Question

What is the pH for a 400 ml solution of 1.6 M Benzoic Acid?

If you added 75 grams of conjugate base (NaC6H5CO2) to the solution, what would the new pH be?

If you now added 75 ml of 1.35 M NaOH, what would be the new pH?

Answer 1

1) What is the pH for a 400 ml solution of 1.6 M Benzoic Acid?

        moles of Benzoic acid = 1.6 x 400/1000= 0.64

C6H5COOH ----------------------> C6H5COO- + H+

0.64                                                 0                  0   -----------------------------initial

0.64-x                                               x                    x ---------------------------equilibrium

Ka = [C6H5COO-][H+ ]/[C6H5COOH]

Ka = x^2/(0.64-x)

6.46 x 10^-5 = x^2/(0.64-x)

x = 6.39 x 10^-3

x = [H+] = 6.39 x 10^-3

pH = -log[H+] = -log(6.39 x 10^-3)

pH = 2.2

2) after addition of 75 g conjugate base

moles of conjugate base = 75/144

                                       =0.52

Buffer

pKa = -log Ka = -log(6.46 x 10-5) =4.19

For acidic buffer

Henderson-Hasselbalch equation

pH = pKa + log[conjugate base /acid]

pH = 4.19+ log[0.52/0.64]

pH =4.09

3) on addition of 75 ml of 1.35 M NaOH,

no of moles of base = 75 x 1.35 /1000

                                     =0.101

on addition of