Question

A 0.100 mole quantity of a monoprotic acid HA is added to 1.00 L of pure water. When equilibrium is reached, the pH of the solution is 3.75. What is the value of Ka for the acid HA

Answer 1

Answer:

$3.167\times 10^{-7}$ is the value of $K_a$for the acid HA.

Explanation:

The pH of the solution = 3.75

The hydrogen ion concentration :

$pH=-log[H^+]$

$3.75=-\log[H^+]$

$[H^+]=10^{-3.75}=0.0001778 M$

Concentration of weak acid = 0.100 mole/L

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

Initually

0.100 M    0   0

At equilibrium

(0.100-x)M     x    x

The expression of dissociation constant can be given as:

$K_a=\frac{x\times x}{(0.100-x)}$

Here the value of x = $[H^+]=0.0001778 M$

$K_a=\frac{0.0001778 M\times 0.0001778 M}{(0.100-0.0001778)M }$

$K_a=3.167\times 10^{-7}$

$3.167\times 10^{-7}$ is the value of $K_a$for the acid HA.