Question

Please solve

What is the percent dissociation of HNO₂ when 0.074 g of sodium nitrite is added to 115.0 mL of a 0.068 M HNO₂ solution? K_a for HNO₂ is 4.0 × 10^–4. (4%)

Answer 1

     NaNO2 --> Na+ + NO2-

Mass of sodium nitrate added = 0.074 g

so moes of sodium nitrate = 69

Moles of soium nitrate = 0.074 /69 =0.0017 moles

                                              HNO2        -->         H+ + NO2-

Initially                                0.0782                   0             0.0017

Let x dissociates then

At equilibrium                        0.0782 - x             x             0.0017+x

Ka = 4X10^-4 = [NO2-[H+] / []HNO2] = [0.0017+x][x] / [0.0782-x] X 0.115

x <<1 so ignore it in denominator

0.00004 X 0.0782 X 0.115 = 0.0017x + x^2

on solving

x = 0.000189

so % dissciation = 0.000189 X 100 / 0.0782 = 0.24%