**Equilibrium III. Solubility and Complex Ion Formation.**

**Questions for quiz #10.**

1. Define the terms "solubility product", "pK_{sp}", "formation constant".

Solubility product is the equilibrium constant for the dissolution of a sparingly soluble salt into water.

pK_{sp} is - log_{10}K_{sp} where K_{sp} is the solublity product.

Formation constant is the equilibrium constant for the formation of complex (coordinated) ions in aqueous solution.

2. Apply the equilibrium analysis I have given you to the following solubility problems. (And, yes, you are supposed to be able to remember the names!)

2a. What is the solubility of iron(II) sulphide (pK_{sp} = 18.8) in pure water?

FeS ¾ Fe^{2+}(aq) + S^{2-}(aq)

x = solubility in mol/L

so [Fe^{2+}] = x and [S^{2-}] = x

K_{sp} = [Fe^{2+}][S^{2-}] = 1.58 x 10^{-19} = x^{2}.

So solubility = 3.98 x 10^{-10} mol/L

2b. What is the solubility of lead(II) chloride (pK_{sp} = 4.8) in pure water?

PbCl_{2} ¾ Pb^{2+}(aq) + 2 Cl^{-}(aq)

x = solublity in mol/L

so [Pb^{2+}] = x and [Cl^{-}] = 2x

K_{sp} = [Pb^{2+}][Cl^{-}]^{2} = 1.58 x 10^{-5} = (x)(2x)^{2} = 4x^{3}.

So solubility = 1.58 x 10^{-2} mol/L.

2c. What is the solubility of copper(II) phosphate (pK_{sp} = 36.89) in pure water?

Cu_{3}(PO_{4})_{2} ¾ 3 Cu^{2+}(aq) + 2 PO_{4}^{3-}(aq)

x = solubility in mol/L

so [Cu^{2+}] = 3x and [PO_{4}^{3-}] = 2x

K_{sp} = [Cu^{2+}]^{3}[PO_{4}^{3-}]^{2} = 1.29 x 10^{-37} = (3x)^{3}(2x)^{2} = 108x^{5}.

So solubility = 1.64 x 10^{-8} mol/L.

2d. What is the solubility of silver chloride (pK_{sp} = 9.74) in 0.100 M NaCl solution?

AgCl ¾ Ag^{+}(aq) + Cl^{-}(aq)

x = solubility in mol/L

so [Ag^{+}] = x and [Cl^{-}] = 0.1 + x (because the NaCl provides the 0.1 mol/L.)

K_{sp} = [Ag^{+}][Cl^{-}] = 1.82 x 10^{-9} = (x)(0.1 + x).

Assume x << 0.1 : 1.82 x 10^{-9} = (0.1x); x = 1.82 x 10^{-8} so approximation is valid.

So solubility = 1.82 x 10^{-8} mol/L.

2e. What is the solubility of barium phosphate (pK_{sp} = 22.47) in 0.100 M magnesium phosphate?

Ba_{3}(PO_{4})_{2} ¾ 3 Ba^{2+}(aq) + 2 PO_{4}^{3-}(aq)

x = solubility in mol/L.

so [Ba^{2+}] = 3x and [PO_{4}^{3-}] = (0.2 + 2x) (because 1 mol Mg_{3}(PO_{4})_{2} produces 2 mol PO_{4}^{3-}.

K_{sp} = [Ba^{2+}]^{3}[PO_{4}^{3-}]^{2} = 3.39 x 10^{-23} = (3x)^{3}(0.2 + 2x)^{2}.

Assume x << 0.2: 3.39 x 10^{-23} = (3x)^{3}(0.2)^{2} = 1.08x^{3}; and x = 3.15 x 10^{-8}. (Approximation valid)

So solubility = 3.15 x 10^{-8} mol/L.

2f. What is the pH of a saturated solution of nickel(II) hydroxide (pK_{sp} = 15.26) in pure water?

Ni(OH)_{2} ¾ Ni^{2+}(aq) + 2 OH^{-}(aq)

If x = solubility, then [Ni^{2+}] = x and [OH^{-}] = 2x.

So pH = 14 - pOH = 14 + log_{10} (2x).

K_{sp} = [Ni^{2+}][OH^{-}]^{2} = 5.50 x 10^{-16} = (x)(2x)^{2} = 4x^{3}.

So x = 5.16 x 10^{-6}.

From which pH = 9.01.

3. Calculate Q_{sp} in the following cases, and decide whether a precipitate will form or not. (Watch
for the dilution when two solutions are mixed, but not otherwise.)

