Question

Ignoring activity, how do you determine the equilibrium pH of a solution containing 10^-3 M NH4CL (pKa = 9.25) using algebraic techniques?

A detailed response please.. not just some expression..

Answer 1

NH4+ concentration = 10^-3 M

NH4+ + H2O <-------------------------> NH4OH + H+

10^-3 0 0 -------------------------> initial

-x +x +x -------------------->chnaged

10^-3-x x x ------------------------> equilibrium

Ka = [NH4+][H+]/[NH4+]

5.62 x 10^-10 = x^2 / 10^-3 -x (Ka = 10^-pKa = 10^-9.25 = 5.62 x 10^-10 )

x^2 + 5.62 x 10^-10 x - 5.62 x 10^-13 = 0

by solving this

x = 7.49 x 10^-7 M

x = [H+] = 7.49 x 10^-7 M

pH = -log [H+]

pH = -log (7.49 x 10^-7)

pH = 6.13