A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Find PH!!!

Question

A 100.0 mL sample of 0.10 M NH₃ is titrated with 0.10 M HNO₃

K_b NH₃ = 1.8 x 10^-5

a) pH of the solution before addition of any HNO_-

b) pH of solution after addition of the 50.0 mL of HNO₃

Answer 1

NH₃ + HNO₃ ==> NH₄⁺ + NO₃^- ... balanced equation for the reaction between NH₃ and HNO₃

This is a titration of a weak base (NH₃) with a strong acid (HNO₃).

Initial moles of NH₃ = 100.0 ml x 1L/1000 ml x 0.10 mol/L = 0.010 mols

(a) pH before addition of HNO₃:

NH₃ + H₂O ==> NH₄⁺ + OH^-

Kb = 1.8x10^-5 = [NH4⁺][OH^-] / [NH₃] = (x)(x) / 0.10 - x assume x is small relative to 0.10 and ignore it

1.8x10^-5 = x²/0.10

x² =1.8x10^-6

x = [OH^-] = 0.00134 M

pOH = -log [OH^-] = 2.87

pH = 14 - 2.87

pH = 11.13

(b) pH after addition of 50.0 ml of 0.10 M HNO3

moles HNO3 added = 50.0 ml x 1L/1000 ml x 0.10 mol/L = 0.005 moles

NH₃ + HNO₃ ==> NH₄⁺ + NO₃^-

0.010...0.005..........0..........0.........Initial

-0.005...-0.005.....+0.005..0.005....Change

0.005.....0.............0.005..0.005.....Equilibrium

After addition of the HNO₃, we have 0.005 mols NH₃ and 0.005 mols NH₄⁺. So we have equal amounts of the weak base (NH₃) and the conjugate acid (NH₄⁺), and thus we have a buffer solution. Therefore, according the the Henderson Hasselbalch equation, the pOH = pKb.

pKb = -log Kb = -log 1.8x10^-5 = 4.74

pOH = pKb = 4.74

pH = 14 - 4.74

pH = 9.26