What is the pH after 0.018 mol of Ba(OH)2 are added to 0.82 L of the solution?

Question

A buffer that contains 0.35 M of acid, HY and 0.3 M of its conjugate base Y^-, has a pH of 4.8. What is the pH after 0.018 mol of Ba(OH)₂ are added to 0.82 L of the solution?

The answer is 4.92

Answer 1

From the data provided, you can find the Ka and pKa for the acid, HY.

Using the Henderson Hasselbalch equation pH = pKa + log [salt]/[acid]

4.8 = pKa + log (0.3/0.35)

4.8 = pKa + (-0.067)

pKa = 4.87

Now that we know the pKa of the acid we can answer the rest of the problem:

moles HY initially present = 0.35 mol/L x 0.82 L = 0.287 moles

moles Y- initially present = 0.3 mol/L x 0.82 L = 0.246 moles

moles OH- added = 0.018 mol/L x 0.82 L x 2 mol OH^-/mol Ba(OH)₂ = 0.0295 moles OH- added

Final moles HY = 0.287 mol - 0.0295 mol = 0.2575 moles HY

Final moles Y- = 0.246 mol + 0.0295 = 0.2755 moles Y-

Final [HY] = 0.2575 mol/0.82 L = 0.314 M

Final [Y-] = 0.2755 mol/0.82 L = 0.336 M

pH = pKa + log [Y-]/[HY] = 4.87 + log (0.336/0.314)

pH = 4.87 + 0.029

pH = 4.90 (difference may be due to a rounding error)