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Chapter 13 Applications 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

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chapter 13 applications of aqueous equilibria
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)

Zumdahl Chapter 8

slide2

The Common Ion Effect (1)

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.

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

NaF (s) → Na+ (aq) + F– (aq)

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)

slide3

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.

human blood is a buffered solution

Applications of Aqueous Equilibrium

Human blood is a buffered solution

Buffered Solutions

The Common Ion Effect on Buffering

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.

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.

slide5

Unbuffered solution

A solution that is buffered by acetic acid/acetate

slide6

How Do Buffers Work?

HA H+ + A –

HA = generic acid

  • 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.
slide7

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

slide8

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

slide9

Buffer Calculation: Add Acid (#1) to a Buffered Solution

Acetic Acid: Ka = 1.8x10-5 pKa = 4.74

HA 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

slide10

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

Acetic Acid: Ka = 1.8x10-5 pKa = 4.74

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

OH– + HA ⇌ A– + H2O

slide12

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.

OH– + HA ⇌ A– + H2O

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

H+ + A–⇌ HA

characteristics of buffered solutions
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.
buffer capacity
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–].
slide16

Buffer Summary

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

[H+] or pH depends on Kaand 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

Zumdahl Chapter 8

slide18

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

HA H+ + A –

  • Assume you want to make a 1L of of 1M buffer at pH 4.74. Use acetic acid, because the pKa of acetic acid is 4.74.
  • To make the Buffer:
  • The easy way (this is how we actually do it in the lab): Weigh out the appropriate amount of Na+Acetate- (70 g). Add it to a volumetric flask. Add water to around 0.9 Liters. Add a stir bar and stir. Slowly add HCl, while monitoring the pH with a pH meter. When the pH reaches the pKa (4.74), stop. Take out the stir bar and add enough water to give 1L.
  • The hard way. Start with two solutions. Solution A is 1M acetic acid. Solution B is 1M Na+Acetate-. Add solution A to solution B. Then check the pH with a pH meter. Adjust the pH with HCl or NaOH as required, to get the pH to 4.74 But this will change the concentration of A- and HA.
slide19

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 consumed

Completion 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]

Indicator phenolphthalein

“X”

Zumdahl Chapter 8

slide20

Titrations and pH Curves

Zumdahl Chapter 8

slide21

Titration of a weak acid with a strong base.

  • The base reacts with the acid.
    • .

OH– + HA ⇌ A– + H2O

Strong acid

Zumdahl Chapter 8

slide22

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

nbase = nacid

number of moles of acid = number of moles of base

Zumdahl Chapter 8

slide23

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.

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

Acid-Base Indicators

slide26

Indicators (Weak Acid Equilibria)

pH = -log10[H3O+]

Zumdahl Chapter 8

slide27

Indicators

A 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.

HIn(aq) + H2O(l)↔ H3O+(aq) + In–(aq)

Acid: HIn (aq)

Conjugate base: In– (aq)

Phenolphtalein

Indicator denoted by In

Zumdahl Chapter 8

slide28

Conjugate base: In- (aq)

Acid: HIn (aq)

HIn(aq) + H2O(l)↔ H3O+(aq) + In-(aq)

[H3O+][OH-] = Kw

pH = -log10[H3O+]

slide29

Methyl Red

Bromothymol blue

Phenolphtalein

Zumdahl Chapter 8

HIn(aq) + H2O(l)↔H3O+(aq) + In-(aq)

the useful ph ranges for several common indicators
The useful pH ranges for several common indicators

HIn(aq) + H2O(l)↔H3O+(aq) + In–(aq)

Zumdahl Chapter 8

slide31

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 point

Strong acid

Weak acid

Could use either indicator

Methyl red changes color to early

Zumdahl Chapter 8

slide32

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)

The solubility expression controls the amount of solid that will dissolve

Ksp =

Ksp =

Zumdahl Chapter 8

slide34

The Solubility of Ionic Solids

The Solubility Product

AgCl(s) ↔Ag+(aq) + Cl-(aq)

excess

Ksp =

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

Ksp

= 1.6  10-10at 25oC

Zumdahl Chapter 8

slide35

The Solubility of Ionic Solids

The Solubility Product

Ag2SO4(s) ↔2Ag+(aq) + SO42-(aq)

excess

Ksp =

Fe(OH)3(s) ↔Fe+3(aq) + 3OH-1(aq)

excess

Ksp =

Zumdahl Chapter 8

slide36

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

Zumdahl Chapter 8

slide37

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

“y” = molar solubility

[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

Gram solubility of

Lead (II) iodate

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

= 0.0223 g L-1

 2.50 L

Molar Mass of lead (II) iodate

Pb(IO3)2 = 557g per mole

Zumdahl Chapter 8

slide38

The Solubility of Salts

Solubility and Ksp

Exercise 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)

Zumdahl Chapter 8

slide39

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)

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

 (1 mol Ag2SO4/311.8 g)

[y] = 2.66  10-2 mol Ag2SO4 L-1

[Ag+]2[SO42-] = Ksp

Ksp =

Zumdahl Chapter 8

slide40

The Nature of Solubility Equilibria

Dissolution and precipitation are reverse of each other.

Dissolution (Solubility)

General reaction

X3Y2 (s) ↔ 3X+2 (aq) + 2Y-3 (aq)

[s]

Ksp =

s = molar solubility

expressed in moles per liter

Zumdahl Chapter 8

relative solubilities
Relative Solubilities

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

Zumdahl Chapter 8

slide42

The Effects of pH on Solubility

Solubility of Hydroxides

Many solids dissolve more readily in more acidic solutions

Zn(OH)2(s) ↔Zn2+(aq) + 2 OH-(aq)

[Zn2+][OH-]2 = Ksp = 4.5  10-17

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.

Make more acidic:

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

[Zn2+][OH-]2 = Ksp

[Zn2+][OH-]2 = Ksp

[Zn2+][OH-]2 = Ksp

Zumdahl Chapter 8

slide43

The Effects of pH on Solubility

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

Fe(OH)3(s) ↔Fe3+(aq) + 3 OH-(aq)

[OH-] = 7.9  10-12 mol L-1

(pH = 2.9 and) pOH = 11.1

[Fe3+][OH-]3 = Ksp

[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

In pure water:

[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]=

[OH-] = 3y = 1.3  10-9 mol L-1

pOH = 8.87

(and pH = 5.13)

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

Zumdahl Chapter 8

slide44

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

Initial concentration

Change in concentration

Equilibrium concentration

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+][IO3-] = Ksp

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

Zumdahl Chapter 8

slide45

The Common Ion Effect

[s]

[s]

[s]

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

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

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

Zumdahl Chapter 8

chapter 8 applications of aqueous equilibria
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)

Zumdahl Chapter 8