What is the pH after 0.022 mol of HCl are added to 0.53 L of the solution?

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?

Answer 1

pH = pKa + log [A^-]/[HA]

4.24 = pKa + log (0.49/0.11) = pKa + 0.65

pKa = 3.59

Since acid, HCl, is being added, it will react with the conjugate base thereby reducing its concentration and increasing the concentration of the weak acid.

A^- + H⁺ ==> HA

Moles H⁺ added = 0.022 moles

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

moles A^- = 0.49 mol/L x 0.53 L = 0.260 moles

After addition of 0.022 moles H⁺.....

[HA] = 0.0583 mol + 0.022 moles = 0.0803 mol/0.53 L = 0.152 M

[A^-] = 0.260 mol -0.022 mol = 0.238 mol/0.53 L = 0.449M

pH = pKa + log (A^-/HA)

pH = 3.59 + log (0.449/0.152)

pH = 3.12