Ch. 16: Ionic EquilibriaBuffer Solution An acid/base equilibrium system that is

capable of maintaining a relatively constant pH even if a small amount of strong acid or base is added.

a) Components of a buffer solution:• a mixture of a weak acid and its conjugate base

e.g. acetic acid & sodium acetate (HA & A–) HA(aq) H+

(aq) + A–(aq)

or, ammonium chloride & ammonia (NH4+ & NH3)

NH4+

(aq) H+(aq) + NH3(aq)

pH of a Buffer Solution(b) pH of buffer solution: (pH ≈ pKa of HA)

HA(aq) H+ + A–(aq)

Initial “M” 0 “A”Change –x +x +xEquil M – x x A + x

Ka =[H+][A–]

[HA] =x(A + x)(M – x)

Since Ka is usually small (< ~ 10–4):

M – x ≈ M and A + x ≈ Athus, Ka ≈ x(A)/M or simply,

[H+] ≈Ka[HA][A–]

(general buffer system)

and [H+] ≈ Kamoles HAmoles A–

Henderson-Hasselbach Eqn

[H+] ≈Ka[HA][A–]

(general buffer system)

Again assuming that x is small, you can take the log of both sides and rearrange to get the Henderson-Hasselbach equation;

pH = pKa + log [acid]

[base ]

The HH equation is optional; if x is small approximation does not work, you need to use the more general method anyway...

Buffering Action (LeChatelier’s Principle)

– The major equilibrium in a buffer system is:

HA(aq) H+(aq) + A–

(aq)

1. What if some acid [H+] is added?Since H+ is a product of above equilibrium, the reaction will

shift in reverse, so that [HA] will increase and [A–] will decrease

But, as long as some A– remains, the system is still a mixture of HA and A–. Therefore, the general equation,

[H+] ≈ Ka[HA]/[A–]

still applies. Thus, [H+] is still ≈ Ka.

Buffering Action (cont)2. What if some base (OH-) is added instead?

The added base (OH-) will neutralize some of the acid HA and produce more of the conjugate base A–.

HA + OH– --> H2O + A–

[HA] will decrease and [A-] will increaseBut, as long as some HA remains, the system is still a mixture

of HA and A–, and ∴ still a buffer solution.

buffer capacity: is the amount of acid or base you can add without causing a large change in pH.

buffer range: the pH values for which a buffer system is the most effective.

(usually ±1 pH unit of either side of pKa)

Sample Buffer Solution Problems1. (a) The pKa of HF is 3.17. Calculate the pH of a 1.00-L

solution that is 1.00 M HF and 1.50 M NaF.(b) What is the pH of this solution after addition of 50.0 mL of 10.0 M HCl?

2. What mass of solid NaCN must be added to 500 mL (3 sig fig) of 0.50 M HCN to prepare a buffer solution with a pH of 9.50? pKa for HCN is 9.21.

Sample Buffer Solution Problems1. (a) The pKa of HF is 3.17. Calculate the pH of a 1.00-L

solution that is 1.00 M HF and 1.50 M NaF.(b) What is the pH of this solution after addition of 50.0 mL of 10.0 M HCl?

Answer: (a) pH = 3.35 (b) pH = 2.99

2. What mass of solid NaCN must be added to 500 mL (3 sig fig) of 0.50 M HCN to prepare a buffer solution with a pH of 9.50? pKa for HCN is 9.21.

Answer: 24 g NaCN

Another Sample Problem• Write the balanced chemical equation for the important

equilibrium that is occurring in an aqueous solution of NaNO2 and HNO2.

• Write the appropriate equilibrium constant expression for the reaction.

• Determine the pH of a solution of 0.100 M of each compound. (Ka HNO2 = 4.6 x 10–4)

Another Sample Problem• Write the balanced chemical equation for the important

equilibrium that is occurring in an aqueous solution of NaNO2 and HNO2.

• HNO2(aq) H+(aq) + NO2

–(aq)

• Write the appropriate equilibrium constant expression for the reaction.

