Question

A weak monobasic acid is $1\%$ ionized at $1M$. Thus, percentage ionisation of the acid at $4M$ is

• A. $4.0\%$
• B. $5.0\%$
• C. $0.2\%$
• D. $0.5\%$

Answer 1

Answer: (D)

$HA+H_{2}O\rightleftharpoons H_3{O}^{+}+A^{-}$

$K_a=\frac{C{x}^2}{(1-x)}$, when $C=1M,x=0.01\dots (i)$

$K_a=\frac{C^{\prime }{y}^2}{(1-y)}$, when $C=4M,y=$? $\dots (ii)$

From Eqs. (i) and (ii)

$\frac{1\times x^2}{(1-x)}=\frac{4y^2}{(1-y)}$

$\frac{(0.01{)}^2}{(1-0.01)}=\frac{4y^2}{(1-y)}$

$\frac{4y^2}{(1-y)}=1.0\times 10^{-4}$

$\frac{y^2}{1-y}=0.25\times 10^{-4}$

$=y^2=0.25\times 10^{-4}$

$y=5\times 10^{-3}$

Thus, percentage $=5\times 10^{-3}\times 100=0.5\%$