Chapter 15: Problem 32

Calculate the molar solubility of \(\mathrm{Cd}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=5.9 \times 10^{-11}\).

Short Answer

Expert verified

The molar solubility of Cd(OH)2 is approximately \(1.17 \times 10^{-4}\,\mathrm{mol/L}\).

Step by step solution

Write the solubility equilibria

Write the solubility equilibria for the dissolution of Cd(OH)2 in water: \[ \mathrm{Cd(OH)_{2}(s)} \rightleftharpoons \mathrm{Cd^{2+}(aq)} + 2\,\mathrm{OH^{-}(aq)} \]

Write the expression for Ksp

Write the expression for the solubility product constant (Ksp) based on the solubility equilibria: \[ K_{\mathrm{sp}} = [\mathrm{Cd^{2+}}]\,[\mathrm{OH^{-}}]^2 \]

Define the molar solubility

Let the molar solubility of Cd(OH)2 be S. As Cd(OH)2 dissolves into Cd²⁺ and OH⁻ ions, the concentration of Cd²⁺ ions is equal to S while the concentration of OH⁻ ions is double that of the Cd²⁺ ions due to the stoichiometry of the reaction: \[ [\mathrm{Cd^{2+}}] = S \] \[ [\mathrm{OH^{-}}] = 2S \]

Substitute the expressions into the Ksp equation

Replace the terms in the Ksp equation with the molar solubility expressions obtained in the previous step: \[ K_{\mathrm{sp}} = (S)(2S)^2 \]

Solve the equation for S

Simplify the equation and solve for the molar solubility (S) using the given value of Ksp: \begin{align*} K_{\mathrm{sp}} &= 4S^3 \\ 5.9 \times 10^{-11} &= 4S^3 \\ S^3 &= \frac{5.9 \times 10^{-11}}{4} \\ S &= \sqrt[3]{\frac{5.9 \times 10^{-11}}{4}} \end{align*}