Question

What is the pH of a solution prepared by mixing 150 mL of 0.1 M HCl with 300 mL of 0.1 M sodium acetate (NaOAc) and diluting the mixture to 1 L? The pKa value of acetic is 4.76

Answer 1

Solve for the final concentration of NaOAc and HCl in the diluted solution.
NaOAc = 0.1 M x (150/1000) = 0.15 M
HCl = 0.1 M x (300/1000) = 0.030 M

...................OAc^- + H^+ ==> HOAc
Initial..........0.03.........0...............0
add........................0.015.................
change.....-0.015..-0.015............+0.015
equil...........0.015.........0.............0.015
Substitute the equilibrium line into the Henderson-Hasslebalch equatiion and solve for pH. The HH equation is pH = pKa + log (base)/(acid). The base is OAc^- and the acid is HOAc.