Question

A monoprotic weak acid, HA , dissociates in water according to the reaction:

Answer 1

Answer: The $pKa$ of the acid is 6.09

Explanation:

For the given chemical reaction:

$HA(aq.)\rightleftharpoons H^+(aq.)+A^-(aq.)$

The expression of equilibrium constant [tex[(K_a)[/tex] for the above equation follows:

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

We are given:

$[HA]_{eq}=0.200M$

$[H^+]_{eq}=4.00\times 10^{-4}M$

$[A^-]_{eq}=4.00\times 10^{-4}M$

Putting values in above expression, we get:

$K_a=\frac{(4.00\times 10^{-4})\times (4.00\times 10^{-4}}{0.200}\\\\K_a=8.0\times 10^{-7}0$

p-function is defined as the negative logarithm of any concentration.

$pKa=-\log(K_a)$

So,

$pKa=-\log(8.0\times 10^{-7})\\\\pKa=6.09$

Hence, the $pKa$ of the acid is 6.09