• Ka = [H+][NO2–]/[HNO2]

• Determine the pH of a solution of 0.100 M of each compound. (Ka HNO2 = 4.6 x 10–4)

• 3.34

TitrationsTitration An unknown amount of one reactant is combined

exactly with a precisely measured volume of a standard solution of the other.

End-point When exactly stoichiometric amounts of two reactants have been combined.

Indicator Substance added to aid in detection of the endpoint (usually via a color change)

Example ProblemVinegar is an aqueous solution of acetic acid, HC2H3O2. A

12.5 mL sample of vinegar was titrated with a 0.504 M solution of NaOH. The titration required 15.40 mL of the base solution in order to reach the endpoint. What is the molar concentration of HC2H3O2 in vinegar?

NaOH(aq) + HC2H3O2(aq) --> NaC2H3O2(aq) + H2O(l)

(15.40 mL NaOH) x x

= 0.007762 mol HC2H3O2

M = = 0.621 M acetic acid

0.504 mol NaOH

1000 mL NaOH soln

1 mole HC2H3O2

1 mole NaOH

0.007762 mol HC2H3O2

0.0125 L vinegar

Titration of a Weak Acid by a Strong Base

e.g. 25.0 mL of 0.200 M HA (acetic acid) is titrated with 0.200 M NaOH (Ka = 1.8 x 10–5)

Examplee.g. 25.0 mL of 0.200 M HA (acetic acid) is titrated with

0.200 M NaOH.

Determine pH in each of four regions:(a) at the start -- before any NaOH is added

Just a simple weak acid: pH = 2.72

(b) before the equivalence point:The acid is partially neutralized by the added OH–

HA + OH– --> H2O + A–

Since both HA and A– are now present, the mixture is a buffer solution !and [H+] ≅ Ka (mole HA left) / (mole A– produced)

e.g. after addition of 10.0 mL of NaOH, pH = 4.57

Example, cont.(c) at the equivalence point:

The acid (HA) is completely neutralized, and only the anion A– is present∴ the solution is now the salt of a weak acid!Total volume is now 50.0 mL,

so [A–] = 0.10 M and pH = 8.88

(d) after the equivalence point:All acid (HA) is gone and excess OH– is present∴ the pH is determined by how much OH– is left after all the acid has been neutralizede.g. after 40.0 mL of NaOH added: pH = 12.66

Titration Curve

16

Indicators: Phenolphthalein

Choosing an Indicator

SUMMARY of Acid-Base Equilibrium Concepts

(Chapters 15 and 16)

1. pH scale and the autoionization of water (Kw)

2. Strong acids and bases3. Weak acids and bases (Ka or Kb)

4. Salts of weak acids and salts of weak bases (conjugate pairs: Kw ≅ KaKb)

5. Polyprotic acids (Ka1 and Ka2 etc)

6. Buffer solutions (Ka)

(mixture of weak acid and its conjugate base)

7. Acid-Base Titrations (especially weak acid by strong base, or weak base by

strong acid)

Sample Problems1. A 0.625 g sample of an unknown weak acid (call it HA for

short) is dissolved in enough water to make 25.0 mL of solution. This weak acid solution is then titrated with 0.100 M NaOH, and 45.0 mL of the NaOH solution is required to reach the equivalence point. Using a pH meter, the pH of the solution at the equivalence point is found to be 8.25. (a) Determine the molecular mass of the unknown acid.(b) Determine the pKa value of the unknown acid.

2. The pKb value for ammonia (NH3) is 4.74. Calculate the mass (in grams) of solid ammonium chloride (NH4Cl, 53.5 g/mole) that must be added to 300 mL of 0.25 M NH3 solution to make a buffer solution with a pH equal to 10.00.

Sample Problems1. A 0.625 g sample of an unknown weak acid (call it HA for

short) is dissolved in enough water to make 25.0 mL of solution. This weak acid solution is then titrated with 0.100 M NaOH, and 45.0 mL of the NaOH solution is required to reach the equivalence point. Using a pH meter, the pH of the solution at the equivalence point is found to be 8.25. (a) Determine the molecular mass of the unknown acid.(b) Determine the pKa value of the unknown acid.

