Question

If degree of ionization is 0.01 of decimolar solution of weak acid ha, then pka of acid is

Answer 1

If level of ionization is 0.01 of deci-molar solution of weak acid i . e . Ha, at that point pKa of acid will be 5 .

As ,  

C a ^ 2 = Ka = 0.1 x ( 0.01 ) ^ 2 = 10 ^ -5

So, the pKa will be 5.

Answer 2

$p_{K_a}$ of acid $HA$ is 4.96

Explanation:

$p_{K_a}$ of $HA$ = ?

The dissociation reaction for HA is given by:

$HA\rightarrow H^++A^-$

c                   0             0

$c-c\alpha$       $c\alpha$          $c\alpha$

So dissociation constant $K_a$ will be:

$K_a=\frac{[H^+][OH^-]}{[HA]}$

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Given :

c= concentration = 0.1 M

$\alpha$ = degree of dissociation = 0.01

Putting in the values we get:

$K_a=\frac{(0.1\times 0.01)^2}{(0.1-0.1\times 0.01}$

$K_a=1.1\times 10^{-5}$

$pK_a=-log[K_a]$

$pK_a=-log(1.1\times 10^{-5})=4.96$

Learn More about dissociation of acids

https://brainly.com/question/14287135

https://brainly.com/question/8862891