Question

A 42.0 mL sample of 0.120 M HNO2 is titrated with 0.214 M KOH. (Ka for HNO2 is 4.57×10⁻⁴.)

Determine the pH at the equivalence point for the titration of HNO2 and KOH.

Answer 1

find the volume of KOH used to reach equivalence point

M(HNO2)*V(HNO2) =M(KOH)*V(KOH)

0.12 M *42.0 mL = 0.214M *V(KOH)

V(KOH) = 23.5514 mL

Given:

M(HNO2) = 0.12 M

V(HNO2) = 42 mL

M(KOH) = 0.214 M

V(KOH) = 23.5514 mL

mol(HNO2) = M(HNO2) * V(HNO2)

mol(HNO2) = 0.12 M * 42 mL = 5.04 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.214 M * 23.5514 mL = 5.04 mmol

We have:

mol(HNO2) = 5.04 mmol

mol(KOH) = 5.04 mmol

5.04 mmol of both will react to form NO2- and H2O

NO2- here is strong base

NO2- formed = 5.04 mmol

Volume of Solution = 42 + 23.5514 = 65.5514 mL

Kb of NO2- = Kw/Ka = 1*10^-14/4.57*10^-4 = 2.188*10^-11

concentration ofNO2-,c = 5.04 mmol/65.5514 mL = 0.0769M

NO2- dissociates as

NO2- + H2O -----> HNO2 + OH-

0.0769 0 0

0.0769-x x x

Kb = [HNO2][OH-]/[NO2-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.188*10^-11)*7.689*10^-2) = 1.297*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.297*10^-6 M

[OH-] = x = 1.297*10^-6 M

use:

pOH = -log [OH-]

= -log (1.297*10^-6)

= 5.887

use:

PH = 14 - pOH

= 14 - 5.887

= 8.113

Answer: 8.11