In: Chemistry
Determine the pH during the titration of 56.0 mL of 0.389 M hydrofluoric acid (Ka = 7.2×10-4) by 0.389 M NaOH at the following points. (Assume the titration is done at 25 °C.)
(a) Before the addition of any NaOH
(b) After the addition of 14.0 mL of NaOH
(c) At the half-equivalence point (the titration midpoint)
(d) At the equivalence point
(e) After the addition of 84.0 mL of NaOH
pKa = -log Ka
pKa = 3.14
(a) Before the addition of any NaOH
pH = 1/2 [pKa - log C]
pH = 1/2 [3.14 - log 0.389]
pH = 1.78
(b) After the addition of 14.0 mL of NaOH
millimoles of HF = 56.0 x 0.389 = 21.8
millimoles of NaOH = 0.389 x 14 = 5.45
HF + NaOH -------------------------> NaF + H2O
21.8 5.45 0 0
16.4 0 5.45
pH = pKa + log [NaF/ HF]
pH = 3.14 + log (5.45 / 16.4)
pH = 2.66
(c) At the half-equivalence point (the titration midpoint)
here pH = pKa
pH = 3.14
(d) At the equivalence point
salt concentration =- 56 x 0.389 / (56 +56)
= 0.504 M
pH = 7 + 12/ [pKa + logC]
pH = 7 + 1/2 [3.14 + log 0.504]
pH = 8.42
(e) After the addition of 84.0 mL of NaOH
answer : 12.9