Question

Determine the pH of a 0.0100 M solution of NH₄NO₃.

Answer 1

NH4NO3 is a strong electrolyte :
NH4NO3 -----> NH4⁺ + NO₃^-
NH4⁺ <----> NH3 + H+
Initial concentration
0.0100 M
at equilibrium

NH4⁺    <---->    NH3 +    H+
0.01-x              x             x
The constant of this reaction is K = Kw/ Kb
K = 1 x 10^-14 / 1.80 * 10^-5 = 5.56 * 10^-10
5.56 x 10^-10 = x^2 / 0.01 -x
x =2.36 * 10^-6 M
pH = - log (2.36 * 10^-6) = 5.63