25.00 mL of a solution containing 0.1571 M ammonia (Kb = 1.76 x 10-5) is 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

Seems like this was answered previously, so this will be abbreviated. If you don't understand or disagree with any part of it, you should mention so in the "comment" section.

NH₃ + HCl → NH₄Cl ... This is a neutralization reaction of an base + acid

moles NH3 present = 0.025 L x 0.1571 mol/L = 3.928x10^-3 moles

moles HCl present = 0.04709 L x 0.1134 mol/L = 5.340x10^-3 moles

Since these 2 react in a 1:1 mol ratio, the HCl is present in excess

moles HCl in excess = 5.340x10^-3 - 3.928x10^-3 = 1.412x10^-3 moles

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

Final [HCl] = 1.412x10^-3 mol/0.07209 L = 0.01959 M

pH = -log 0.01959

pH = 1.708

The contribution of H+ from the NH4+ that is formed is negligible compared to that provided by the HCl so can be ignored in calculating the pH. If you must include it, see my previous answer to this same question.