Question

What is the percent ionization $\left(\alpha \right)$ of a $0.01M$ $HA$ solution?
(Given that $K_a={10}^{-4}$)

• A. $9.5\text{\%}$
• B. $1\text{\%}$
• C. $10.5\text{\%}$
• D. $10\text{\%}$

Answer 1

Answer: (A)

$9.5\text{\%}$

As we know, $\frac{c{\alpha }^2}{1-\alpha }=K_a$, where $\alpha$ is the degree of dissociation and $K_a$ is the dissociation constant.

Given,

$c=0.01M$ and $K_a={10}^{-4}$.

Therefore, $k_a=\frac{0.01\times {\alpha }^2}{1-\alpha }={10}^{-4}$,

$0.01{\alpha }^2+{10}^{-4}\alpha -{10}^{-4}=0$

or, $\alpha =0.095$.

$\text{\%}$ dissocation is $0.095\times 100=9.5\text{\%}$.