Buffers and

Acid/Base Titration

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Common Ion

• Suppose we have a solution containing hydrofluoric acid (HF) and its salt sodium fluoride (NaF).

• Major species (HF, Na+, F-, and H2O)• HF slightly dissociated, therefore common ion

is F-

• How does this effect equilibrium of HF dissociation?

Common Ion Effect

HF(aq) H+(aq) + F-

(aq)

Added fluoride ions from NaFShifts reaction left.Less H+ ions present.

• This effect makes a solution of NaF and HF less acidic than a solution of HF alone.

Buffered Solutions

• A solution that resists a change in pH when either hydroxide ions or protons are added.

• Buffered solutions contain either:• A weak acid and its salt• A weak base and its salt

Acid/Salt Buffering Pairs

Weak AcidFormula

of the acidExample of a salt of the

weak acid Hydrofluoric  HF   KF – Potassium fluoride 

 Formic   HCOOH   KHCOO – Potassium formate 

 Benzoic   C6H5COOH   NaC6H5COO – Sodium benzoate

 Acetic   CH3COOH   NaH3COO – Sodium acetate 

 Carbonic   H2CO3   NaHCO3 - Sodium bicarbonate

 Propanoic   HC3H5O2    NaC3H5O2  - Sodium propanoate

 Hydrocyanic   HCN   KCN - potassium cyanide 

The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

Base/Salt Buffering Pairs

The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)

BaseFormula of the base

Example of a salt of the weak acid

Ammonia   NH3  NH4Cl - ammonium chloride

 Methylamine

 CH3NH2 CH3NH3Cl – methylammonium

chloride

 Ethylamine  C2H5NH2 C2H5NH3NO3 -  ethylammonium

nitrate

 Aniline  C6H5NH2  C6H5NH3Cl – aniline hydrochloride

 Pyridine  C5H5N    C5H5NHCl – pyridine hydrochloride

Henderson-Hasselbalch Equation

][

][log

][

][log

acid

basepK

HA

ApKpH aa

][

][log

][

][log

base

acidpK

B

BHpKpOH bb

This is an exceptionally powerful tool, and it’s use will be emphasized in our problem solving.

Calculate the pH of the following mixtures:• 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium

lactate (Ka = 1.4 x 10-4)

• 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)

Buffer capacity

• The pH of a buffered solution is determined by the ratio [A-]/[HA].

• As long as this doesn’t change much the pH won’t change much.

• The more concentrated these two are the more H+ and OH- the solution will be able to absorb.

• Larger concentrations bigger buffer capacity.

Buffer Capacity

• Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following:

• 5.00 M HC2H3O2 and 5.00 M NaC2H3O2

• 0.050 M HC2H3O2 and 0.050 M NaC2H3O2

• Ka= 1.8x10-5

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0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)

pH

Weak Acid/Strong Base Titration

A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

Equivalence is above pH 7

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m illiliters NaOH (0.10 M)

pH

Strong Acid/Strong Base Titration

A solution that is 0.10 M HCl is titrated with 0.10 M NaOH

Equivalance is at pH 7

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m illiliters HCl (0.10 M)

pH

Strong Acid/Strong Base Titration

A solution that is 0.10 M NaOH is titrated with 0.10 M HCl

Equivalence is at pH 7 It is important to

recognize that titration curves are not always increasing from left to right.

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m illiliters NH3 (0.10 M)

pH

Strong Acid/Weak Base Titration

A solution that is 0.10 M HCl is titrated with 0.10 M NH3

Equivalence is below pH 7

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0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00 45.00m illiliters NaOH (0.10 M)

pH

Titration of an Unbuffered Solution

A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

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m illiliters NaOH (0.10 M)

pH

Titration of a Buffered Solution

A solution that is 0.10 M CH3COOH and 0.10 M NaCH3COO is titrated with 0.10 M NaOH

Comparing ResultsGraph

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mL 0.10 M NaOH

pH

Buffered

Unbuffered

Selection of Indicators

The equivalence point of a titration, defined by the stoichiometry, is not necessarily the same as the end point, careful selection of the indicator will ensure that the error is negligible.

Indicator TransitionsIndicator Low pH color Transition pH range High pH colorGentian violet (Methyl violet 10B) yellow 0.0–2.0 blue-violet

Leucomalachite green (first transition) yellow 0.0–2.0 greenLeucomalachite green (second transition) green 11.6–14 colorless

Thymol blue (first transition) red 1.2–2.8 yellow

Thymol blue (second transition) yellow 8.0–9.6 blue

Methyl yellow red 2.9–4.0 yellowBromophenol blue yellow 3.0–4.6 purpleCongo red blue-violet 3.0–5.0 redMethyl orange red 3.1–4.4 orangeBromocresol green yellow 3.8–5.4 blueMethyl red red 4.4–6.2 yellowMethyl red red 4.5–5.2 greenAzolitmin red 4.5–8.3 blueBromocresol purple yellow 5.2–6.8 purpleBromothymol blue yellow 6.0–7.6 bluePhenol red yellow 6.8–8.4 redNeutral red red 6.8–8.0 yellowNaphtholphthalein colorless to reddish 7.3–8.7 greenish to blueCresol Red yellow 7.2–8.8 reddish-purplePhenolphthalein colorless 8.3–10.0 fuchsiaThymolphthalein colorless 9.3–10.5 blueAlizarine Yellow R yellow 10.2–12.0 redLitmus red 4.5-8.3 blue

Source: Wikipedia

Indicators• Weak acids that change color when they

become bases.• weak acid written HIn• Weak base• HIn H+ + In- clear

red• Equilibrium is controlled by pH• End point - when the indicator changes color.

