Question

The solubility product of Fe(OH)₃ is \(1.1 \times 10^{-36}\) at 25 C. What is the solubility of ferric hydroxide?

• A. \(3.7 \times 10^{-8}\ \text{g L}^{-1}\)
• B. \(4.8 \times 10^{-9}\ \text{g L}^{-1}\)
• C. \(4.8 \times 10^{-8}\ \text{g L}^{-1}\)
• D. \(3.7 \times 10^{-9}\ \text{g L}^{-1}\)

Hint:

Write the Ksp expression for Fe(OH)3 in terms of solubility s, then convert mol/L to g/L.

Answer 1

The Correct Option is C

Step 1: Start from $K_{sp} = 27s^4$ and take logarithms on both sides: $\log K_{sp} = \log 27 + 4\log s$.

Step 2: $\log(1.1\times10^{-36}) = \log 1.1 + (-36) = 0.0414 - 36 = -35.9586$

Step 3: $\log 27 = 1.4314$

Step 4: $4\log s = -35.9586 - 1.4314 = -37.39$, so $\log s = -9.3475$, giving $s = 10^{-9.3475} = 4.49\times10^{-10}$ mol/L.

Step 5: Multiply by the molar mass of Fe(OH)3, 107 g/mol: $s = 4.49\times10^{-10} \times 107 = 4.8\times10^{-8}$ g/L. \[\boxed{s = 4.8\times10^{-8}\ \text{g L}^{-1}}\]