3a. 0.100 g of sodium chloride is added to 100 mL of 0.100 silver nitrate solution. (pK_{sp} of
silver chloride = 9.74)

AgCl ¾ Ag^{+}(aq) + Cl^{-}(aq)

K_{sp} = [Ag^{+}][Cl^{-}]

Q_{sp} = 0.100*(0.100 g /58.5 g/mol )/0.1 L = 1.71 x 10^{-3}.

Since Q_{sp} exceeds K_{sp} a ppt will form.

3b. 100 mL of a 0.00200 M solution of nickel(II) nitrate is added to 1.00 L of 0.000200 M
solution of sodium sulphide. (pK_{sp} of nickel(II) sulphide = 20.97.)

NiS(s) ¾ Ni^{2+}(aq) + S^{2-}(aq)

Q_{sp} = [Ni^{2+}][S^{2-}] = 0.002*(0.1/1.1) * 0.0002*(1/1.1) = 3.31 x 10^{-8}.

Since this exceeds K_{sp} a ppt will form.

3c. 100 mL of a 0.0100 M sodium phosphate solution is added to 10.00 L of a 0.000100 M
solution of calcium nitrate. (pK_{sp} of calcium phosphate = 28.70.)

Ca_{3}(PO_{4})_{2}(s) ¾ 3 Ca^{2+}(aq) + 2 PO_{4}^{3-}(aq)

Q_{sp} = [Ca^{2+}]^{3}[PO_{4}^{3-}]^{2} = 0.0001 * 0.01*3*(.01/10) = 3 x 10^{-9}.

Since this exceeds K_{sp} a ppt will form.

4. A solution is 0.0100 M in lead ion and 0.000100 M in silver ion. Upon the very slow addition
of iodide ion, which of the metal ions will precipitate out first, and how much of this first ion is
left in solution when the second ion just starts to precipitate. Ignore any dilution effects. pK_{sp} for
lead iodide = 8.19; pK_{sp} for silver iodide = 16.08.

PbI_{2} ¾ Pb^{2+}(aq) + 2 I^{-}(aq)

AgI ¾ Ag^{+}(aq) + I^{-}(aq)

The [I^{-}] when PbI_{2} starts to ppt can be obtained:

K_{sp} = 6.46 x 10^{-9} = 0.01 * [I^{-}]^{2}

From which [I^{-}] = 8.04 x 10^{-4} mol/L.

The [I^{-}] when AgI starts to ppt can be calculated:

K_{sp} = 8.32 x 10^{-17} = 0.0001 * [I^{-}]

From which [I^{-}] = 8.32 x 10^{-13} mol/L.

Since the [I^{-}] is increasing from 0, the AgI will ppt out first.

When the PbI_{2} starts to ppt, the [I^{-}] = 8.04 x 10^{-4} mol/L. The [Ag^{+}] at this point can be calculated:

K_{sp} = 8.32 x 10^{-17} = [Ag^{+}] * 8.04 x 10^{-4}.

From which [Ag^{+}] = 1.03 x 10^{-9} mol/L.

5a. Apply the common ion effect to show how the solubility of copper(II) sulphide (pK_{sp} = 47.6)
is greater in hydrochloric acid than it is in pure water?

CuS(s) ¾ Cu^{2+}(aq) + S^{2-}(aq)

And, since S^{2-} is a weak base:

S^{2-}(aq) + H_{2}O ¾ HS^{-}(aq) + OH^{-}(aq).

Thus when HCl is added it reacts with the OH^{-} produced in this second equilibrium, thereby
removing S^{2-}. This then disturbs the first equilibrium: removal of S^{2-} leads to this equilibrium
moving to the right increasing the solubility of the copper sulfide as more dissolves.

5b. Would this also apply to silver acetate (pK_{sp }= 2.7)?

Yes. Acetate ion is a weak base (the conjugate of the weak acid, acetic acid).

6. The solubility of silver carbonate (pK_{sp} = 11.09) increases dramatically when dissolved in dilute
nitric acid, but hardly changes when dilute hydrochloric acid is used. What is happening?

Actually, the solubility of the silver carbonate is increasing dramatically with both acids:

Ag_{2}CO_{3}(s) ¾ 2 Ag^{+}(aq) + CO_{3}^{2-}(aq)

and

CO_{3}^{2-} + 2 H^{+} Ž CO_{2}(g) + H_{2}O(l)

Thus the loss of CO_{2} means the removal of CO_{3}^{2-} and its replacement with the acid anion.

With nitric acid this anion is NO_{3}^{-}, and silver nitrate is soluble.

With hydrochloric acid this anion is Cl^{-} and silver chloride is itself insoluble.

So in the case of nitric acid insoluble silver carbonate is replaced by soluble silver nitrate, and in the case of hydrochloric acid, the insoluble silver carbonate is replaced by insoluble silver chloride.

