Question

What is the pH of the solution that results from the combination of 50.00mL of 0.0200M HOCl and 25.00mL of 0.0200M KOH? Ka= 3.162x10^-8 for HOCl

Answer 1

millimoles of HOCl = 50 x 0.02 = 1

millimoles of KOH = 25 x 0.02 = 0.5

Ka = 3.162 x 10^-8

pKa = -log Ka

pKa = 7.50

HOCl + KOH -------------------------> KOCl + H2O

1            0.5                                        0           0

0.5         0.0                                     0.5         0.5

millimoles of HOCl = millimoles of KOCl . so it is half equivalence point.

here pH = pKa

pH = 7.50