Question

Calculate the change in pH when 67.0 mL of a 0.700 M solution of NaOH is added to 1.00 L of a solution that is 1.00 M  in sodium acetate and 1.00 M  in acetic acid.

Answer 1

initial pH = pKa + log [CH3COONa / CH3COONa]

pH = 4.74 + log (1 /1))

pH = 4.74

moles of NaOH = 67 x 0.7 / 1000 = 0.0469 = C

new pH :

pH = pKa + log [CH3COONa + C / CH3COONa -C]

pH = 4.74 + log (1 +0.0469 /1 -0.0469)

pH   = 4.78

change in pH = final pH - initial pH

                       = 4.78 - 4.74

                     = 0.041