Calcium Lactate is a salt of weak acid and represented as . A saturated solution of contains of salt — Ionic Equilibrium Chemistry Question
Question
Calcium Lactate is a salt of weak acid and represented as $\text{Ca(Lac)}_2$. A saturated solution of $\text{Ca(Lac)}_2$ contains $0.125\text{ mole}$ of salt in $0.50\text{ L}$ solution. The pOH of this is 5.60. Assuming complete dissociation of salt, calculate $K_a$ of lactate acid. ($\log 2.5 = 0.4$)
💡 Solution & Explanation
Salt concentration $C = 0.125 / 0.50 = 0.25\text{ M}$. Since it's $\text{Ca(Lac)}_2$, the concentration of lactate ion $[\text{Lac}^-] = 2 \times 0.25 = 0.50\text{ M}$. Given $\text{pOH} = 5.60$, $[\text{OH}^-] = 10^{-5.6} = 10^{0.4} \times 10^{-6} = 2.5 \times 10^{-6}\text{ M}$. For anionic hydrolysis, $[\text{OH}^-] = \sqrt{\frac{K_w}{K_a} \times [\text{Lac}^-]}$. Substituting the values: $(2.5 \times 10^{-6})^2 = \frac{10^{-14}}{K_a} \times 0.50 \implies 6.25 \times 10^{-12} = \frac{5.0 \times 10^{-15}}{K_a}$. Thus, $K_a = \frac{5.0 \times 10^{-15}}{6.25 \times 10^{-12}} = 8.0 \times 10^{-4}$. Therefore, correct answer is B.