Chapter 13 Applications of Aqueous Equilibria

04/22/23 Zumdahl Chapter 8 1

Chapter 13Applications of Aqueous Equilibria

13.1 Solutions of Acids or Bases Containing a Common Ion

13.2 Buffered Solutions 13.3 Exact Treatment of Buffered Solutions (skip)13.4 Buffer Capacity 13.5 Titrations and pH Curves 13.6 Acid-Base Indicators 13.7 Titration of Polyprotic Acids (skip)13.8 Solubility Equilibria and the Solubility Product 13.9 Precipitation and Qualitative Analysis (skip)13.10 Complex Ion Equilibria (skip)

The Common Ion Effect (1)

HF (aq) H+ (aq) + F- (aq) Ka = 7.2 x 10-4

When F- is added (from NaF), then [H+] must decrease (Le Chatelier’s principle). The pH increases. See example 13.1 in the text to see how to quantitatively determine this effect, with an ICE box)

This applies to weak acids, weak bases and solubility of salts when a common ion is added to the equilibrium reaction. You can change the pH, but adding salt.

NaF (s) → Na+ (aq) + F– (aq)

€

Ka =[H3O

+][F−]

[HF]

The Common Ion Effect (2)

If a solution and a salt to be dissolved in it have an ion in common, then the solubility of the salt is depressed relative to pure.

General equation

AB (s) A+ (aq) + B- (aq)

If you add BH(aq), which dissociates into B- and H+, then the [A+]

decreases, and AB is driven out of solution.

Buffers: resist change in pH when acid or base is added.

Buffer Solutions: contain a common ion and are important in biochemical and physiological processes

Organisms (and humans) have built-in buffers to protect them against changes in pH.

Applications of Aqueous Equilibrium

Buffered Solutions

Human blood is a buffered solution

The Common Ion Effect on Buffering

Blood: (pH 7.4)Death = 7.0 <pH > 7.8 = Death

Human blood is maintained by a combination of CO3-2, PO4

-3 and protein buffers.

A solution that is buffered by acetic acid/acetate

Unbuffered solution

How Do Buffers Work?

• Ka = [H+][A-]/[HA] => [H+] = Ka[HA]/[A-]

•If Ka is small (weak acid) then [H+] does not change much when [HA] and [A-] change.

If [HA] and [A-] are large, and [HA]/[A-] ≈ 1,then small additions of acid ([H+]) or base ([OH-]) don’t change the ratio much.

HA H+ + A –

HA = generic acid

Example: A solution of 0.5 M acetic acid plus 0.5 M acetate Ka = 1.8x10-5 pKa = 4.7

HA H+ + A-

Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA]

Use an ICE box to calculate the pH[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0

=> => pH = 4.74 i.e., pH=pKa

Example: A solution of 0.5 M acetic acid plus 0.5 M acetate Ka = 1.8x10-5 pKa = 4.74

HA H+ + A-

Ka = [H+][A-]/[HA] pH = pKa + log[A-]/[HA]

Use an ICE box to calculate the pH[AH]ini = 0.5, [A-]ini = 0.5, [H+]ini = 0

=> => pH = 4.74 i.e., pH=pKa

Now add NaOH to 0.01 M(in an unbuffered solution this would give a pOH of 2 and pH of 12)

Use HA + OH- → H2O + A- Redo the ICE Box[AH]ini = 0.49, [A-]ini = 0.51

=> => pH = 4.76

Buffer Calculation: Add Acid (#1) to a Buffered Solution Acetic Acid: Ka = 1.8x10-5 pKa = 4.74HA H+ + A-

Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] => pH = pKa + log[A-]/[HA]

Case 1) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 1.0M pH = pKa => [CH3COOH] = [CH3COO-] = 0.5

Now add 0.01 M HCl (strong acid)

[CH3COOH] = 0.51 [CH3COO-] = 0.49

pH = pKa + log[A-]/[HA] = 4.74 + log(0.49/0.51) = 4.74 – 0.02 = 4.72

ΔpH = -0.02

Buffer Calculation: Add Acid (#2) to a dilute Buffered Solution

Acetic Acid: Ka = 1.8x10-5 pKa = 4.74HA H+ + A-

Ka = [H+][A-]/[HA] => -pKa = -pH + log[A-]/[HA] => pH = pKa + log[A-]/[HA]

Case 2) [CH3COOH]tot = [CH3COOH] + [CH3COO-] = 0.10 M pH = pKa => [CH3COOH] = [CH3COO-] = 0.05

Now add 0.01 M HCl (strong acid)

[CH3COOH] = 0.06 [CH3COO-] = 0.04

pH = pKa + log[A-]/[HA] = 4.74 + log(0.04/0.06) = 4.74 – 1.5 = 3.54

ΔpH = -1.5

Adding Base to a Buffered Solution:OH- ions do not accumulate but are replaced

by A- ions.

OH– + HA A⇌ – + H2O

When the OH- is added, the concentrations of HA and A- change, but only by small amounts. Under these conditions the

[HA]/[A-] ratio and thus the [H+] stay virtually constant.

