Question

Determine the pH during the titration of 25.7 mL of 0.304 M hydrofluoric acid (K_a = 7.2×10^-4) by 0.478 M KOH at the following points.

(a) Before the addition of any KOH _____

(b) After the addition of 4.20 mL of KOH _____

(c) At the half-equivalence point (the titration midpoint) _____

(d) At the equivalence point _____

(e) After the addition of 24.5 mL of KOH _____

Answer 1

HF + KOH ----------> KF + H2O

25.7x 0.304= 7.8128 0 0 0 initial mmoles

a) before titration

pH = 1/2 [pKa + logC]

= 1/2 [3.14 + log 0.304]

= 1.311

b) - 4.2 x0.478 - - change

5.8052 0 2.0076 - after reaction

It is a buffer

pH = pKa + log [conjugate base]/[acid]

= 3.14 + log 2.0076/5.8052

= 2.678

c) at the half equivalence point

[conjugate base] = [acid]

thus pH = pKa = 3.14

d) At the equivalene point

0 0 7.8128 - after reaction volume of NaOH = 7.8128/0.478=16.344mL

Thus [salt] = 7.8128/42.04 =0.1858 M

At equivalenc ethe pH is given by

pH = 1/2 [pkw + pKa + log C]

= 1/2[14 + 3.14 + log 0.1858]

= 8.204

e) 0 3.8982 7.8128 - after reaction.

The solution now has a strong base ad salt.

[salt] = 3.8982/50.2

thus pH = 14 - pOh

= 14 - {-log 3.8982/50.2}

= 12.89