Question

In a titration of 41.42 mL of 0.3164 M nitrous acid with 0.3164 M aqueous sodium hydroxide, what is the pH of the solution when 41.42 mL of the base have been added?

Answer 1

HNO2 + NaOH -----------> NaNO2 + H2O

millimoles of HNO2 = 41.42 x 0.3164 = 13.1053

millimoles of NaOH added = 41.42 x 0.3164 = 13.1053

means all acid convert to salt.

[salt] = 13.1053 / 82.84 = 0.1582 M

for salt of weak acid

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

pKa of HNO2 = 3.15 standard value

pH = 1/2 [14 + 3.15 + log 0.1582]

pH = 1/2 [16.35]

pH = 8.18