Question

A solution has 0.001 mole/L zinc ions with pH = 6. The solubility product of zinc hydroxide is \[ K_{sp} = 8 \times 10^{-18}\ (\text{mol/L})^{3}. \] Ignoring activity corrections, find the ratio (rounded to two decimals) of the reaction quotient to the solubility product.

Hint:

Reaction quotient for metal hydroxides follows $Q = [\text{M}^{2+}][\text{OH}^-]^2$. Comparing $Q$ with $K_{sp}$ tells whether precipitation will occur.

Answer 1

Correct Answer: 0.01

For zinc hydroxide: \[ \text{Zn(OH)}_2 \rightleftharpoons \text{Zn}^{2+} + 2\ \text{OH}^- \] The reaction quotient is: \[ Q = [\text{Zn}^{2+}] [\text{OH}^-]^2. \] Given: \[ [\text{Zn}^{2+}] = 0.001\ \text{M}. \] pH = 6 ⇒ \[ \text{pOH} = 14 - 6 = 8 \] \[ [\text{OH}^-] = 10^{-8}\ \text{M}. \] Hence: \[ Q = (0.001)(10^{-8})^2 = 10^{-3} \times 10^{-16} = 10^{-19}. \] Now compute the ratio: \[ \frac{Q}{K_{sp}} = \frac{10^{-19}}{8 \times 10^{-18}} = \frac{1}{8} \times 10^{-1} = 0.0125. \] Rounded to two decimals: \[ \boxed{0.01} \]