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?

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 3.2

please explain your answer, thank you :)

Answer 1

Prior to any addition, we have a known pH with known concentrations of weak acid and conjugate base. We can thus calculate the pKa as follows using the Henderson Hasselbalch equation:

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

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

pKa = 3.032

Addition of a base, Ba(OH)₂ to this weak acid buffer solution will result in reaction with the acid, HY to produce the conjugate base, Y^-, as follows:

2HY + Ba(OH)₂ ===> BaY₂ + H₂O

First we should find moles of acid and conjugate base originally present:

moles acid = 0.3 moles/L x 0.68 L = 0.204 moles acid

moles salt = 0.32 moles/L x 0.68 L = 0.2176 moles salt

0.02 mol Ba(OH)₂ will react with 0.04 moles of HY to produce 0.04 moles of BaY₂ (the salt)

Final moles HY = 0.204 - 0.04 = 0.164 moles HY

Final moles Y^- = 0.2176 - 0.04 = 0.2576 moles Y^-

Final [HY] = 0.164 mol/0.68 L = 0.2412 M

Final [Y^-] = 0.2576 mol/0.68 L = 0.3788 M

Now we can plug these values back into the HH equation and solve for pH:

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

pH = 3.032 + log [0.3788]/[0.2412] = 3.032 + 0.1960 = 3.228

pH = 3.2