Acid-Base Equilibria andSolubility Equilibria

Chapter 16Semester 2/2014

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Chapter 16

Semester 2 / 2013

16.1 Homogeneous versus Heterogeneous Solution Equilibria

16.3 Buffer Solution

16.6 Solubility Equilibria

16.8 The Common Ion Effect and Solubility

16.1 Homogeneous Versus Heterogeneous Solution Equilibria

At equilibrium a weak acid solution contains nonionized acid as well as H+ ions and the conjugate base. All of these species are dissolved so the system is said to be Homogeneous.

If we consider equilibrium of the dissolution and precipitation of slightly soluble substances, the system is said to be Heterogeneous.

16.3 Buffer SolutionsA buffer solution is a solution of:

1. A weak acid or a weak base and

2. The salt of the weak acid or weak base (Both must be present!)

Definition: A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base.

16.3

Add strong acid H+ (aq) + CH3COO- (aq) CH3COOH (aq) CH3COO- combines with H+ ions from strong acid to produce weak acid (weak dissociation)

Add strong baseOH- (aq) + CH3COOH (aq) CH3COO- (aq) + H2O (l)CH3COOH combines with OH- ions from strong base to produce H2O(weak dissociation)

Consider an equal molar mixture of CH3COOH and CH3COONa

HCl H+ + Cl-

HCl + CH3COO- CH3COOH + Cl-

16.3

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) HF is a weak acid and F- is its conjugate basebuffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is it conjugate acidbuffer solution

16.3

= 9.20

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log [NH3][NH4

+]pKa = 9.25 pH = 9.25 + log [0.30]

[0.36]= 9.17

NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)

start (moles)

end (moles)

0.029 0.001 0.024

0.028 0.0 0.025

pH = 9.25 + log [0.25][0.28]

[NH4+] =

0.0280.10

final volume = 80.0 mL + 20.0 mL = 100 mL

[NH3] = 0.0250.10

16.3

moles of NH4+ =

0.029=(0.36*80)/1000

moles of NH3 =0.024= (0.30*80)/1000

moles of OH- =0.001=(0.05*20)/1000

Chemistry In Action: Maintaining the pH of Blood

16.3

16.6 Solubility Equilibria

16.6

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

Ksp = [Ag+][Cl-] Ksp is the solubility product constant

MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp = [Mg2+][F-]2

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+ (aq) + 2PO43- (aq) Ksp = [Ca2+]3[PO3

3-]2

Dissolution of an ionic solid in aqueous solution:

Q = Ksp Saturated solutionQ < Ksp Unsaturated solution No precipitate

Q > Ksp Supersaturated solution Precipitate will form

No precipitate

16.6

Molar solubility (mol/L) is the number of moles of solute dissolved in 1 L of a saturated solution.Solubility (g/L) is the number of grams of solute dissolved in 1 L of a saturated solution.

16.6

What is the solubility of silver chloride in g/L ?

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

Ksp = [Ag+][Cl-]Initial (M)

Change (M)

Equilibrium (M)

0.00

+s

0.00

+s

s s

Ksp = s2

s = Ksps = 1.3 x 10-5

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl = 1.3 x 10-5 mol AgCl1 L soln

143.35 g AgCl1 mol AgClx

= 1.9 x 10-3 g/L

Ksp = 1.6 x 10-10

16.6

16.6

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form?

16.6

The ions present in solution are Na+, OH-, Ca2+, Cl-.Only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?

[Ca2+]0 = 0.100 M [OH-]0 = 4.0 x 10-4 M

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

16.8 The Common Ion Effect and Solubility

Consider a solution containing two dissolved substances that share a common ion (AgCl and AgNO3)

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

AgNO3 (s) Ag+ (aq) + NO3- (aq)

Ag+ ions come from two sources: i.e AgCl & AgNO3

The total increase in [Ag+] ion concentration will make the ion product

greater than the solubility product:Q = [Ag+]total[Cl-] > Ksp, To reestablish equilibrium, some

AgCl will precipitate out of the solution, until the ion product

is equal to Ksp Adding a common ion decreases the solubility of the

salt (AgCl) in solution.

16

16.12

Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10-3 M silver nitrate solution.

17

Let s be the molar solubility of AgCl in AgNO3 solution. We summarize the changes in concentrations as follows:

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

Initial (M): 6.5 x 10-3 0.00Change (M): -s +s +sEquilibrium (M): (6.5 x 10-3 +s) s

16.12

Step 3: Ksp = [Ag+][Cl-]

1.6 x 10-10 = (6.5 x 10-3 + s)(s)

18

Because AgCl is quite insoluble and the presence of Ag+ ions from AgNO3 further lowers the solubility of AgCl, s must be very small compared with 6.5 x 10-3. Therefore, applying the approximation 6.5 x 10-3 + s ≈ 6.5 x 10-3 , we obtain

1.6 x 10-10 = (6.5 x 10-3 )ss = 2.5 x 10-8 M

Step 4: At equilibrium

[Ag+] = (6.5 x 10-3 + 2.5 x 10-8 ) M ≈ 6.5 x 10-3 M[Cl+] = 2.5 x 10-8 M

16.12

19

and so our approximation was justified in step 3. Because all the Cl- ions must come from AgCl, the amount of AgCl dissolved in AgNO3 solution also is 2.5 x 10-8 M. Then, knowing the molar mass of AgCl (143.4 g), we can calculate the solubility of AgCl as follows:

16.12

-8

32.5 × 10 mol AgCl 143.4 AgClsolubility of AgCl in AgNO solution = ×

1 L soln 1 mol AgCl

= -6 3.6 ×10 g / L