Question

If the K a of a monoprotic weak acid is 1.0 × 10 − 6 , what is the [ H + ] of a 0.22 M solution of this acid?

Answer 1

Answer: The concentration of hydrogen ions is $4.7\times 10^{-4}M$

Explanation:

We are given:

Concentration of acid = 0.22 M

The chemical equation for the dissociation of monoprotic weak acid follows:

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

Initial:             0.22

At eqllm:       0.22-x    x     x

The expression of $K_a$ for the above equation follows:

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

We are given:

$K_a=1.0\times 10^{-6}$

Putting values in above equation, we get:

$1.0\times 10^{-6}=\frac{x\times x}{(0.22-x)}\\\\x=-0.00047,0.00047$

Neglecting the negative value of 'x', because concentration cannot be negative

So, equilibrium concentration of hydrogen ions = x = 0.00047 M = $4.7\times 10^{-4}M$

Hence, the concentration of hydrogen ions is $4.7\times 10^{-4}M$