Question

The sodium salt of a certain weak monobasic organic acid is hydrolysed to an extent of $3\%$ in its $0.1M$ solution at ${25}^{\circ }\text{C}$. Given that the ionic product of water is ${10}^{-14}$ at this temperature, what is the dissociation constant of the acid ?

• A. $1\times {10}^{-10}$
• B. $1\times {10}^{-9}$
• C. $3.33\times {10}^{-9}$
• D. $3.33\times {10}^{-10}$

Answer 1

The correct option is A $1\times {10}^{-10}$

$\begin{array}{ccccccccc}& NaX& +& H_{2}O& \rightleftharpoons & & XH& +& NaOH\\ \text{Before hydrolysis}& 1& & & & & 0& & 0\\ \text{After hydrolysis}& C(1-h)& & & & & Ch& & Ch\end{array}$
$h=0.03$
From salt hydrolysis, we know,
$K_h=C{h}^2=9\times {10}^{-5}$
$K_h=\frac{K_w}{K_a}=9\times {10}^{-5}$
$\Rightarrow K_a=\frac{{10}^{-14}}{9\times {10}^{-5}}=1.11\times {10}^{-10}=1\times {10}^{-10}$