Question

What is the pH of a 0.124 M aqueous solution of ammonium iodide, NH₄I ?

pH =

This solution is _________ (acidic, basic, or neutral)

Answer 1

   We know that Kb of NH₃ = 1.8 x 10^-5

                       Ka = Kw/Kb

                            = 1.0x 10^-14 / 1.8 x 10^-5

                            = 0.55 x 10^-9

                       Ka = 0.55 x 10^-9

            

      NH₄I     -------------------> NH₄⁺ + I^-

   NH₄⁺ --------------->   NH3 + H+

    0.124 M                     0       0

0.124-X                      X       X

Then,

       Ka = X.X / (0.124-X)

0.55 x 10^-9 = X²/ (0.124-X)

X² + (0.55 x 10^-9)X - 0.0682 x 10^-9 = 0

On solving,

X = 8.25 x 10^-6 M

[H+] = 8.25 x 10^-6 M

pH = -log [H+] = -log [ 8.25 x 10^-6] = 5.08

pH = 5.08

Therefore,

pH of 0.124 M aqueous solution of ammonium iodide, NH₄I = 5.08

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Since pH = 5.08 < 7,

the solution is acidic.