Question

Calcium lactate is a salt of a weak organic acid and represented as $Ca{\left(Lac\right)}_2$. A saturated solution of $Ca{\left(Lac\right)}_2$ contains $0.13$ mole of this salt in $0.50$l litre solution. The pOH of this solution is $5.60$. Assuming a complete dissociation of salt, calculate $K_a$ of the lactic acid.

Answer 1

$Ca{\left(Lac\right)}_2\rightleftharpoons {Ca}^{2+}+2{Lac}^{-}$

$0.13\times 2M$ $2\times 2\times 0.13M$

${Lac}^{-}+H_{2}O\rightleftharpoons HLac+{OH}^{-}$

At equi. $(0.52-x)$ $x$ $x$

$K_h=\frac{x^2}{(0.52-x)}=\frac{x^2}{0.52}$

$\left[{OH}^{-}\right]={10}^{-5.6}=2.51\times {10}^{-6}=x$

$K_h=\frac{251\times {10}^{-6}\times 2.51\times {10}^{-6}}{0.52}=12.12\times {10}^{-12}$

$K_a=\frac{K_w}{K_h}=\frac{{10}^{-14}}{12.12\times {10}^{-12}}=8.26\times {10}^{-4}$