Question

Exactly 47.0 mL of 0.2 M HNO₂ are titrated with a 0.2 M NaOH solution. What is the pH at the equivalence point ?

Answer 1

find the volume of KOH used to reach equivalence point
M(HNO2)*V(HNO2) =M(KOH)*V(KOH)
0.2 M *47.0 mL = 0.2M *V(KOH)
V(KOH) = 47 mL
Given:
M(HNO2) = 0.2 M
V(HNO2) = 47 mL
M(KOH) = 0.2 M
V(KOH) = 47 mL


mol(HNO2) = M(HNO2) * V(HNO2)
mol(HNO2) = 0.2 M * 47 mL = 9.4 mmol

mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.2 M * 47 mL = 9.4 mmol


We have:
mol(HNO2) = 9.4 mmol
mol(KOH) = 9.4 mmol

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

NO2- here is strong base
NO2- formed = 9.4 mmol
Volume of Solution = 47 + 47 = 94 mL
Kb of NO2- = Kw/Ka = 1*10^-14/4.5*10^-4 = 2.222*10^-11
concentration ofNO2-,c = 9.4 mmol/94 mL = 0.1M

NO2- dissociates as

NO2-        + H2O   ----->     HNO2 +   OH-
0.1                        0         0
0.1-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 ((2.222*10^-11)*0.1) = 1.491*10^-6

since c is much greater than x, our assumption is correct
so, x = 1.491*10^-6 M



[OH-] = x = 1.491*10^-6 M

use:
pOH = -log [OH-]
= -log (1.491*10^-6)
= 5.8266


use:
PH = 14 - pOH
= 14 - 5.8266
= 8.1734
Answer: 8.17