The solubility product () of at is . A 500 ml of saturated solution of is mixed with equal volume of — Ionic Equilibrium Chemistry Question
Question
The solubility product ($K_{sp}$) of $\text{Ca(OH)}_2$ at $25^\circ\text{C}$ is $3.2 \times 10^{-5}$. A 500 ml of saturated solution of $\text{Ca(OH)}_2$ is mixed with equal volume of 0.36 M – NaOH. How much $\text{Ca(OH)}_2$ (in milligrams) is precipitated?
💡 Solution & Explanation
For a saturated $\text{Ca(OH)}_2$ solution, $4S^3 = 3.2 \times 10^{-5} \implies S = 0.02\text{ M}$. The initial $500\text{ ml}$ solution contains $0.01\text{ moles}$ of $\text{Ca}^{2+}$ and $0.02\text{ moles}$ of $\text{OH}^-$. Adding $500\text{ ml}$ of $0.36\text{ M NaOH}$ adds $0.18\text{ moles}$ of $\text{OH}^-$. The total volume is $1\text{ L}$. Without precipitation, $[\text{Ca}^{2+}] = 0.01\text{ M}$ and $[\text{OH}^-] = 0.20\text{ M}$. Since $(0.01)(0.20)^2 = 4 \times 10^{-4} > K_{sp}$, precipitation occurs. Let $x$ moles precipitate. $[\text{Ca}^{2+}] = 0.01 - x$ and $[\text{OH}^-] = 0.20 - 2x \approx 0.20\text{ M}$. At equilibrium: $(0.01 - x)(0.20)^2 = 3.2 \times 10^{-5} \implies (0.01 - x)(0.04) = 3.2 \times 10^{-5} \implies 0.01 - x = 8 \times 10^{-4}$. Thus, $x = 0.0092\text{ moles}$. The molar mass of $\text{Ca(OH)}_2 = 74\text{ g/mol}$. Mass precipitated = $0.0092 \times 74 = 0.6808\text{ g} = 680.8\text{ mg}$. Formatted to four digits, this is 0681.