What is the pH after 0.02 mol of Ba(OH)2 are added to 0.68 L of the solution?

Question

A buffer that contains 0.3 M of acid, HY and 0.32 M of its conjugate base Y^-, has a pH of 3.06. What is the pH after 0.02 mol of Ba(OH)₂ are added to 0.68 L of the solution?

The answer is 3.23

Answer 1

First, we should find the pKa of the weak acid so that we can later use it in the Henderson Hasselbalch equation. You have posted several other questions of this nature, and it seems that you should be "getting the hang" of it. We won't be there to help you on your test. Having said that, let us find the pKa.

pH = pKa + log [salt]/[acid]

3.06 = pKa + log (0.32/0.30) = pKa + 0.028

pKa = 3.03

Now we can look at the addition of Ba(OH)₂. It will react with the acid to form the conjugate base (salt). So moles of HY will decrease and moles Y- will increase.

moles Ba(OH)₂ = 0.02 moles

moles OH- = 2 x 0.02 = 0.04 moles OH-

Initial moles HY = 0.3 mol/L x 0.68 L = 0.204 moles

Final moles HY = 0.204 - 0.04 = 0.164 M HY

Initial moles Y- = 0.32 mol/L x 0.68 L = 0.2176 moles

Final moles Y- = 0.2176 + 0.04 = 0.258 moles Y-

Final [HY] = 0.164 mol/0.68 L = 0.241 M HY

Final [Y-] = 0.2176 mol/0.68 L = 0.32 M Y-

pH = 3.03 + log (0.32/0.241)

pH = 3.03 + 0.12

pH = 3.15 (disagrees with your given answer. check my math)