7. The solubility of silver salts increases dramatically in ammonia solutions due to the formation
of the complex ion Ag(NH_{3})_{2}^{+}, (pK_{f} = -7.2). Show that silver chloride (pK_{sp} = 9.74) and silver
bromide (pK_{sp} = 12.30) will dissolve in 1.00 M NH_{3} solution, but silver iodide (pK_{sp} = 16.08) will
not.

The increase in solubility of silver salts in ammonia is due to the formation of a soluble silver
ammine complex ion: Ag(NH_{3})_{2}^{+}.

Thus the three reactions in ammonia are:

AgCl(s) + 2 NH_{3}(aq) ¾ Ag(NH_{3})_{2}^{+}(aq) + Cl^{-}(aq)

AgBr(s) + 2 NH_{3}(aq) ¾ Ag(NH_{3})_{2}^{+}(aq) + Br^{-}(aq)

AgI(s) + 2 NH_{3}(aq) ¾ Ag(NH_{3})_{2}^{+}(aq) + I^{-}(aq)

The K values for these reactions can be calculated from K_{sp} and K_{f}(Ag(NH_{3})_{2}^{+}) since:

AgCl ¾ Ag^{+}(aq) + Cl^{-}(aq) (K_{sp}) and

Ag^{+}(aq) + 2 NH_{3}(aq) ¾ Ag(NH_{3})_{2}^{+}(aq) (K_{f})

So:

for AgCl in ammonia, the combined K = 2.88 x 10^{-3};

for AgBr in ammonia, the combined K = 7.94 x 10^{-6}; and

for AgI, in ammonia, the combined K = 1.32 x 10^{-9}.

The difference between these is the decreasing solubility of the silver halide in ammonia. Silver iodide with the smallest K for this process is therefore the least soluble.

**Electrochemistry.**

**Questions for Quiz #10.**

1. Sketch, give the shorthand notation and calculate Eº for voltaic cells using the reactions :

i. Cl_{2}(g) + 2 I^{-}(aq) Ž 2 Cl^{-}(aq) + I_{2}(s)

The two half reactions are:

Cl_{2}(g) + 2 e^{-} Ž 2 Cl^{-}(aq) (reduction) and

2 I^{-}(aq) Ž I_{2}(s) + 2 e^{-} (oxidation).

Thus the shorthand notation for the cell could be:

I_{2}(s) / Pt | I^{-}(aq) || Cl^{-}(aq) | Cl_{2}(g)/Pt

(Use inert Pt electrodes when a reaction does not involve a metal)

ii. H_{2}(g) + Br_{2}(l) Ž 2 H^{+}(aq) + 2 Br^{-}(aq)

The two half reactions are:

H_{2}(g) Ž 2 H^{+}(aq) + 2 e^{-} (oxidation) and

Br_{2}(l) + 2 e^{-} Ž 2 Br^{-}(aq) (reduction).

Thus the shorthand notation for the cell could be:

H_{2}(g)/Pt | H^{+}(aq) || Br^{-}(aq) | Br_{2}(l)/Pt.

2. For the cell Cu | Cu^{2+ }|| Pd^{2+} | Pd, Eº is +0.650 V.

Calculate the standard reduction potential for

Pd ^{2+} + 2 e^{-} Ž Pd.

The Eº_{cell} = Eº(reduction) + Eº(oxidation)

The cell half reactions are:

Pd ^{2+} + 2 e^{-} Ž Pd and

Cu Ž Cu^{2+} + 2 e^{-}.

From tables Eº(reduction) for the Cu/Cu^{2+} reaction is +0.153 V

so the Eº(oxidation) for this process is -0.153 V.

So Eº_{cell} = 0.650 V = -0.153 V + Eº(reduction)

from which Eº(reduction) = 0.803 V.

(According to this calculation!)

3. Use electrode potentials to calculate K and DGº_{298} for the reactions

i 2 H_{2}O_{2}(l) Ž O_{2}(g) + 2 H_{2}O(l).

The two half reactions are:

H_{2}O_{2}(l) Ž O_{2}(g) + 2 H^{+} + 2 e^{-} (oxidation); Eº = -0.68 V

H_{2}O_{2}(l) + 2 H^{+} + 2 e^{-} Ž 2 H_{2}O (reduction); Eº = 1.77 V

So Eº_{cell} = 1.09 V.

DGº = -nFEº = -RTlnK

so ln K = nFEº / RT = 2 * 96500 * 1.09 / (8.314 * 298) = 84.9

and K =7.5 x 10^{36}.

ii I_{2}(s) + H_{2}S(aq) Ž S(s) + 2 H^{+}(aq) + 2 I^{-}(aq).

The two half reactions are:

S^{2-}(aq) Ž S(s) + 2 e^{-} (oxidation); Eº = -0.14 V

I_{2}(s) + 2 e^{-} Ž 2 I^{-}(aq) (reduction); Eº = 0.535 V

So Eº_{cell} = 0.395 V

and K = 2.31 x 10^{13}.

