Chemistry Buffer pH question

Question

A buffer that contains 0.124 M acid, HY, and 0.222 M of its conjugate base, Y⁻, has a pH of 8.72. What is the pH after 0.0016 mol of Ba(OH)₂ is added to 0.340 L of this solution?

Answer 1

From the data given for the original buffer (before any additions), we can determine the pKa of the weak acid, HY. We do this using the Henderson Hasselbalch equation pH = pKa + log [conj.base[/[acid]

8.72 = pKa + log (0.222/0.124)

pKa = 8.72 - 0.25 = 8.47

Now that we know the pKa for the weak acid, we can use that and the HH equation again to solve for the pH of the solution after the addtion of Ba(OH)2.

Looking at what happens when Ba(OH)2 is added, we see it will react with the weak acid HY as follows:

HY + OH- ==> Y- + H2O

Adding 0.0016 moles Ba(OH)2 = 0.0032 moles of OH- b/c Ba(OH)2 => Ba2+ + 2OH-

Using an ICE table we have...

HY + OH- ==> Y- + H2O

0.124....0.0032.........0.222.............Initial

-0.0032...-0,0032.....+0.0032.........Change

0.1208........0...........0.2252...........Equilibrium

pH = pKa + log [conj.base]/[acid]

pH = 8.47 + log (0.2252/0.1208)

pH = 8.47 + 0.27

pH = 8.74