When protons are added to a buffered solution, the conjugate base (A–) reacts

H+ + A– ⇌ HA

OH– + HA A⇌ – + H2O

Characteristics of Buffered solutions

• Contain relatively large concentrations of a weak acid and its conjugate base.

• When acid is added, it reacts with the conjugate base

• When base is added, it reacts with the acid.

• pH is determined by the ratio of the base and acid.

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

−

[HA]

][Alog pKpH a

Buffer Capacity

• Buffer capacity is the amount of protons or hydroxide ions that can be absorbed without a significant change in pH.

• pH is determined by the ratio of [A–]/[HA] and pKa

• Capacity is determined by the magnitudes of [HA] and [A–].

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pH = pKa + log[A−]

[HA]

⎛

⎝ ⎜ ⎞

⎠ ⎟

Equivalence Point


Buffer Summary

Buffer Design: Add known amount of HA (weak acid) and salt of HA (its conjugate base, A─)

[H+] or pH depends on Ka and the ratio of acid to salt or [A─].

Thus if both conc. HA and A- are large then small additions of acid or base don’t change the ratio much

How to Actually Make a Buffer (a buffered solution in a lab)

HA H+ + A –

Zumdahl Chapter 8

Acid-Base Titrations

A controlled addition of measured volumes of a solution of known concentration (the Titrant) from a buret to a second solution of unknown concentration under conditions in which the solutes react cleanly (without side reactions), completely, and rapidly.

Titration:

A titration is complete when the second solute is fully consumedCompletion is signaled by a change in some physical property, such as the color of the reacting mixture or the color of an indicator that has been added to it.

[NaOH]

“X”

Indicator phenolphthalein


Titrations and pH Curves

Zumdahl Chapter 8 21

Strong acid

Titration of a weak acid with a strong base.

The base reacts with the acid.

.

OH– + HA A⇌ – + H2O


Equivalence point determined defined by the Stoichiometry, not by the pKa.

nbase = nacid

number of moles of acid = number of moles of base

For a variety of weak acids:

The Equivalence point occurs at the same stoichiometric amount of base added

The weaker the acid, the greater the pH value for the equivalent point.


Similar for titration of weak bases with strong acids

The indicator phenolphthalein is pink in basic solution and colorless in acidic solution.

Acid-Base Indicators


Indicators (Weak Acid Equilibria)

pH = -log10[H3O+]


IndicatorsA soluble compound, generally an organic dye, that changes its color noticeably over a fairly short range of pH.

Typically, Indicators are a weak organic acid that has a different color than its conjugate base.

Acid: HIn (aq)

HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)

Phenolphtalein

Indicator denoted by In

Conjugate base: In– (aq)

28

Acid: HIn (aq) Conjugate base: In- (aq)

HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)

pH = -log10[H3O+]

[H3O+][OH-] = Kw


Methyl Red

Bromothymol blue

Phenolphtalein

HIn(aq) + H2O(l) ↔ H3O+(aq) + In-(aq)


The useful pH ranges for several common indicators

HIn(aq) + H2O(l) ↔ H3O+(aq) + In–(aq)


Could use either indicator Methyl red changes color to early

Weak acid

Indicator Selection

• Want indicator color change and titration equivalence point to be as close as possible

• Easier with a large pH change at the equivalence pointStrong acid


Solubility Product Ksp

Describes a chemical equilibrium in which an excess solid salt is in equilibrium with a saturated aqueous solution of its separated ions.

General equation

AB (s) ↔ A+ (aq) + B- (aq)

Ksp =

Ksp =The solubility expression controls the amount of solid that will dissolve


Ksp Values at 25°C for Common Ionic Solids


The Solubility of Ionic Solids

The Solubility Product

AgCl(s) ↔Ag+(aq) + Cl-(aq)

= 1.6 10-10 at 25oC

Ksp =

Ksp

The solid AgCl, which is in excess, is understood to have a concentration of 1 mole per liter.

excess


The Solubility of Ionic SolidsThe Solubility Product

Ag2SO4(s) ↔2Ag+(aq) + SO42-(aq)

Ksp =

Fe(OH)3(s) ↔Fe+3(aq) + 3OH-1(aq)

Ksp =

excess

excess


Solubility and Ksp

Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6 10-13.

Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq)

Let “y” = molar solubility in mol/L


Determine the mass of lead(II) iodate dissolved in 2.50 L of a saturated aqueous solution of Pb(IO3)2 at 25oC. The Ksp of Pb(IO3)2 is 2.6 10-13.

Pb(IO3)2(s) ↔ Pb2+(aq) + 2 IO3-(aq)

[Pb2+][IO3-]2 = Ksp

[Pb2+][IO3-]2 =

y = 4.0 10-5 [Pb(IO3)2] = [Pb2+] = y = 4.0 10-5 mol L-1

[IO3-] = 2y = 8.0 10-5 mol L-1

= (4.0 10-5 mol L-1) (557 g mol-1)

= 0.0223 g L-1 2.50 LMolar Mass of lead (II) iodate

Pb(IO3)2 = 557g per mole

Gram solubility of

Lead (II) iodate

“y” = molar solubility


The Solubility of SaltsSolubility and KspExercise 9-3

Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1.

