remove oxidizable substances.

Complete and balance the reaction for the removal of sulfite ion.

MnO4-(aq) + SO3

2-(aq) MnO2(s) + SO42-(aq)


(+7) (+4) (+6) (+4)


ox: SO32-(aq) SO4

2-(aq)

red: MnO4-(aq) MnO2(s)

O.N. S = +6 – (+4) = +2 oxidation

O.N. Mn = +4 – (+7) = -3 reduction

(-2) (-2) (-2) (-2)




continue as if in acidic solution, but at the end each H+ ion will be neutralized

by adding OH- ions

ox: SO32-(aq) SO4

2-(aq)

red: MnO4-(aq) MnO2(s)

S and Mn are

already balanced

ox: H2O + SO32-(aq) SO4

2-(aq)

red: MnO4-(aq) MnO2(s) + 2H2O


2-(aq) + 2H+(aq)

red: 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O

41



left




2-(aq) + 2H+(aq)

red: 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O

+ 2 e

3 e +


2-(aq) + 2H+(aq) + 2 e

red: 4H+(aq) + MnO4-(aq) MnO2(s) + 2H2O 3 e +

x 3

x 2

Net: 3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4

-(aq)

3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O


as H+ or OH-

3H2O + 3SO32-(aq) + 8H+(aq) + 2MnO4

-(aq) 3SO42-(aq) + 6H+(aq) + 2MnO2(s) + 4H2O

3SO32-(aq) + 8H+(aq) + 2MnO4

-(aq) 3SO42-(aq) + 6H+(aq) + 2MnO2(s) + H2O

3SO32-(aq) + 2H+(aq) + 2MnO4

-(aq) 3SO42-(aq) + 2MnO2(s) + H2O

2H2O

3SO32-(aq) + H2O(l) + 2MnO4

-(aq) 3SO42-(aq) + 2MnO2(s) + 2OH-(aq)

+ 2OH-

+ 2OH-

42


Mass: S = 3 S = 3

O = 18 O = 18

H = 2 H = 2

Mn = 2 Mn = 2

Charge: 3(-2) + 2(-1) = -8 3(-2) +2(-1) = -8

3SO32-(aq) + H2O(l) + 2MnO4

-(aq) 3SO42-(aq) + 2MnO2(s) + 2OH-(aq)

Chapter Outline


8.2 Dilutions





8.7 Titrations

8.8 Ion Exchange

43

Titrations In a titration a solution of accurately known concentration is

added gradually added to another solution of unknown

concentration until the chemical reaction between the two

solutions is complete.

Equivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the

equivalence point

Slowly add base

to unknown acid

UNTIL

the indicator

changes color

Titration Example, p. 344-345 H2SO4 + 2NaOH → Na2SO4 + 2H2O

(titrant)

H2SO4

(unknown)

End Point

44

Titration Calculations, p. 344

Suppose it takes 22.40 mL of 0.00100 M sodium hydroxide to react completely with

the sulfuric acid in a 100.0 mL sample of acidic mine drainage. What was the

molarity of the sulfuric acid?

2NaOH(aq) + H2SO4(aq) 2H2O + Na2SO4

0.00100 M

22.40 mL

= 0.02240 L

100.0 mL

= 0.1000 L

M = ?

Plan: M x V = mol NaOH mol H2SO4 M

1 mol H2SO4

2 mol NaOH V

mol H2SO4 = 0.00100 mol NaOH x 0.02240 L NaOH x 1 mol H2SO4

L 2 mol NaOH

= 1.120 x 10-5 mol

M = 1.120 x 10-5 mol

0.1000 L = 1.120 x 10-4 M

Sample Exercise 8.13: Apple cider vinegar is an aqueous solution of acetic acid (MW = 60.05) made from fermented

apple juice. Commercial vinegar must contain at least 4 grams of acetic acid per 100 mL of

vinegar. Suppose the titration of a 25.00 mL sample of vinegar requires 12.15 mL of 1.885 M

sodium hydroxide. What is the molarity of acetic acid in the vinegar sample? Does it meet the

4 g/100 mL standard?

HC2H3O2(aq) + NaOH(aq) H2O + NaC2H3O2

25.00 mL

M = ?

mg/L = ?

12.15 mL

= 0.01215 L

M = 1.885

Plan: M x V = mol NaOH mol HC2H3O2 M g/mL

1 mol HC2H3O2

1 mol NaOH V

= 0.05501 g/mL

= 0.9161 M

MW

M HC2H3O2 = 1.885 mol NaOH

L 0.02500 L

1 x x 0.01215 L NaOH

1 mol NaOH

x 1 mol HC2H3O2

g = 0.9161 mol

L L

60.05 g 1mol

x 1 L 1000 mL

x = 5.501 g/100 mL

Ch 8: Aqueous Solutions: Chemistry of the Hydrosphere · 4 Concentration of Solutions - Molarity...

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Transcript of Ch 8: Aqueous Solutions: Chemistry of the Hydrosphere · 4 Concentration of Solutions - Molarity...