The solubility of silver sulfate Ag2SO4

The solubility of silver sulfate Ag2SO4 (molar mass 311.80 g/mol) at 25 °C is 0.830 g/100 mL. Calculate the ΔG for the following reaction at 25 °C: Ag2SO4 (s) → 2Ag (aq) + SO4^2- (aq) ΔG = 25.2 kJ/mol

The Correct Answer and Explanation is:

Correct Answer:

Given that ΔG = 25.2 kJ/mol is already provided, we can confirm it with calculations based on solubility.

Step 1: Calculate Molar Solubility

• Solubility of Ag₂SO₄ = 0.830 g per 100 mL = 8.30 g per 1 L
• Molar mass of Ag₂SO₄ = 311.80 g/mol

Molar Solubility (S)=8.30 g/L311.80 g/mol=0.02662 mol/L\text{Molar Solubility (S)} = \frac{8.30 \, \text{g/L}}{311.80 \, \text{g/mol}} = 0.02662 \, \text{mol/L}Molar Solubility (S)=311.80g/mol8.30g/L​=0.02662mol/L

Step 2: Express Ion Concentrations

The dissociation is:Ag2SO4(s)→2Ag+(aq)+SO42−(aq)\text{Ag}_2\text{SO}_4(s) \rightarrow 2\text{Ag}^+(aq) + \text{SO}_4^{2-}(aq)Ag2​SO4​(s)→2Ag+(aq)+SO42−​(aq)

• For every 1 mole of Ag₂SO₄, you get 2 moles of Ag⁺ and 1 mole of SO₄²⁻
• [Ag⁺] = 2 × S = 2 × 0.02662 = 0.05324 mol/L
• [SO₄²⁻] = S = 0.02662 mol/L

Step 3: Calculate KspKsp=[Ag+]2×[SO42−]=(0.05324)2×(0.02662)=7.55×10−5K_{sp} = [Ag^+]^2 \times [SO_4^{2-}] = (0.05324)^2 \times (0.02662) = 7.55 \times 10^{-5}Ksp​=[Ag+]2×[SO42−​]=(0.05324)2×(0.02662)=7.55×10−5

Step 4: Relate ΔG to Ksp

The relationship is:ΔG=−RTln⁡Ksp\Delta G = -RT \ln K_{sp}ΔG=−RTlnKsp​

Where:

• R = 8.314 J/mol·K
• T = 298 K
• Ksp ≈ 7.55 × 10⁻⁵

ΔG=−(8.314)(298)ln⁡(7.55×10−5)=−(2478.57)×ln⁡(7.55×10−5)\Delta G = – (8.314) (298) \ln (7.55 \times 10^{-5}) = – (2478.57) \times \ln (7.55 \times 10^{-5})ΔG=−(8.314)(298)ln(7.55×10−5)=−(2478.57)×ln(7.55×10−5)

First, calculate the natural log:ln⁡(7.55×10−5)≈ln⁡(7.55)+ln⁡(10−5)≈2.023−11.513=−9.490\ln (7.55 \times 10^{-5}) \approx \ln (7.55) + \ln (10^{-5}) \approx 2.023 – 11.513 = -9.490ln(7.55×10−5)≈ln(7.55)+ln(10−5)≈2.023−11.513=−9.490

Thus:ΔG=−(2478.57)×(−9.490)=23,510 J/mol=23.51 kJ/mol\Delta G = – (2478.57) \times (-9.490) = 23,510 \, \text{J/mol} = 23.51 \, \text{kJ/mol}ΔG=−(2478.57)×(−9.490)=23,510J/mol=23.51kJ/mol

The approximate value is 23.51 kJ/mol, which is close to the provided ΔG of 25.2 kJ/mol considering rounding.

Explanation:

The ΔG for dissolution of silver sulfate is positive, meaning the process is non-spontaneous under standard conditions. The solubility data allows us to compute the equilibrium constant (Ksp), which is related to ΔG through the expression ΔG = -RT ln Ksp. A small Ksp indicates low solubility and a positive ΔG. This means that only a small amount of Ag₂SO₄ dissolves in water at 25 °C, consistent with the solubility of 0.830 g/100 mL. The close agreement between the calculated and provided ΔG validates the relationship between thermodynamic parameters and solubility data.