Question

In a titration of 44.50 mL of 0.3082 M nitrous acid with 0.3082 M aqueous sodium hydroxide, what is the pH of the solution when 44.50 mL of the base have been added?

Answer 1

we have:

Molarity of HNO2 = 0.3082 M

Volume of HNO2 = 44.5 mL

Molarity of NaOH = 0.3082 M

Volume of NaOH = 44.5 mL

mol of HNO2 = Molarity of HNO2 * Volume of HNO2

mol of HNO2 = 0.3082 M * 44.5 mL = 13.7149 mmol

mol of NaOH = Molarity of NaOH * Volume of NaOH

mol of NaOH = 0.3082 M * 44.5 mL = 13.7149 mmol

We have:

mol of HNO2 = 13.7149 mmol

mol of NaOH = 13.7149 mmol

13.7149 mmol of both will react to form NO2- and H2O

NO2- here is strong base

NO2- formed = 13.7149 mmol

Volume of Solution = 44.5 + 44.5 = 89 mL

Kb of NO2- = Kw/Ka = 1*10^-14/4.5*10^-4 = 2.222*10^-11

concentration ofNO2-,c = 13.7149 mmol/89 mL = 0.1541M

NO2- dissociates as

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

0.1541 0 0

0.1541-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.222*10^-11)*0.1541) = 1.851*10^-6

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

so, x = 1.851*10^-6 M

[OH-] = x = 1.851*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.851*10^-6)

= 5.73

we have below equation to be used:

PH = 14 - pOH

= 14 - 5.73

= 8.27

Answer: 8.27