In: Chemistry
NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.034 M in NH4Cl at 25 °C?
first find the
Ka of the NH3
Ka x Kb = 1 x 10-14
Ka x 1.8 × 10^–5 = 1 x 10-14
Ka = 1 x 10-14 / 1.8 × 10^–5
Ka = 5.56 x 10-10
NH4Cl (aq) <====> HCl (aq) + NH3 (aq)
I 0.034 0 0
C -x +x +x
E 0.034 -x +x +x
Ka = [NH3] [HCl] / [NH4Cl]
5.56 x 10-10 = [x] [x] / [0.034 - x]
x2 + x 5.56 * 10-10 - 1.8904 x 10-11 = 0
solve the quadratic equation
x = 4.35 x 10-6 = [HCl]
since HCl is strong acid concentration of HCl = concentration of H+
pH = -log[4.35 x 10-6 ] = 5.36