What is the concentration/pH of ammonia after titration?

Question

40.00 mL of NH_3(aq) are titrated with 25.00 mL of 0.50 mol/L HCl_(aq). K_b = 1.8x10^-5

a) What is the concentration of the ammonia solution? What is the pH at the beginning?

b) What is the pH at the equivalence point?

Answer 1

NH3(aq) + HCl(aq) ==> NH4Cl(aq)

moles HCl used = 25.00 ml x 1 L/1000 ml x 0.50 mol/L = 0.0125 moles HCl (assuming this is equivalence)

moles NH3 present in 40.00 ml = 0.0125 moles

Initial [NH3] = 0.0125 moles / 0.0400 L = 0.3125 M

NH₃(aq) + H₂O(l) ==> NH₄⁺(aq) + OH^-(aq)

Kb = 1.8x10^-5 = [NH₄⁺][OH^-] / [NH₃]

1.8x10^-5 = (x)(x) / 0.3125

x2 = 5.6x10^-6

x = 2.4x10^-3 M = [OH-] in initial solution

pOH = -log 2.4x10-3 = 2.6

pH = 14 - pOH

pH = 11.4

After equivalence, all NH3 is converted to NH₄⁺ so there will be 0.0125 moles NH₄⁺ in a total volume of 65 ml (0.065L). [NH₄⁺] = 0.0125 mol / 0.065 L = 0.192 M

To find pH at equivalence, we look at the hydrolysis of NH₄⁺ as follows:

NH₄⁺ + H2O ==> NH3 + H3O⁺ (NH₄⁺ acts as an acid)

Ka = [NH3][H3O⁺] / [NH₄⁺] and Ka = 1x10^-14 / Kb = 1x10^-14 / 1.8x10^-5 = 5.6x10^-10

5.6x10^-10 = (x)(x) / 0.192

x² = 1.08x10^-10

x = 1.04x10^-5 = H3O⁺

pH = -log 1.04x10^-5

pH = 4.98 @ equivalence