Calculate the pH of a solution prepared by adding 20.0 mL of 0.100 M HCl

Question

Calculate the pH of a solution prepared by adding 20.0 mL of 0.100 M HCl to 80.0 mL of a buffer that is comprised of 0.25 M NH₃ and 0.25 M NH₄Cl. K _b of NH₃ = 1.8 x 10 ^-5.

The correct answer is 9.17

Answer 1

Using the Henderson Hassselbalch equation:

pOH = pKb + log [salt]/[base]

Initial moles NH₃ = 0.08L x 0.25 mol/L = 0.02 moles NH₃

Initial moles NH₄Cl = 0.08L x 0.25 mol/L = 0.02 moles NH₄⁺

Addition of HCl reacts with NH₃ to decrease [NH₃] and increase [NH₄⁺]

NH₃ + H⁺ == NH₄⁺

moles H⁺ added = 0.02L x 0.1 mol/L = 0.002 moles

Final moles NH₃ = 0.02 mol - 0.002 mol = 0.018 moles NH₃

Final moles NH₄⁺ = 0.02 mol + 0.002 mol = 0.022 moles NH₄⁺

Final volume = 20 ml + 80 ml = 100 ml = 0.1 L

Final [NH₄⁺] = 0.022 mol/0.1 L = 0.22 M

Final [NH₃] = 0.018 mol/0.1 L = 0.18 M

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

pOH = pKb + log [NH₄⁺]/[NH₃] = 4.74 + log (0.22/0.18)

pOH = 4.74 + 0.087

pOH = 4.83

pH = 14 - 4.83 = 9.17