Equilibrium III. Solubility and Complex Ion Formation.

Questions for quiz #10.

1. Define the terms "solubility product", "pKsp", "formation constant".

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

pKsp is - log10Ksp where Ksp 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 (pKsp = 18.8) in pure water?

FeS ¾ Fe2+(aq) + S2-(aq)

x = solubility in mol/L

so [Fe2+] = x and [S2-] = x

Ksp = [Fe2+][S2-] = 1.58 x 10-19 = x2.

So solubility = 3.98 x 10-10 mol/L

2b. What is the solubility of lead(II) chloride (pKsp = 4.8) in pure water?

PbCl2 ¾ Pb2+(aq) + 2 Cl-(aq)

x = solublity in mol/L

so [Pb2+] = x and [Cl-] = 2x

Ksp = [Pb2+][Cl-]2 = 1.58 x 10-5 = (x)(2x)2 = 4x3.

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

2c. What is the solubility of copper(II) phosphate (pKsp = 36.89) in pure water?

Cu3(PO4)2 ¾ 3 Cu2+(aq) + 2 PO43-(aq)

x = solubility in mol/L

so [Cu2+] = 3x and [PO43-] = 2x

Ksp = [Cu2+]3[PO43-]2 = 1.29 x 10-37 = (3x)3(2x)2 = 108x5.

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

2d. What is the solubility of silver chloride (pKsp = 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.)

Ksp = [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 (pKsp = 22.47) in 0.100 M magnesium phosphate?

Ba3(PO4)2 ¾ 3 Ba2+(aq) + 2 PO43-(aq)

x = solubility in mol/L.

so [Ba2+] = 3x and [PO43-] = (0.2 + 2x) (because 1 mol Mg3(PO4)2 produces 2 mol PO43-.

Ksp = [Ba2+]3[PO43-]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 = 1.08x3; 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 (pKsp = 15.26) in pure water?

Ni(OH)2 ¾ Ni2+(aq) + 2 OH-(aq)

If x = solubility, then [Ni2+] = x and [OH-] = 2x.

So pH = 14 - pOH = 14 + log10 (2x).

Ksp = [Ni2+][OH-]2 = 5.50 x 10-16 = (x)(2x)2 = 4x3.

So x = 5.16 x 10-6.

From which pH = 9.01.

3. Calculate Qsp 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. (pKsp of silver chloride = 9.74)

AgCl ¾ Ag+(aq) + Cl-(aq)

Ksp = [Ag+][Cl-]

Qsp = 0.100*(0.100 g /58.5 g/mol )/0.1 L = 1.71 x 10-3.

Since Qsp exceeds Ksp 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. (pKsp of nickel(II) sulphide = 20.97.)

NiS(s) ¾ Ni2+(aq) + S2-(aq)

Qsp = [Ni2+][S2-] = 0.002*(0.1/1.1) * 0.0002*(1/1.1) = 3.31 x 10-8.

Since this exceeds Ksp 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. (pKsp of calcium phosphate = 28.70.)

Ca3(PO4)2(s) ¾ 3 Ca2+(aq) + 2 PO43-(aq)

Qsp = [Ca2+]3[PO43-]2 = 0.0001 * 0.01*3*(.01/10) = 3 x 10-9.

Since this exceeds Ksp 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. pKsp for lead iodide = 8.19; pKsp for silver iodide = 16.08.

PbI2 ¾ Pb2+(aq) + 2 I-(aq)

AgI ¾ Ag+(aq) + I-(aq)

The [I-] when PbI2 starts to ppt can be obtained:

Ksp = 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:

Ksp = 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 PbI2 starts to ppt, the [I-] = 8.04 x 10-4 mol/L. The [Ag+] at this point can be calculated:

Ksp = 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 (pKsp = 47.6) is greater in hydrochloric acid than it is in pure water?

CuS(s) ¾ Cu2+(aq) + S2-(aq)

And, since S2- is a weak base:

S2-(aq) + H2O ¾ HS-(aq) + OH-(aq).

Thus when HCl is added it reacts with the OH- produced in this second equilibrium, thereby removing S2-. This then disturbs the first equilibrium: removal of S2- 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 (pKsp = 2.7)?

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

6. The solubility of silver carbonate (pKsp = 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:

Ag2CO3(s) ¾ 2 Ag+(aq) + CO32-(aq)

and

CO32- + 2 H+ Ž CO2(g) + H2O(l)

Thus the loss of CO2 means the removal of CO32- and its replacement with the acid anion.

With nitric acid this anion is NO3-, 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(NH3)2+, (pKf = -7.2). Show that silver chloride (pKsp = 9.74) and silver bromide (pKsp = 12.30) will dissolve in 1.00 M NH3 solution, but silver iodide (pKsp = 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(NH3)2+.

