A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 150.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

Question

I've been working on this for a while but can't seem to figure out what I am doing wrong, so far, here are my steps below:

moles of NH3: 0.100L * 0.10M = 0.01 moles

moles of HNO3: 0.150L * 0.10M = 0.015 moles ----> excess: 0.015 - 0.01 = 0.005 moles of HNO3

total V = 0.250 L

M of HNO3 = 0.005/0.250 = 0.02 M

1*10^-14 = [H+][0.02] ----> [H+] = 5*10^-13

pH = -log[5*10^-13] = 12.30

however, 12.30 is shown to be incorrect. what did i do wrong?

Answer 1

Almost there... after adding 150 mL of 0.10 M nitric acid (a strong acid), what is the concentration of hydronium H3O+ in the titration mixture? Then pH = - log [H3O+].