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* 10^(-4)M`

Step-by-step 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=([H^+][A^-])/([HA])`

We are given:

`K_a=1.0* 10^(-6)`

Putting values in above equation, we get:

`1.0* 10^(-6)=(x* 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* 10^(-4)M`

Hence, the concentration of hydrogen ions is
`4.7* 10^(-4)M`