Given that the solubility product constant K_{sp} for Cr(OH)_3 is 6.70 imes 10^{-31}, what is the molar solubility of Cr(OH)_3 at pH 11.10?

• A. 2.0 imes 10^{-10} M
• B. 1.4 imes 10^{-8} M
• C. 6.7 imes 10^{-31} M
• D. 4.5 imes 10^{-6} M

Answer: (B)

Verified step by step guidance

1. Write the dissociation equation for chromium(III) hydroxide: \(\mathrm{Cr(OH)_3 (s) \rightleftharpoons Cr^{3+} (aq) + 3 OH^- (aq)}\).
2. Express the solubility product constant \(K_{sp}\) in terms of the molar solubility \(s\): \(K_{sp} = [Cr^{3+}][OH^-]^3 = s \times (3s)^3 = 27 s^4\) if \(s\) is the molar solubility and \([OH^-] = 3s\) in pure water. However, since the pH is given, the hydroxide concentration is controlled by pH, not by solubility alone.
3. Calculate the hydroxide ion concentration \([OH^-]\) from the given pH using the relation \(pH + pOH = 14\), so \(pOH = 14 - pH\), and then \([OH^-] = 10^{-pOH}\).
4. Set up the \(K_{sp}\) expression using the known \([OH^-]\) from the pH and the unknown \([Cr^{3+}] = s\): \(K_{sp} = s \times [OH^-]^3\). Rearrange to solve for \(s\): \(s = \frac{K_{sp}}{[OH^-]^3}\).
5. Substitute the values of \(K_{sp}\) and \([OH^-]\) into the equation to find the molar solubility \(s\) of \(\mathrm{Cr(OH)_3}\) at the given pH.