4. Given the following standard reduction potentials calculate DGº_{f} for H_{3}BO_{3}(s).

H_{3}BO_{3}(s) + 3 H^{+}(aq) + 3 e^{-} Ž B(s) + 3 H_{2}O(l) Eº = -0.869 V

4 H^{+}(aq) + O_{2}(g) + 4 e^{-} Ž 2 H_{2}O(l) Eº = +1.229 V

Adding these two half reactions to give a balanced equation gives:

4 B(s) + 6H_{2}O(l)^{} + 3 O_{2}(g) Ž 4 H_{3}BO_{3}(s)^{} Eº = 2.098 V

And DGº_{298} can be calculated ( = -nFEº) = 12 * 96500 * 2.098 = 2430 kJ.

From thermodynamic tables:

DGº = Sm_{i}DGº_{f}(i) = (-4)(0) + (-6)(-237.2) + (-3)(0) + (+4)DGº_{f}(H_{3}BO_{3}) = 2430

So DGº_{f}(H_{3}BO_{3}) = 251.7 kJ.

5a A voltaic cell is made using the Mg^{2+}/Mg half cell with the Ni^{2+}/Ni half cell. If [Ni^{2+}] = 1.500
M and [Mg^{2+}] = 0.0500 M, calculate the cell potential.

The reaction is:

Mg + Ni^{2+} Ž Mg^{2+} + Ni; Eº_{cell} = 2.37 + (-0.25) = 2.12 V

The Nernst Equation applies: E_{cell} = Eº_{cell} - (RT/nF) ln Q.

Q = [Mg^{2+}] / [Ni^{2+}]

So E_{cell} = 2.12 - (8.314*298/(2*96500)) ln ( 0.5 / 1.5) = 2.13 V

5b. What is the concentration of Cd^{2+} in the cell

Zn | Zn^{2+}(0.0900 M) || Cd^{2+}( ? M) | Cd

if the cell potential is 0.4000 V ?

The cell reaction is: Zn + Cd^{2+} Ž Zn^{2+} + Cd and Eº_{cell} = 0.763 + (-0.403) = 0.36 V

Q = [Zn^{2+}] / [Cd^{2+}]

and E_{cell} = Eº_{cell} - (RT/nF) ln Q

So 0.400 = 0.36 - (RT/nF) ln Q

from which Q = 0.0443 = 0.090 / [Cd^{2+}],

from which [Cd^{2+}] = 2.03 M.

5c. Consider the cell Sn | Sn^{2+} || Pb^{2+} | Pb.

The cell reaction is:

Sn + Pb^{2+} Ž Sn^{2+} + Pb; Eº_{cell} = 0.14 + (-.126) = 0.014 V.

Starting from the standard cell :

i. which ion concentration will increase as the cell discharges ?

The Sn^{2+} concentration will increase.

ii. what will be the concentration of the two ions when the cell potential becomes zero?

When the cell potential is 0 V, the reaction is at equilibrium.

So ln K = nFEº / RT = 2 * 96500 * 0.014 / (8.314 * 298) = 1.091

and K = 2.98.

K = [Sn^{2+}] / [Pb^{2+}] which, from an ICE analysis gives

K = (1 + x) / (1 - x) = 2.98

from which x = 0.497

so [Sn^{2+}] = 1.497 M and [Pb^{2+}] = 0.503 M at equilibrium.

6. An acidic solution containing Pb^{2+} ions is electrolysed and PbO_{2} plates out at the anode.

a. Write out the chemical half reaction for this process.

Pb^{2+}(aq) + 2 H_{2}O(l) Ž PbO_{2}(s) + 4 H^{+}(aq) + 2 e^{-}.

b. If a current of 0.500 A is used for 15.0 min., how many grams of PbO_{2} plate out ?

1 C = 1 A.s

So the charge passed = 0.500 A x 15.0*60 = 450 C.

1 mole of charge = 96500 C, so 450 C gives 4.66 x 10^{-3} moles of electrons, which (ffrom the
above equation) will produce 2.33 x 10^{-3} moles of PbO_{2}

Mass of PbO_{2} formed = 2.33 x 10^{-3} mol * 239.2 g/mol = 0.557 g.

c. If a solution contains 1.50 g of Pb2+, how many minutes will it take to plate out all of
the lead as PbO_{2}?

Since 0.557 g plate out in 15 minutes (at 0.500 amps) it will take 15 min * 1.5 g / 0.557 g = 40.4 minutes.

7a. Define the following terms with an example:

primary cell;

secondary cell;

fuel cell.

7b. Discuss some of the problems facing the designers of an efficient fuel cell.