Question

What is the pH at the equivalence point in the titration of of a 29.5 mL sample of a 0.460 M aqueous hydroflouric acid solution with a 0.358 M aqueous sodium hydroxide solution?

pH - ___

Answer 1

Ka of HNO2 = 4.0*10^-4

find the volume of NaOH used to reach equivalence point

M(HNO2)*V(HNO2) =M(NaOH)*V(NaOH)

0.46 M *29.5 mL = 0.358M *V(NaOH)

V(NaOH) = 37.905 mL

Given:

M(HNO2) = 0.46 M

V(HNO2) = 29.5 mL

M(NaOH) = 0.358 M

V(NaOH) = 37.905 mL

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

mol(HNO2) = 0.46 M * 29.5 mL = 13.57 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.358 M * 37.905 mL = 13.57 mmol

We have:

mol(HNO2) = 13.57 mmol

mol(NaOH) = 13.57 mmol

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

NO2- here is strong base

NO2- formed = 13.57 mmol

Volume of Solution = 29.5 + 37.905 = 67.405 mL

Kb of NO2- = Kw/Ka = 1*10^-14/4.4*10^-4 = 2.273*10^-11

concentration ofNO2-,c = 13.57 mmol/67.405 mL = 0.2013M

NO2- dissociates as

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

0.2013 0 0

0.2013-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 ((2.273*10^-11)*0.2013) = 2.139*10^-6

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

so, x = 2.139*10^-6 M

[OH-] = x = 2.139*10^-6 M

use:

pOH = -log [OH-]

= -log (2.139*10^-6)

= 5.6698

use:

PH = 14 - pOH

= 14 - 5.6698

= 8.3302

Answer: 8.33