Calculate the pH of a 0.190 M solution of NaNO2?

Answer 1

Ka (HNO2) = 4.5*10^-4, so Kb (NO2-) = 2.22*10^-11

Kb = [HNO2][OH-]/[NO2-]
2.22*10^-11 = x^2/0.190-x

x <<< 0.190, so it will drop out of the denominator.

2.22*10^-11 = x^2/0.190
4.22*10^-12 = x^2
x = [OH-] = 2.05*10^-6
[H+] = Kw/[OH-] = 1*10^-14/2.05*10^-6 = 4.88*10^-9
pH = 8.31

Answer 2

Wow, orange, thanks for copying my answer from the other two times this question was asked.

So, plan to make a career out of passing off someone else's work as your own?