Question

1) 30.0 mL of 0.44M hydrazoic acid; (HN3; Ka=2.6 x 10^-5) is titrated with a 0.22 M KOH solutions . What's pH of solution after adding 60.0 mL of base?

2) 25.0mL of 0.800M HClO4 is titrated with a 0.300M KOH solution. What is the H3O+ concentration after 10.0 mL of KOH are added?

Answer 1

KOH will react with HNO3 to form some conjugate base
HN3 + OH- ---> N3- + H2O

moles of N3- formed = moles of KOH added = 0.22 M * 60 mL = 13.2 mmol
moles of HN3 remaining = 0.44M * 30 mL - 13.2 mmol = 0

so finally it will contain N3- only
[N3-] = 13.2 mmol / (30 mL + 60 mL) = 0.147 M

the equilibrium will be set up as :
N3- + H2O ----> HN3 + OH-
0.147-x x x

Kb of N3- = 10^-14 / ka
= 10^-14 / (2.6*10^-5)
= 3.85*10^-10

Kb = x*x/ (0.147-x)
since kb is small
Kb = x*x/ (0.147)
3.85*10^-10 = x*x/ (0.147)
x = 7.52*10^-6 M

So,[OH-] = x = 7.52*10^-6 M

pOH = -log [OH-]
= -log (7.52*10^-6)
= 5.12

pH = 14 - pOH
= 14 - 5.12
=8.88
Answer: 8.88

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