Answer: (a) 139 g/mol HA(b) pKa = 3.70

2. The pKb value for ammonia (NH3) is 4.74. Calculate the mass (in grams) of solid ammonium chloride (NH4Cl, 53.5 g/mole) that must be added to 300 mL of 0.25 M NH3 solution to make a buffer solution with a pH equal to 10.00.

Answer: 0.73 g NH4Cl

Solubility and Complex Ion Equilibria

Solubility Equilibria(a) “Insoluble” salts are actually sparingly soluble

equilibrium: solid salt ions in solution

e.g. in a saturated solution of Ag2CO3, the following equilibrium is occurring.

Ag2CO3(s) 2 Ag+(aq) + CO3

2–(aq)

Solubility Product Constant = Ksp

Ksp = [Ag+]2[CO32–] = 8.1 x 10–12

Molar Solubility

Ag2CO3(s) 2 Ag+(aq) + CO3

2–(aq)

Ksp = [Ag+]2[CO32–] = 8.1 x 10–12

• Given the Ksp value, determine the “molar solubility”

Let x = molar solubility ! = moles Ag2CO3 dissolved/liter

[CO32–] = x and [Ag+] = 2x

Ksp = 8.1 x 10–12 = [Ag+]2[CO32–] = (2x)2(x)

Ksp = 4x3 ∴ x = 1.3 x 10–4 M

Common Ion EffectCommon ion effect:

Sparingly soluble salts are even less soluble when a “common” ion is present.

Le Châtelier’s Principle applies here:salt cation + anion

If an excess of one of the ions is present, the equilibrium will shift in reverse and less salt will dissolve!

Example Problem• Molar solubility of Ag2CO3 in 0.10 M Na2CO3?

Ag2CO3(s) 2 Ag+(aq) + CO3

2–(aq)

Ksp = 8.1 x 10–12

Let x = molar solubility of Ag2CO3

[Ag+] = 2x and [CO3

2–] = 0.10 + x

Ksp = 8.1 x 10–12 = [Ag+]2[CO32–]

= (2x)2(0.10 + x)

• But since Ag2CO3 is not very soluble,

assume x << 0.1Ksp ≈ (2x)2(0.10) ≈ 8.1 x 10–12

∴ x ≈ 4.5 x 10–6 M

PrecipitationWhen will a precipitate form?

The solution must be saturated for a ppt to form. Thus, the initial ion concentrations must be greater than the equilibrium values.

e.g. Ag2CO3(s) 2 Ag+(aq) + CO3

2–(aq)

If initial concentrations of the ions are given, calculate the ion product:

Q = [Ag+]2[CO32–] using the initial conc values

If Q > Ksp then a precipitate will form!

In some cases where Q > Ksp there is no precipitate unless it is disturbed; these are supersaturated.

precipitation occurs if Q > Ksp

a supersaturated solution will precipitate if a seed crystal is added

Precipitation

Sample ProblemA saturated solution of Mg3(PO4)2 has a concentration of 0.939 mg per liter. Calculate the solubility product constant for Mg3(PO4)2.

Sample ProblemA saturated solution of Mg3(PO4)2 has a concentration of 0.939 mg per liter. Calculate the solubility product constant for Mg3(PO4)2.

Answer: Ksp = 6.28 x 10–26

Qualitative analysis• A solution containing several different cations can often

be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others

• an analytical scheme that utilizes selective precipitation to identify the ions present in a solution is called a qualitative analysis scheme

• a sample containing several ions is subjected to the addition of several precipitating agents

• addition of each reagent causes one of the ions present to precipitate out

30

Qualitative Analysis

Multiple Equilibria(d) “Insoluble” salts of weak acids are more soluble in

acidic solutionse.g. Ag2S(s) 2 Ag+

(aq) + S2–(aq)

• If the solution is acidic, some S2– will be converted to HS– and H2S, thus causing more Ag2S to dissolve

The process can be viewed as a combination of multiple equilibria, occurring simultaneously:

Ag2S(s) 2 Ag+(aq) + S2–

(aq)

H+(aq) + S2–

(aq) HS–(aq)

H+(aq) + HS–

(aq) H2S(aq)

Net: Ag2S(s) + 2 H+(aq) 2 Ag+

(aq) + H2S(aq)

Complex Ion Equilibria2. Complex Ion Equilibria (increases solubility)• Metal ions form complex ions with Lewis bases, such as:

Cu2+(aq) + 4 NH3(aq) Cu(NH3)4

2+(aq)

The reaction is generally very favorable so…K is very large!

