Multiple Choice
Given that the solubility product constant (Ksp) for PbBr2 is 6.6 × 10^{-6}, what is the molar solubility of PbBr2 in a 0.500 M Pb(NO3)2 solution?
70
views
18. Aqueous Equilibrium
Solubility Product Constant: Ksp
Multiple Choice
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
Verified step by step guidance1
Write the dissociation equation for chromium(III) hydroxide: \(\mathrm{Cr(OH)_3 (s) \rightleftharpoons Cr^{3+} (aq) + 3 OH^- (aq)}\).
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.
Calculate the hydroxide ion concentration \([OH^-]\) from the given pH using the relation \(pH + pOH = 14\), so \(pOH = 14 - pH\), and then \([OH^-] = 10^{-pOH}\).
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 \(s\): \(s = \frac{K_{sp}}{[OH^-]^3}\).
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.
Related Videos
Related Practice
