Question

Calcium lactate is a salt of weak acid i.e., lactic acid having general formula formula $Ca ( LaC )_{2}$. Aqueous solution of salt has $0.3 M$ concentration. pOH of solution is $5.60 .$ If $90 \%$ of the salt is dissociated then what will be the value of $pK _{ a }$ ?

• A. 2.8 - log (0.54)
• B. 2.8 + log (0.54)
• C. 2.8 + log (0.27)
• D. None of these

Answer 1

Answer: (A)

$Ca ( LaC )_{2} \rightarrow Ca ^{2+}+2 LaC ^{-}$

Concentration of salt $=\frac{0.6}{2}=0.3$ M

Concentration of $\left(L a C^{-}\right)$ after dissociation $=2\times 0.3\times 0.9=0.54$ M

pH after salt hydrolysis may be calculated as,

$\text{pH} \, \text{=} \, \frac{\text{1}}{\text{2}} \left(\text{[(pK}\right)_{\text{w}} \, \text{+} \, \left(\text{pK}\right)_{\text{a}} \, \text{+} \, \left(\text{log(LaC}\right)^{-} \text{)]}$

$14-5.6=\frac{1}{2}\left[\right.14+pK_{a}+log 0.54\left]\right.$

$pK_{a}=2.8-log 0.54$