CHM 112 Homework 28 Answers

Calculate K_sp using data from the Table of Thermodynamic Quantities for the following ionic compounds. How does your calculated result compare to K_sp found from the Table of Solubility Product Constants? Choose the data set corresponding to the seventh digit of your Student ID Number.

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Use ΔG°_f values for the components in the solubilization reaction to find ΔG°, then use the relation ΔG° = –RTln K_eq, where K_eq = K_sp for the reaction. R = 8.314 J/mol·K and T = 25 °C = 298 K.

0) BaCO₃

BaCO₃(s) → ← Ba²⁺(aq) + CO₃^2–(aq)

ΔG° = [–560.8 + –527.8] – [–1138] = 49 kJ/mol = 49000 J/mol

K_sp = exp(–(49000)/(8.314)(298)) = 2.6×10^–9

K_sp(table) = 5.1×10^–9; the agreement is adequate.

1) CaCO₃

CaCO₃(s) → ← Ca²⁺(aq) + CO₃^2–(aq)

ΔG° = [–553.04 + –527.8] – [–1128] = 47 kJ/mol = 47000 J/mol

K_sp = exp(–(47000)/(8.314)(298)) = 5.8×10^–9

K_sp(table) = 2.8×10^–9; the agreement is adequate.

2) BaSO₄

BaSO₄(s) → ← Ba²⁺(aq) + SO₄^2–(aq)

ΔG° = [–560.8 + –744.5] – [–1362] = 57 kJ/mol = 57000 J/mol

K_sp = exp(–(57000)/(8.314)(298)) = 1.0×10^–10

K_sp(table) = 1.1×10^–10; the agreement excellent.

3) CaSO₄

CaSO₄(s) → ← Ca²⁺(aq) + SO₄^2–(aq)

ΔG° = [–553.04 + –744.5] – [–1322] = 24 kJ/mol = 24000 J/mol

K_sp = exp(–(24000)/(8.314)(298)) = 6.2×10^–5

K_sp(table) = 9.1×10^–6; the agreement is adequate.

4) FeCO₃

FeCO₃(s) → ← Fe²⁺(aq) + CO₃^2–(aq)

ΔG° = [–78.90 + –527.8] – [–666.7] = 60.0 kJ/mol = 60000 J/mol

K_sp = exp(–(60000)/(8.314)(298)) = 3.0×10^–11

K_sp(table) = 3.2×10^–11; the agreement is good.

5) PbSO₄

PbSO₄(s) → ← Pb²⁺(aq) + SO₄^2–(aq)

ΔG° = [–24.43 + –744.5] – [–813.2] = 44.3 kJ/mol = 44300 J/mol

K_sp = exp(–(44300)/(8.314)(298)) = 1.7×10^–8

K_sp(table) = 1.6×10^–8; the agreement is excellent.

6) MgCO₃

MgCO₃(s) → ← Mg²⁺(aq) + CO₃^2–(aq)

ΔG° = [–454.8 + –527.8] – [–1012] = 29 kJ/mol = 29000 J/mol

K_sp = exp(–(29000)/(8.314)(298)) = 8.3×10^–6

K_sp(table) = 3.5×10^–8; the agreement is poor.

7) Ag₂SO₄

Ag₂SO₄(s) → ← 2 Ag⁺(aq) + SO₄^2–(aq)

ΔG° = [2(77.11) + –744.5] – [–618.5] = 28.2 kJ/mol = 28200 J/mol

K_sp = exp(–(28200)/(8.314)(298)) = 1.1×10^–5

K_sp(table) = 1.4×10^–5; the agreement is good.

8) SrCO₃

SrCO₃(s) → ← Sr²⁺(aq) + CO₃^2–(aq)

ΔG° = [–557.3 + –527.8] – [–1138] = 53 kJ/mol = 53000 J/mol

K_sp = exp(–(53000)/(8.314)(298)) = 5.1×10^–10

K_sp(table) = 1.1×10^–10; the agreement is adequate.

9) SrSO₄

SrSO₄(s) → ← Sr²⁺(aq) + SO₄^2–(aq)

ΔG° = [–557.3 + –744.5] – [–1334] = 32 kJ/mol = 32000 J/mol

K_sp = exp(–(32000)/(8.314)(298)) = 2.5×10^–6

K_sp(table) = 3.2×10^–7; the agreement is poor.

Homework Assignment #28

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