Solubility

Equilibria

Lead (II) iodide precipitates when potassium iodide is mixed with lead (II) nitrate.

Graphic: Wikimedia Commons user PRHaney

Ksp Values for Some Salts at 25C

Name Formula

Ksp

 Barium carbonate   BaCO3   2.6 x 10-9 

 Barium chromate   BaCrO4   1.2 x 10-10 

 Barium sulfate   BaSO4   1.1 x 10-10 

 Calcium carbonate   CaCO3   5.0 x 10-9 

 Calcium oxalate   CaC2O4   2.3 x 10-9 

 Calcium sulfate   CaSO4   7.1 x 10-5 

 Copper(I) iodide   CuI   1.3 x 10-12 

 Copper(II) iodate   Cu(IO3)2   6.9 x 10-8 

 Copper(II) sulfide   CuS   6.0 x 10-37 

 Iron(II) hydroxide   Fe(OH)2   4.9 x 10-17 

 Iron(II) sulfide   FeS   6.0 x 10-19 

 Iron(III) hydroxide   Fe(OH)3   2.6 x 10-39 

 Lead(II) bromide   PbBr2   6.6 x 10-6 

 Lead(II) chloride   PbCl2   1.2 x 10-5 

 Lead(II) iodate   Pb(IO3)2   3.7 x 10-13 

 Lead(II) iodide   PbI2   8.5 x 10-9 

 Lead(II) sulfate   PbSO4   1.8 x 10-8 

Name Formula

Ksp

 Lead(II) bromide   PbBr2   6.6 x 10-6 

 Lead(II) chloride   PbCl2   1.2 x 10-5 

 Lead(II) iodate   Pb(IO3)2   3.7 x 10-13 

 Lead(II) iodide   PbI2   8.5 x 10-9 

 Lead(II) sulfate   PbSO4   1.8 x 10-8 

 Magnesium carbonate   MgCO3   6.8 x 10-6 

 Magnesium hydroxide   Mg(OH)2   5.6 x 10-12 

 Silver bromate   AgBrO3   5.3 x 10-5 

 Silver bromide   AgBr   5.4 x 10-13 

 Silver carbonate   Ag2CO3   8.5 x 10-12 

 Silver chloride   AgCl   1.8 x 10-10 

 Silver chromate   Ag2CrO4   1.1 x 10-12 

 Silver iodate   AgIO3   3.2 x 10-8 

 Silver iodide   AgI   8.5 x 10-17 

 Strontium carbonate   SrCO3   5.6 x 10-10 

 Strontium fluoride   SrF2   4.3 x 10-9 

 Strontium sulfate   SrSO4   3.4 x 10-7 

 Zinc sulfide   ZnS   2.0 x 10-25 

Solving Solubility ProblemsFor the salt AgI at 25C, Ksp = 1.5 x 10-16

AgI(s) Ag+(aq) + I-(aq)

I

C

E

OO

+x +x

x x

1.5 x 10-16 = x2

x = solubility of AgI in mol/L = 1.2 x 10-8 M

Solving Solubility ProblemsFor the salt PbCl2 at 25C, Ksp = 1.6 x 10-5

PbCl2(s) Pb2+(aq) + 2Cl-(aq)

I

C

E

OO

+x +2x

x 2x

1.6 x 10-5 = (x)(2x)2 = 4x3

x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

Solving Solubility with a Common IonFor the salt AgI at 25C, Ksp = 1.5 x 10-16

What is its solubility in 0.05 M NaI?

AgI(s) Ag+(aq) + I-(aq)

I

C

E

0.05O

+x 0.05+x

x 0.05+x

1.5 x 10-16 = (x)(0.05+x) (x)(0.05)

x = solubility of AgI in mol/L = 3.0 x 10-15 M

Precipitation

• Ion Product, Q =[M+]a[Nm-]b • If Q>Ksp a precipitate forms.• If Q<Ksp No precipitate.• If Q = Ksp equilibrium.

Precipitation

A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3

is added to 300.0 mL of 2.00 x 10-2M KIO3. Will

Ce(IO3)3 (Ksp= 1.9 x 10-10) precipitate and if so, what is the concentration of the ions?

Precipitation and Qualitative Analysis

Complex IonsA Complex ion is a charged species composed of:

1. A metallic cation

2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

NH3, CN-, and H2O are Common Ligands

O

H

HNH

HH

C N-

Coordination Number Coordination number refers to the number of ligands attached to the cation 2, 4, and 6 are the most common coordination numbers

Coordination

number

Example(s)

2 Ag(NH3)2+

4 CoCl42- Cu(NH3)4

2+

6 Co(H2O)62+ Ni(NH3)6

2+

Complex Ions and Solubility

AgCl(s) Ag+ + Cl- Ksp = 1.6 x 10-10

Ag+ + NH3 Ag(NH3)+ K1 = 2.1 x 103

Ag(NH3)+ NH3 Ag(NH3)2+ K2 = 8.2 x 103

AgCl + 2NH3 Ag(NH3)2+ + Cl- K = KspK1K2

23

233

][

]][)([108.2

NH

ClNHAgxK

• Calculate the concentrations of Ag+, Ag(S2O3)-,

and Ag(S2O3)2-3 in a solution made by mixing

150.0 mL of 1.00x10-3M AgNO3 with 200.0 mL of 5.00 M Na2S2O3.

Ag+ + S2O3-2 Ag(S2O3)-

K1=7.4 x 108

Ag(S2O3)- + S2O3-2 Ag(S2O3)2

-3

K2=3.9 x 104