Question

BUFFER PROBLEMS

3. Start with either the conjugate acid or base and add a strong base or strong acid (conjugate acid plus strong base forms the conjugate base and water; conjugate base plus strong acid forms the conjugate acid): What is the pH of 0.5 L of a 0.1 M acetic acid solution to which 0.73 g of NaOH are added? The pKa of acetic acid is 4.76.

Answer 1

moles of CH3COOH = molarity x volume

                                = 0.1 x 0.5

                                = 0.05

moles of NaOH = mass / molar mass

                         = 0.73 / 40

                         = 0.01825

CH3COOH + NaOH ---------------------> CH3COONa + H2O

0.05              0.01825                                0                 0 -----------------> initial

-0.001825 - 0.01825                     + 0.01825          + 0.01825   ---------------> change

0.03175        0                                    0.01825             0.01825      --------------->equilibrium

pH = pKa + log [CH3COONa / CH3COOH]

pH = 4.76 + log ( 0.01825 / 0.03175)

pH = 4.52