Question

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 0.011 M in NH4Cl at 25 °C?

Answer 1

Let α be the dissociation of the weak base(NH³)   where c is the initial concentration.
                            BOH <---> B ⁺ + OH^-

initial conc.            c               0         0

Equb. conc.         c(1-α)        cα      cα

Dissociation constant , K_b = cα x cα / ( c(1-α)                             

              = c α² / (1-α)

In the case of weak bases α is very small so 1-α is taken as 1

So K_b = cα²

   α = √ ( K_b / c )

Given K_b = 1.8x10^-5

          c = concentration = 0.011M

Plug the values we get α = √ ( K_b / c ) = √ ( 1.8x10^-5 / 0.011 ) = 0.040
So the concentration of [OH-] = cα

                                               = 0.011 x 0.040
                                               = 4.45x10 ^-4 M

pOH = - log [OH-]

        = - log(4.45x10 ^-4 )

        = 3.35

So pH = 14 - pOH

          = 14 - 3.35

         = 10.65

Therefore the pH of the solution is 10.65