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
           Â
                               Â
HNO2Â Â Â Â Â Â Â
-->Â Â Â Â Â Â Â Â H+ +
NO2-
Initially                              Â
0.0782Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â Â
0.0017
Let x dissociates then
At
equilibrium                        0.0782
-
x           Â
x           Â
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%
