Question

You are given 10.00 mL of a solution of an unknown acid. The pH of this solution is exactly 2.95. You determine that the concentration of the unknown acid was 0.1224 M. You also determined that the acid was monoprotic (HA). What is the K_a and pK_a of your unknown acid

Answer 1

Answer:

$1.038\times 10^{-5}$ is dissociation constant and the value of $pK_a$ is 4.98.

Explanation:

The pH of the solution = 2.95 M

$pH=-\log[H^+]$

$2.95=-\log[H^+]$

$[H^+]=10^{-2.95}=0.001122 M$..[1]

Concentration of unknown monoprotic acid = C = 0.1224 M

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

Initially

C                  0         0

At equilibrium

(C-x)              x         x

The expression of a dissociation reaction can be written as;

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

$K_a=\frac{x^2}{(C-x)}$

$[H^+]=x =0.001122 M$  ( from [1])

$K_a=\frac{(0.001122 M)^2}{(0.1224 M-0.001122 M)}$

$K_a=1.038\times 10^{-5}$

The value of $pK_a$ :

$pK_a=-\log[K_a]$

$=-\log[1.038\times 10^{-5}]=4.98$

$1.038\times 10^{-5}$ is dissociation constant and the value of $pK_a$ is 4.98.