Calculate the pH of a 1.33 M aqueous solution of ammonium chloride (NH4Cl).
(For ammonia, NH3, Kb = 1.80×10-5.)
NH4Cl hydrolyzes. Actually, only the NH4^+ hydrolyzes this way.
...................NH4^+ + H2O --> NH3 + H3O^+
I....................1.33......................0............0
C....................-x.........................x.............x
E................1.33-x.......................x............x
Ka for NH4^+ = (Kw/Kb for NH3) = (H3O^+)(NH3)/(NH4+)
Ka = (1E-14/1.80E-5) = (x)(x)/(1.33 - x)
Solve for x = (H3O^+) and convert to pH. Post your work if you get stuck.