Thus the three reactions in ammonia are:

AgCl(s) + 2 NH3(aq) ¾ Ag(NH3)2+(aq) + Cl-(aq)

AgBr(s) + 2 NH3(aq) ¾ Ag(NH3)2+(aq) + Br-(aq)

AgI(s) + 2 NH3(aq) ¾ Ag(NH3)2+(aq) + I-(aq)

The K values for these reactions can be calculated from Ksp and Kf(Ag(NH3)2+) since:

AgCl ¾ Ag+(aq) + Cl-(aq) (Ksp) and

Ag+(aq) + 2 NH3(aq) ¾ Ag(NH3)2+(aq) (Kf)

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. Cl2(g) + 2 I-(aq) Ž 2 Cl-(aq) + I2(s)

The two half reactions are:

Cl2(g) + 2 e- Ž 2 Cl-(aq) (reduction) and

2 I-(aq) Ž I2(s) + 2 e- (oxidation).

Thus the shorthand notation for the cell could be:

I2(s) / Pt | I-(aq) || Cl-(aq) | Cl2(g)/Pt

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

ii. H2(g) + Br2(l) Ž 2 H+(aq) + 2 Br-(aq)

The two half reactions are:

H2(g) Ž 2 H+(aq) + 2 e- (oxidation) and

Br2(l) + 2 e- Ž 2 Br-(aq) (reduction).

Thus the shorthand notation for the cell could be:

H2(g)/Pt | H+(aq) || Br-(aq) | Br2(l)/Pt.

2. For the cell Cu | Cu2+ || Pd2+ | Pd, Eº is +0.650 V.

Calculate the standard reduction potential for

Pd 2+ + 2 e- Ž Pd.

The Eºcell = Eº(reduction) + Eº(oxidation)

The cell half reactions are:

Pd 2+ + 2 e- Ž Pd and

Cu Ž Cu2+ + 2 e-.

From tables Eº(reduction) for the Cu/Cu2+ reaction is +0.153 V

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

So Eºcell = 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 D298 for the reactions

i 2 H2O2(l) Ž O2(g) + 2 H2O(l).

The two half reactions are:

H2O2(l) Ž O2(g) + 2 H+ + 2 e- (oxidation); Eº = -0.68 V

H2O2(l) + 2 H+ + 2 e- Ž 2 H2O (reduction); Eº = 1.77 V

So Eºcell = 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 1036.

ii I2(s) + H2S(aq) Ž S(s) + 2 H+(aq) + 2 I-(aq).

The two half reactions are:

S2-(aq) Ž S(s) + 2 e- (oxidation); Eº = -0.14 V

I2(s) + 2 e- Ž 2 I-(aq) (reduction); Eº = 0.535 V

So Eºcell = 0.395 V

and K = 2.31 x 1013.

4. Given the following standard reduction potentials calculate Df for H3BO3(s).

H3BO3(s) + 3 H+(aq) + 3 e- Ž B(s) + 3 H2O(l) Eº = -0.869 V

4 H+(aq) + O2(g) + 4 e- Ž 2 H2O(l) Eº = +1.229 V

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

4 B(s) + 6H2O(l) + 3 O2(g) Ž 4 H3BO3(s) Eº = 2.098 V

And D298 can be calculated ( = -nFEº) = 12 * 96500 * 2.098 = 2430 kJ.

From thermodynamic tables:

DGº = SmiDf(i) = (-4)(0) + (-6)(-237.2) + (-3)(0) + (+4)Df(H3BO3) = 2430

So Df(H3BO3) = 251.7 kJ.

5a A voltaic cell is made using the Mg2+/Mg half cell with the Ni2+/Ni half cell. If [Ni2+] = 1.500 M and [Mg2+] = 0.0500 M, calculate the cell potential.

The reaction is:

Mg + Ni2+ Ž Mg2+ + Ni; Eºcell = 2.37 + (-0.25) = 2.12 V

The Nernst Equation applies: Ecell = Eºcell - (RT/nF) ln Q.

Q = [Mg2+] / [Ni2+]

So Ecell = 2.12 - (8.314*298/(2*96500)) ln ( 0.5 / 1.5) = 2.13 V

5b. What is the concentration of Cd2+ in the cell

Zn | Zn2+(0.0900 M) || Cd2+( ? M) | Cd

if the cell potential is 0.4000 V ?

The cell reaction is: Zn + Cd2+ Ž Zn2+ + Cd and Eºcell = 0.763 + (-0.403) = 0.36 V

Q = [Zn2+] / [Cd2+]

and Ecell = Eºcell - (RT/nF) ln Q

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

from which Q = 0.0443 = 0.090 / [Cd2+],

from which [Cd2+] = 2.03 M.

5c. Consider the cell Sn | Sn2+ || Pb2+ | Pb.

The cell reaction is:

Sn + Pb2+ Ž Sn2+ + Pb; Eºcell = 0.14 + (-.126) = 0.014 V.

Starting from the standard cell :

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

The Sn2+ 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 = [Sn2+] / [Pb2+] which, from an ICE analysis gives

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

from which x = 0.497

so [Sn2+] = 1.497 M and [Pb2+] = 0.503 M at equilibrium.

6. An acidic solution containing Pb2+ ions is electrolysed and PbO2 plates out at the anode.

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

Pb2+(aq) + 2 H2O(l) Ž PbO2(s) + 4 H+(aq) + 2 e-.

b. If a current of 0.500 A is used for 15.0 min., how many grams of PbO2 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 PbO2

Mass of PbO2 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 PbO2?

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.