Calcium lactate is a salt of a weak organic a | vedclass

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Question

(D) Calcium lactate is represented as $Ca(Lac)_2$. It dissociates as: $Ca(Lac)_2 \rightarrow Ca^{2+} + 2Lac^-$. Concentration of salt $C = \frac{0.13

Vedclass · Accepted Answer

$8.28 	imes 10^{-4}$

Vedclass · Answer

(D) Calcium lactate is represented as $Ca(Lac)_2$. It dissociates as: $Ca(Lac)_2 ightarrow Ca^{2+} + 2Lac^-$. Concentration of salt $C = \frac{0.13 \ mol}{0.5 \ L} = 0.26 \ M$. Concentration of lactate ion $[Lac^-] = 2 	imes 0.26 = 0.52 \ M$. Given $p^{OH} = 5.6$,so $p^H = 14 - 5.6 = 8.4$. $[OH^-] = 10^{-p^{OH}} = 10^{-5.6} = 2.51 	imes 10^{-6} \ M$. For the hydrolysis of the salt of a weak acid and strong base: $[OH^-] = \sqrt{\frac{K_w \cdot C_{salt}}{K_a}}$. Here,$C_{salt}$ is the concentration of the anion,which is $0.52 \ M$. $(2.51 	imes 10^{-6})^2 = \frac{10^{-14} 	imes 0.52}{K_a}$. $6.3 	imes 10^{-12} = \frac{5.2 	imes 10^{-15}}{K_a}$. $K_a = \frac{5.2 	imes 10^{-15}}{6.3 	imes 10^{-12}} \approx 8.25 	imes 10^{-4}$. Comparing with the given options,the closest value is $8.28 	imes 10^{-4}$.

Calcium lactate is a salt of a weak organic acid. $A$ $0.5 \ L$ saturated solution of calcium lactate contains $0.13 \ mol$ of the salt. If the $p^{OH}$ of the solution is $5.6$,what will be the $K_a$ of lactic acid,assuming complete dissociation of the salt?

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