In: Chemistry
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?
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