Question

What is the [h3o+] at equilibrium of a 0.50 m weak acid (ha) solution if the ka of the acid is 4.6 × 10−4?

Answer 1

Answer : The concentration of $H_3O^+$ at equilibrium is, 0.015 M

Solution :

The balanced equilibrium reaction will be,

$HA+H_2O\rightleftharpoons H_3O^++A^-$

The expression for dissociation constant of weak aciod will be,

$k_a=\frac{[H_3O^+]\times [A^-]}{[HA]}$

where,

$k_a$ = dissociation constant of weak acid

Let the concentration of $H_3O^+$ and $A^-$ be 'x'

Now put all the given values in this expression, we get

$4.6\times 10^{-4}=\frac{(x)\times (x)}{0.50}$

$x=0.015M$

The concentration of $H_3O^+$ = $A^-$ = x = 0.015 M

Therefore, the concentration of $H_3O^+$ at equilibrium is, 0.015 M