Question

If $0.1M$ 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 $\alpha =\sqrt{\frac{K_a}{C}}$
$C=0.1M,\alpha =1.34\%=0.0134$
$K_a=0.1\times {\left(1.34\times 10^{-2}\right)}^2$
$=1.79\times 10^{-5}$