Calculate the percent ionization of nitrous acid in a solution that is 0.241 M in nitrous...

Calculate the percent ionization of nitrous acid in a solution that is 0.241 M in nitrous acid (HNO₂) and 0.195 M in potassium nitrite (KNO₂). The acid dissociation constant of nitrous acid is 4.50 ⋅ 10^-4.

Expert Solution

[HNO2] = 0.241 M
[NO2-] = 0.195 M

HNO2    <--------->   H + NO2-
0.241 M                        0      0.195 M    (initial)
0.241-x                         x       0.195+x   (at equilibrium)

Ka = [H+] [NO2-] / [HNO2]
4.5*10^-4 = x * (0.195+x) / (0.241-x)
Since Ka is small, x will be small and it can be ignored as compared to 0.241 and 0.195
So, above expression becomes,
4.5*10^-4 = x * (0.195) / (0.241)
x = 5.56*10^-4 M

Percent dissociation = x*100 / 0.241
=(5.56*10^-4)*100 / 0.241
= 0.23 %
Answer: 0.23 %

queen_honey_blossom answered 2 years ago

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