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Question: The solubility of Fe(OH)3 in a buffer solution of pH = 4 is 4.32 × 10-2 mol/L. How many times is thi...

The solubility of Fe(OH)3 in a buffer solution of pH = 4 is 4.32 × 10-2 mol/L. How many times is this solubility greater than its solubility in pure water. (Ignore the hydrolysis of Fe 3+ ions) Given: 4.32/0.4\sqrt{0.4} = 6.83

A

109

B

6.83 × 106

C

2.16 × 109

D

none of these

Answer

109

Explanation

Solution

For the dissolution reaction in acid:

Fe(OH)3(s)+3H+→Fe3++3H2O\text{Fe(OH)}_3(s) + 3\text{H}^+ \to \text{Fe}^{3+} + 3\text{H}_2\text{O}

The solubility ss is given by:

s=K [H+]3s = K\,[\text{H}^+]^3

where KK is a constant (ignoring Fe³⁺ hydrolysis). In a buffer of pH 4, [H+]=10−4[\text{H}^+] = 10^{-4} M and the solubility is:

sacid=K (10−4)3=K×10−12s_{\text{acid}} = K\,(10^{-4})^3 = K \times 10^{-12}

Given that sacid=4.32×10−2 s_{\text{acid}} =4.32 \times 10^{-2} M, we find:

K=4.32×10−210−12=4.32×1010K = \frac{4.32 \times 10^{-2}}{10^{-12}} = 4.32 \times 10^{10}

In pure water, [H+]=10−7[\text{H}^+] = 10^{-7} M, so the solubility is:

swater=K (10−7)3=4.32×1010×10−21=4.32×10−11 Ms_{\text{water}} = K\,(10^{-7})^3 = 4.32 \times 10^{10}\times 10^{-21} = 4.32 \times 10^{-11}\text{ M}

The ratio of the solubilities is:

sacidswater=4.32×10−24.32×10−11=109\frac{s_{\text{acid}}}{s_{\text{water}}} = \frac{4.32 \times 10^{-2}}{4.32 \times 10^{-11}} = 10^9