Question

Calculate the pH of a 0.40 M solution of NH ₄NO ₃. (K _b(NH ₃) = 1.8 x 10 ^-5)

Answer 1

NH₄NO₃ ( Ammonium nitrate ) is a salt formed by combination of weak base ( NH₄OH ) and strong acid ( HNO₃ ) . Hence its pH must  lie < 7 . As for the salt which consists of weak base and strong acid in it . We know that pH is given by

pH = 1/2 [ pKw - pKb - log c]

. We are given that concentration of weak base solution , c = 0.40 M

and Kb ( NH₃ ) = 1.8 x 10^-5 So , pKb = - log Kb = - log ( 1.8 x 10^-5 ) = 4.74

and at 298.15 K temperature , pKw = 14

So , substituting the values in the pH expression ,then we will get ,

pH = 1/2 [ 14 - 4.74 - log 0.40] = 9.657 / 2 = 4.828

Hence clearly it verifies our answer as pH of this salt solution must lies below than 7 due to presence of strong acid.