Question

Find the pH of the equivalence point and the volume (mL) of 0.0358 M KOH needed to reach the equivalence point in the titration of 23.4 mL of 0.0390 M HNO2.

Answer 1

M1V1 = M2V2

0.0358*V1 = 0.0390*23.4

V1 = 25.5 ml

volume of 0.0358 KOH is 25.5 ml

..........................................................................................

no.of moles of HNO2 = 0.9126

Conc. of NO2^-1 = 0.9126/(0.0234+0.0255) = 18.66 M

.........................NO2^- + H2O ............> HNO2 + OH-

before eq.     18.66 .............................0...........0

at equilibrium (18.66-X)....................X.............X

   Kb   = x^2/(18.66-x)

Ka of HNO2 = 5.1*10^-4

Kb = Kw/Ka = 1.0*10^-14 / 5.1*10^-4 = 1.96*10^-11

1.96*10^-11 = x^2/(18.66-x)

x^2 + 1.96*10^-11x - 3.66*10^-10 = 0

x = 1.913*10^-5

[OH-] = 1.913*10^-5 M

[H3O+] = 1*10^-14/ 1.913*10^-5 = 5.23*10^-10

Ph = -log[H3O+] = -log[ 5.23*10^-10] =

Ph = 9.28