1 Ag2SO4 (s) ↔ 2 Ag+(aq) + 1 SO42-(aq)


Compute the Ksp of silver sulfate (Ag2SO4) at 25oC if its mass solubility is 8.3 g L-1.

[y] = (8.3 g Ag2SO4 L-1)

[Ag+]2[SO42-] = Ksp

Ksp =

(1 mol Ag2SO4/311.8 g)

[y] = 2.66 10-2 mol Ag2SO4 L-1

1 Ag2SO4 (s) ↔ 2 Ag+(aq) + 1 SO42-(aq)


The Nature of Solubility Equilibria

Dissolution and precipitation are reverse of each other.

General reaction

X3Y2 (s) ↔ 3X+2 (aq) + 2Y-3 (aq)

Ksp =

Dissolution (Solubility)

[s]

s = molar solubility

expressed in moles per liter


A salt’s Ksp value gives us information about its solubility.

Salt ↔ Cation + Anion

If the salts being compared produce the same number of ions, eg., AgI, CuI, CaSO4

s = [cation]s = [anion]

Ksp = [cation] [anion] = s2

Salt molar solubility = s = (Ksp)1/2

Salt Ksp Solubility

AgI 1.5 x 10-16 1.2 x 10-8

CuI 5.0 x 10-12 2.2 x 10-6

CaSO4 6.1 x 10-5 7.8 x 10-3

Solubility CaSO4 > CuI > AgI

Relative Solubilities


The Effects of pH on SolubilitySolubility of Hydroxides

Zn(OH)2(s) ↔Zn2+(aq) + 2 OH-(aq)

[Zn2+][OH-]2 = Ksp = 4.5 10-17

Make more acidic:

[Zn2+][OH-]2 = Ksp[Zn2+][OH-]2 = Ksp[Zn2+][OH-]2 = Ksp

Zinc hydroxide is more soluble in acidic solution than in pure water.

Many solids dissolve more readily in more acidic solutions

If pH decreases (or made more acidic), the [OH-] decreases. In order to maintain Ksp the [Zn2+] must increase and consequently more solid Zn(OH)2 dissolves.


The Effects of pH on SolubilityEstimate the molar solubility of Fe(OH)3 in a solution that is buffered to a pH of 2.9. Lookup Ksp = 1.1x10-36

In pure water:

[OH-] = 3y = 1.3 10-9 mol L-1

pOH = 8.87 (and pH = 5.13)

(pH = 2.9 and) pOH = 11.1 [OH-] = 7.9 10-12 mol L-1

[Fe3+] = Ksp/[OH-]3 = 1.1 10-36 / (7.9 10-12)3 [Fe3+] = [Fe(OH)3] = 2.2 10-3 mol L-1 answer

[Fe3+] = y [OH-] = 3y

y(3y)3 = 27y4 = Ksp = 1.1 10-36

y = 4.5 10-10 mol L-1 = [Fe3+] = [Fe(OH)3]=

[Fe3+][OH-]3 = Ksp

In pure water, Fe(OH)3 is 5 x 10 6 less soluble than at pH = 2.9

Fe(OH)3(s) ↔Fe3+(aq) + 3 OH-(aq)


The Common Ion Effect

The Ksp of thallium(I) iodate (TlO3) is 3.1 10-6 at 25oC. Determine the molar solubility of TlIO3 in 0.050 mol L-1 KIO3 at 25oC.

TlIO3(s) ↔ Tl+(aq) + IO3-(aq)

[Tl+] (mol L-1) [IO3-] (mol L-1)

Initial concentration

Equilibrium concentration Change in concentration

[Tl+][IO3-] = Ksp

s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility


The Common Ion Effect

The Ksp of thallium(I) iodate (TlO3) is 3.1 10-6 at 25oC. Determine the molar solubility of TlIO3 in 0.050 mol L-1 KIO3 at 25oC.

TlIO3(s) ↔ Tl+(aq) + IO3-(aq)

With common ion (from previous calculation)

s = [TlIO3] = 6.2 × 10-5 mol L-1 = molar solubility

With common ion, s = [TlIO3]= 6.2 × 10-5 mol L-1

[s] [s] [s]

What if no common ion is added? i.e., dissolve thallium iodate in pure water

s= 1.76x10-3 mol L-1 = molar solubility

[Tl+][IO3-] = Ksp


Chapter 8Applications of Aqueous Equilibria

8.1 Solutions of Acids or Bases Containing a Common Ion

8.2 Buffered Solutions 8.3 Exact Treatment of Buffered Solutions (skip)8.4 Buffer Capacity 8.5 Titrations and pH Curves 8.6 Acid-Base Indicators 8.7 Titration of Polyprotic Acids (skip)8.8 Solubility Equilibria and the Solubility Product 8.9 Precipitation and Qualitative Analysis (skip)8.10 Complex Ion Equilibria (skip)

Chapter 13 Applications of Aqueous Equilibria

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