`HNO_2 harr H^++NO_2^-`

Since `HNO_2` is not a strong acid it's ionization is not complete. So in the final mixture we have both `H^+` and `HNO_2` .

We know that;

`P_H = -log[H^+]`

`1.83 = -log[H^+]`

`[H^+] = 10^(-1.83)`

`[H^+] = 0.0148M`

*So the concentration of `H^+` is 0.0148M.*

...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

`HNO_2 harr H^++NO_2^-`

Since `HNO_2` is not a strong acid it's ionization is not complete. So in the final mixture we have both `H^+` and `HNO_2` .

We know that;

`P_H = -log[H^+]`

`1.83 = -log[H^+]`

`[H^+] = 10^(-1.83)`

`[H^+] = 0.0148M`

*So the concentration of `H^+` is 0.0148M.*

We had 0.495M `HNO_2` solution. But only 0.0148M has ionized.

% of ionization `= (0.0148/0.495)xx100% = 29.9%`

*So the percentage ionization of `HNO_2` is 29.9%.*

**Further Reading**