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)
IÂ Â Â Â Â Â 0.034
0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â
-x                           Â
+x                 Â
+x
EÂ Â 0.034
-x                     Â
+x                 Â
+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
