Question

Calculate pH for a weak acid/strong base titration. Determine the pH during the titration of 59.3 mlL of 0.335 M nitrous acid (K-4.5x104) by 0.335 M KOH at the following points. (a) Before the addition of any KOH (b) After the addition of 15.0 mL of KOH (c) At the half-equivalence point (the titration midpoint) (d) At the equivalence point (e) After the addition of 89.0 mL of KOH

Answer 1

a) before the addition of any KOH

Ka = 4.5 x 10^-4

pKa = -log Ka

pKa = 3.35

pH = 1/ 2[pKa - log C]

pH = 1/2 [3.35 -log 0.335]

pH = 1.91

b)

millimoles = 59.3 x 0.335 = 19.9

millimoles of KOH = 0.335 x 15 = 5.025

HNO2    + KOH ---------------> KNO2 + H2O

19.9          5.025                      0

14.875                                   5.025

pH = 3.35 + log (5.025 / 14.875)

pH = 2.88

c) here pH = pKa

pH = 3.35

d)

salt only remains = C = 59.3 x 0.335 / 59.3+59.3

                                    = 0.1675 M

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/2 [3.35 + log 0.1675]

pH = 8.29

e)

pH = 12.83