A 150.0 ml sample of 0.100 M HNO2 is titrated with 0.250 M KOH. The Ka for HNO2 is 4.57*10^-4. Calculate the pH at equivalence point

Answer 1

The equivalence point is when the moles of HNO₂ = moles KOH and all of the HNO₂ is converted to KNO₂. The pH will then be determined by the hydrolysis of the KNO₂.

HNO₂ + KOH ==> KNO₂ + H₂O .. balanced equation

First, we will find the final volume of the solution @ the equivalence point

moles HNO₂ present = 150.0 ml x 1 L / 1000 ml x 0.100 mol / L = 0.0150 moles

moles KOH required for equivalence = 0.0150 mols

mls KOH needed = 0.0150 mols x 1 L / 0.250 mols = 0.060 L = 60 mls

Final volume = 60 mls + 150.0 mls = 210 mls = 0.210 L

Final [KNO₂] = 0.0150 mols / 0.210 L = 0.0714 M

Now, we must consider the hydrolysis of KNO₂, or NO₂^- (K⁺ is a spectator ion):

NO₂^- + H₂O ==> HNO₂ + OH^- ... hydrolysis of NO₂^-

NO₂^- is acting as a base in this reaction, so we want to use Kb for NO₂^-

KaKb = Kw = 1x10^-14

Kb = 1x10^-14 / 4.57x10^-4

Kb = 2.19x10^-11

Kb = 2.19x10^-11 = [HNO₂][OH^-] / [NO₂^-]

2.19x10^-11 = (x)(x) / 0.0714 - x (since Kb is so small, we can ignore x in the denominator)

2.19x10^-11 = x² / 0.0714

x² = 1.56x10^-12

x = [OH^-] = 1.25x10^-6 (note: this is very small compared to 0.0174 M so above assumption was valid)

pOH = -log [OH^-] = -log 1.25x10^-6

pOH = 5.90

pH = 14 - pOH

pH = 14 - 5.90

pH = 8.10