Calculate the pH of the solution?
Part A -
Calculate the pH of a solution prepared by dissolving 0.750 mol of NH3 and 0.250 mol of NH4Cl in water sufficient to yield 1.00 L of solution. The Kb of ammonia is 1.8 x 10^(-5).
pOH = pKb + log[(conj. acid) / (base)]
pOH = -log(1.8*10^(-5)) + log[0.750 / 0.250]
pOH = 5.22
pH = 14 - pOH = 14 - 5.22 = 9.78
Answer 1: pH = 9.74
Part B -
Calculate the pH of the above solution after addition of 0.25 mol of HCl, assuming there is no change in volume.
Answer 2: 9.56
This part I don't understand how to do. Could someone show or explain it to me? I tried assuming that HCl completely dissociates; I tried writing it with an initial concentration for each: [NH4] = 0.250, [NH3] = 0.750, [H+] = 0.250. But this didn't give me any sort of definite answer, and I was unsure how to proceed. I have a test on Monday, so it would be great if someone could help me on this today. Thanks in advance.