Question

What is the pH at the equivalence point of the titration of 60.0 mL of 0.0100 M HN3 with 0.0400 M LiOH? (Ka = 1.9 x 10−5)

Answer 1

find the volume of LiOH used to reach equivalence point

M(HN3)*V(HN3) =M(LiOH)*V(LiOH)

0.01 M *60.0 mL = 0.04M *V(LiOH)

V(LiOH) = 15 mL

Given:

M(HN3) = 0.01 M

V(HN3) = 60 mL

M(LiOH) = 0.04 M

V(LiOH) = 15 mL

mol(HN3) = M(HN3) * V(HN3)

mol(HN3) = 0.01 M * 60 mL = 0.6 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.04 M * 15 mL = 0.6 mmol

We have:

mol(HN3) = 0.6 mmol

mol(LiOH) = 0.6 mmol

0.6 mmol of both will react to form N3- and H2O

N3- here is strong base

N3- formed = 0.6 mmol

Volume of Solution = 60 + 15 = 75 mL

Kb of N3- = Kw/Ka = 1*10^-14/1.9*10^-5 = 5.263*10^-10

concentration ofN3-,c = 0.6 mmol/75 mL = 0.008M

N3- dissociates as

N3- + H2O -----> HN3 + OH-

0.008 0 0

0.008-x x x

Kb = [HN3][OH-]/[N3-]

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 ((5.263*10^-10)*8*10^-3) = 2.052*10^-6

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

so, x = 2.052*10^-6 M

[OH-] = x = 2.052*10^-6 M

use:

pOH = -log [OH-]

= -log (2.052*10^-6)

= 5.6878

use:

PH = 14 - pOH

= 14 - 5.6878

= 8.3122

Answer: 8.31