Question

A 25.0 mL sample of 0.150 M hydrazoic acid (HN3) is titrated with a 0.150 M NaOH solution. Calculate the pH at equivalence and the pH after 26.0 mL of base is added. The Ka of hydrazoic acid is 1.9x10^-5.

Answer 1

i) pH at equivalence point

Reaction between HN3 and NaOH is

HN3 + NaOH ------> NaN3 + H2O

this is 1:1 molar reaction

So, 25ml of 0.150M HN3 react with 25ml of 0.150M NaOH

Equivalence point is 25ml

at equivalence point all the HN3 is converted into NaN3

Total Volume = 25ml + 25ml = 50ml

Dilution of NaN3 concentration = 2 time

Concentration of NaN3 = 0.150M/2 = 0.075M

N3^- is partly hydrolysed by water

N3^- + H2O <-------> HN3 + OH-

Kb = [HN3][OH-]/[N3^-] = 1.9×10^-5

Kb = Kw/Ka

= 1.00 ×10^-14/1.9 ×10^-5

= 5.26×10^-10

at equillibrium

[HN3] = X

[OH-] = X

[ N3^-] = 0.075 - X

X^2/(0.075 - X) = 5.26 ×10^-10

we can assume 0.075 - X = 0.075 because x is small value

X^2/0.075 = 5.26 ×10^-10

X^2 = 3.95 ×10^-11

X= 6.28×10^-6

[OH-] = 6.28×10^-6M

pOH = 5.20

pH = 14 - pOH

= 14 - 5.20

= 8.80

ii) pH at 26ml of NaOH solution addition

out of 26 ml , 25ml used for neutralization of HN3

excecess volume of NaOH = 1ml

No of mole of excess OH- added = (0.150mole/1000ml)×1ml = 0.00015

Total volume = 51ml

[OH-] = (0.00015mol/51ml)×1000ml = 0.00294M

pOH = 2.53

pH = 14 - 2.53

= 11.47