Question

Calcium lactate is a salt of weak acid and represented as $Ca(LaC{)}_2$. A saturated solution of $Ca(LaC{)}_2$ contains 0.13 mole of salt in 0.50 litre solution. The $pOH$ of solution is 5. Assuming complete dissociation of salt, calculate $K_a$ of lactic acid.

• A. $8.6\times {10}^{-5}$
• B. $5.2\times {10}^{-5}$
• C. $7.8\times {10}^{-6}$
• D. $4\times {10}^{-4}$

Answer 1

The correct option is B $5.2\times {10}^{-5}$

$Ca(LaC{)}_2+2H_{2}O\rightleftharpoons Ca(OH{)}_2+2HLAC$
$2La{C}^{-}+2H_{2}O\rightleftharpoons 2O{H}^{-}+2HLaC$
$100\text{(Before hydrolysis)}$
$(1-h)hh\text{(After hydrolysis)}$
$\left[Ca\right(LaC{)}_2]=\frac{0.13}{0.5}=0.26M$
$\left[La{C}^{-}\right]=0.26\times 2=0.52M$
(as 1 mole $Ca(LaC{)}_2$ gives 2 mole $La{C}^{-}$)
$\left[O{H}^{-}\right]=Ch=C.\sqrt{\left(\frac{K_h}{C}\right)}=\sqrt{(K_h.C)}$
$=\frac{\sqrt{(K_w\times C)}}{\sqrt{K_a}}$
Here $C$ is the concentration of anion which undergoes hydrolysis
${10}^{-5}=\frac{\sqrt{({10}^{-14}\times 0.52)}}{\sqrt{K_a}}$
$K_a=5.2\times {10}^{-5}$