A buffer that contains 0.11 M of an acid, HA and 0.49 M of its conjugate base

Question

A buffer that contains 0.11 M of an acid, HA and 0.49 M of its conjugate base A^-, has a pH of 4.24. What is the pH after 0.022 mol of HCl are added to 0.53 L of the solution?

The answer is 4.06

Answer 1

When you add and acid, HCl to this buffer, the acid will react with the conjugate base, A^- to produce HA.

A^- + H⁺ ==> HA so the concentration of HA will increase and the concentration of A^- will decrease.

Initial moles HA = 0.11 mol/L x 0.53 L = 0.0583 moles HA

initial moles A^- = 0.49 mol/L x 0.53 L = 0.2597 moles A^-

Final moles HA = 0.0583 mol + 0.022 mol = 0.0803 moles HA

Final moles A^- = 0.2597 mol - 0.022 mol = 0.2377 moles A^-

Final [A-] = 0.2377 mol/0.53 L = 0.448 M

Final [HA] = 0.0803 mol/0.53 L = 0.152 M

Now we can use the Henderson Hasselbalch equation to find the pH. But to do this, we need to know the Ka or pKa of the weak acid, HA. We can find that because the pH of the original buffer is given to us as 4.24.

Henderson Hasselbalch equation:

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

4.24 = pKa + log (0.49/0.11)

4.24 = pKa + 0.65

pKa = 3.59

Going back to our original problem...

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

pH = 3.59 + log (0.448/0.152)

pH = 3.59 + 0.47

pH = 4.06