25.00 mL of a solution containing 0.1571 M ammonia (Kb = 1.76 x 10-5) is to be titrated with 0.1134 M HCl. NH3 + HCl → NH4Cl Calculate pH after the addition of 47.09 mL of acid.

Question

25.00 mL of a solution containing 0.1571 M ammonia (K_b = 1.76 x 10^-5) is to be titrated with 0.1134 M HCl.

NH₃ + HCl → NH₄Cl

Calculate pH after the addition of 47.09 mL of acid. (Enter your answer to three significant figures. Ignore the dissociation of water.)

Answer 1

NH₃ + HCl ===> NH₄Cl

Initial moles NH₃ = 0.025 L x 0.1571 mol/L = 0.0039275 moles NH₃

Initial moles HCl = 0.04709 L x 0.1134 mol/L = 0.005340 moles HCl

All of the NH₃ will be converted to NH4+ and some HCl will be left over (in excess)

moles NH₄⁺ formed = 0.0039275

moles HCl left over = 0.005340 - 0.0039275 = 0.001313 moles HCl remaining

Final volume = 25 ml + 47.09 ml = 72.09 ml = 0.07209 L

The major contribution to pH will be the H⁺ from HCl, and can be calculated as follows:

[HCl] = 0.001313 mol/0.07209 L = 0.01821 M

pH = -log [H⁺] = -log 0.01821

pH = 1.74

If you want to include the contribution from the NH₄^+, you can do so as follows:

NH₄⁺ ==> NH₃ + H⁺ where [NH₄⁺] = 0.0039275 mol/0.07209 L = 0.0545 M

Ka = 1x10^-14/1.76x10^-5 = 5.68x10^-10

5.68x10^-10 = (x)(x)/0.0545

x = [H+] = 5.6x10^-6 (compare this to H⁺ from HCl and you see it is minor so doesn't contribute to pH)