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 $0.3M$ concentration. pOH of solution is $\mathrm{5.60.}$ If $90\%$ of the salt is dissociated then what will be the value of $p{K}_a$ ?

• 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)

$Ca(LaC{)}_2\to C{a}^{2+}+2La{C}^{-}$

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

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

pH after salt hydrolysis may be calculated as,

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

$14-5.6=\frac{1}{2}[14+p{K}_a+log0.54]$

$p{K}_a=2.8-log0.54$