Question

Calculate the pH of a 0.500 L solution of 0.00375 moles of nitrous acid in water, and the percent ionization of the solution.

pH = ?

percent ionization = ?

Answer 1

note : please check HNO2 , Ka = 5.6 x 10^-4 . this is the value i am using from standard table. before start this problem you need to check whether this value match with your site value or not.

molarity = 0.00375 / 0.500 = 7.5 x 10^-3

HNO2 <-------------------------------> H+ + NO2-

0.0075                                       0             0

-x                                              +x            + x

0.0075 -x                                  x               x

Ka = [H+][NO2-]/[HNO2]

5.6 x 10^-4 = x^2 / 0.0075 -x

x^2 + 5.6 x 10^-4 x - 4.2 x 10^-6 = 0

x = 1.79 x 10^-3

[H+] = 1.79 x 10^-3 M

pH = - log [H+]

pH = 2.75

percent ionization = 1.79 x 10^-3 x 100 / 0.0075

                            = 23.9 %