Question

HNO₂ (aq) + NaOH (aq) → NaNO₂ (aq) + H₂O (l)                          K_a of HNO₂ = 4.5 x 10^-4

H₂O (l) + NO₂^- (aq) ↔ HNO₂ (aq) + OH^- (aq)                                  K_b of NO₂^- = 2.2 x 10^-11

Exactly 100 mL of 0.10 M nitrous acid (HNO₂) are titrated with a 0.10 M NaOH solution. Calculate the pH for:

A:The initial solution

B:At the half-equivalence point

C:The point at which 80 mL of the base has been added

D:The equivalence point

The point at which 105 mL of the base has been added

Answer 1

K_a of HNO₂ = 4.5 x 10^-4

pKa = 3.35

A:)The initial solution

pH = 1/2 (pKa - log C)

     = 1/2 (3.35 - log 0.1)

    = 2.17

pH = 2.17

B) At the half-equivalence point

At half - equivalence point pH = pKa

pH = 3.35

C):The point at which 80 mL of the base has been added

millimoles of NaOH = 80 x 0.1 = 8

millimoles of HNO2 = 100 x 0.1 = 10

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

10              8                                 0            0

2                0                                 8

pH = pKa + log [salt / acid]

     = 3.35 + log [8/2]

pH = 3.95

D:)The equivalence point :

here salt only remains.

salt concentration = 10 / (100 + 100) = 0.05

pH = 7 + 1/2 (pKa + log C)

     = 7 + 1/2 (3.35 + log 0.05)

pH = 8.02

E)The point at which 105 mL of the base has been added

millimoles of NaOH = 105 x 0.1 = 10.5

here NaOH remins = 0.5 / (100 + 105) = 2.44 x 10^-3

pOH = -log (2.44 x 10^-3 ) = 2.61

pH = 11.39