Question

If $0.1\, M$ of a weak acid is taken and its percentage of degree of ionization is $1.34 \%$, then its ionization constant will be :

• A. $ 0.8\times 10^{-5}$
• B. $ 1.79\times 10^{-5}$
• C. $ 0.182\times 10^{-5}$
• D. None of these

Answer 1

Answer: (B)

According to Ostwald's dilution law
$K_{a}=\frac{C \alpha^{2}}{1-\alpha}$
where, $K_{a}=$ equilibrium constant for weak acid
$C =$ concentration
$\alpha=$ degree of ionisation of weak acid $(HA)$
For very weak acid $\alpha<<<<1$ or $1-\alpha \simeq 1$
$K_{a}=C \alpha^{2}$
or $\alpha=\sqrt{\frac{K_{a}}{C}}$
$C=0.1\, M, \alpha=1.34\% =0.0134$
$K_{a}=0.1 \times\left(1.34 \times 10^{-2}\right)^{2}$
$=1.79 \times 10^{-5}$