`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%.