The equilibrium constant is called a Formation Constant

Kf =[Cu(NH3)4

2+]

[Cu2+][NH3]4= 1.1 x 1013 (large!)

Another example: Ag+

(aq) + 2 NH3(aq) Ag(NH3)2+

(aq)

Kf =[Ag(NH3)2

+]

[Ag+][NH3]2= 1.6 x 107

• If a complex ion can form, then an “insoluble” salt will be much more soluble (Le Châtelier’s Principle again!).

Example ProblemGiven the above formation constant for Ag(NH3)2

+ and the Ksp value for AgBr of 5.0 x 10–13, calculate the molar solubility of AgBr in (a) water, and (b) 0.10 M NH3.

(a) in water: AgBr(s) Ag+(aq) + Br–

(aq)

Ksp = [Ag+][Br–] = x2 = 5.0 x 10–13

∴ x = molar solubility = 7.1 x 10–7 M

Example Problem, cont.(b) in NH3 solution, two equilibria must be considered:

Ksp: AgBr(s) Ag+(aq) + Br–

(aq)

Kf: Ag+(aq) + 2 NH3(aq) Ag(NH3)2

+(aq)

Knet: AgBr(s) + 2 NH3(aq) Ag(NH3)2+

(aq) + Br–(aq)

Knet = [Ag(NH3)2+][Br–]/[NH3]2

= KspKf = (5.0 x 10–13)(1.6 x 107) = 8.0 x 10–6

• Let x = molar solubility of AgBr

[Ag(NH3)2+] = [Br–] = x [NH3] = 0.10 – 2x

Knet = x2/(0.10 – 2x)2 = 8.0 x 10–6

Since Knet is small in this case, assume 2x << 0.10

x2 / (0.10)2 ≈ 8.0 x 10–6 ∴ x ≈ 2.8 x 10–4

Thus, AgBr is about 1,000 times more soluble in NH3 solution than in water alone.

Amphoteric Metal Hydroxides• All metal hydroxides become more soluble in acidic

solution (like Ag2S), but some are also more soluble in basic solution, e.g.

Al(H2O)63+

(aq) + H2O(l) Al(H2O)5(OH)2+(aq) + H3O+

(aq)

• addition of OH− drives the equilibrium to the right and continues to remove H+ from the molecules

Al(H2O)5(OH)2+(aq) + OH−

(aq) Al(H2O)4(OH)2+

(aq) + H2O

(l)

Al(H2O)4(OH)2+

(aq) + OH−(aq) Al(H2O)3(OH)3(s) +

H2O (l)

• Others include Cr3+, Zn2+, Pb2+, Sb2+…

Sample ProblemSilver chromate, Ag2CrO4, is an “insoluble” substance with a

Ksp value of 1.2 x 10–12. Silver ion forms a stable complex ion with cyanide ion that has the formula Ag(CN)2

– and a formation constant (Kf) of 5.3 x 1018. Calculate the molar solubility of Ag2CrO4 in each of the following solutions. Write balanced chemical equations for any important equilibrium reactions that are occurring. (a) in water(b) in 2.00 M Na2CrO4

(c) in 2.00 M NaCN

Answers(a) Ag2CrO4 2 Ag+

(aq) + CrO42–

(aq)

6.7 x 10–5 M(b) Ag2CrO4 2 Ag+

(aq) + CrO42–

(aq)

3.9 x 10–7 M(c) Ag2CrO4(s) 2 Ag+

(aq) + CrO42–

(aq)

2 Ag+(aq) + 4 CN–

(aq) 2 Ag(CN)2–(aq)

Ag2CrO4(s) + 4 CN–(aq) 2 Ag(CN)2

–(aq) + CrO4

2–

(aq)

molar solubility ≈ 0.50 M