Question

40.0 mL of 0.10 M HA(aq) was titrated against 0.10 M NaOH(aq). HA (aq) is a weak acid with K_aof 6.3 x 10^-6.

            (a)        Calculate the initial pH of the 0.10 M HA (i.e. before adding NaOH).

            (b)        Calculate the pH at the equivalence point.

Answer 1

(a)        Calculate the initial pH of the 0.10 M HA (i.e. before adding NaOH).

6.3 x 10^-6

HA(aq) H⁺(aq) + A^-(aq)

From ICE table we get

Ka = x²/0.1-x    (x in denominator can be ignored)

6.3 x 10^-6 = x²/0.1

x = 7.93 x 10^-4 This is the H⁺ concentration

pH = -log H⁺

pH = 3.1

Since we will need 1:1 HA and NaOH to reach equivalnec point we will need 40mL of 0.1M NaOH

at the equivalence point all HA has been converetd to A-. So A- concentration would be 0.04 x 0.1 = 0.004 moles in 80 mL which is 0.05 M. A- in water displays the following equilibrium

                      A^-(aq) + H₂O HA + OH^-(aq)

initial                0.05                  0        0

change              -x                    +x       +x

equilibrium        0.05-x               x          x

Kb = Kw/Ka

Kb = 1 x 10^-14/6.3 x 10^-6 = 1.58 x 10^-9

Kb = x²/0.05-x

1.58 x 10^-9 = x²/0.05-x

7.9 x 10^-11 - 1.58 x 10^-9 x= x²

x² + 1.58 x 10^-9 x - 7.9 x 10^-11 =0 solving this quadratic equation we get

x = 8.88 x 10^-6 This is OH- concentration

pOH = - log OH^-

pOH = -log 8.88 x 10^-6

pOH = 5.05

pH = 14-pOH

pH = 14-5.05

pH = 8.95