Question

You have 500.0 mL of a buffer solution containing 0.20 M acetic acid and 0.30 M sodium acetate. What will the pH of the solution be after the addition of 20.0 mL of 1.00 M NaOH solution. K_a (CH₃COOH) = 1.8x10⁵

Please answer the question using ICE tables, I am unsure of whether to use 1 ICE table or 2 ICE Tables. Thank you.

Answer 1

CH₃COOH + NaOH <====> CH₃COONa + H₂O

ICE table

                                      [CH₃COOH]                           [NaOH]                              [CH₃COONa]

Initial                               0.1                                           0.02                                        0.15

Change                         -x                                                -x                                               +x

Equilibrium                 0.1 - x                                0.02 - x 0.15 + x        

0.5 L of 0.20 M acetic acid = 0.1 moles acetic acid
0.5 L of 0.30 M sodium acetate = 0.15 moles sodium acetate
The addition of 0.02 L of 1.00 M NaOH (0.02 moles of NaOH) will convert 0.02 moles of acetic acid to sodium acetate.

The new solution will contain 0.08 moles of acetic acid (0.1 - 0.02 = 0.08) and 0.17 moles of sodium acetate (0.15 + 0.02 = 0.17)
pH before addition of NaOH-
pH = pKa + log (Ac-/HAc)

pH = 4.74 + log(0.15/0.1) = 4.74 + 0.18 = 4.92

pH after the addition of NaOH-
pH = 4.74 + log(0.17/0.08) = 4.74 + 0.33 = 5.07