Question

A 10.00 mL sample of 0.150 M sodium azide, NaN3, a weak base, is titrated with 0.100 M HCl. Calculate the pH after the addition of the following volumes of acid:

(a) 0.00 mL; (b) 9.00 mL; (c) 15.00 mL; (d) 27.00 mL

The Ka of HN3 is 1.9 x 10–5

Answer 1

before HCl added only NaN3 present

pKb of NaN3 = 1.0 x 10^-14 / 1.9 x 10^-5  = 5.26 x 10^-10  

pKb = -logKb = -log [5.26 x 10^-10] = 8.28

pOH = 1/2 [pKb - logC]

pOH = 1/2 [8.28 - log 0.15]

pOH = 4.55

pH = 14 - 4.55

pH = 9.45

b) millimoles NaN3 = 10 x 0.15 = 1.5

millimoles of HCl added = 9 x 0.1 = 0.9

1.5 - 0.9 = 0.6 millimoles HNaN3 left

0. 9 millimoles HN3 will form

[NaN3] = 0.6 / 19 = 0.031 M

[HN3] = 0.9 / 19 = 0.047 M

pH = pKa + log [NaN3] / [HN3]

pKa = -log Ka = -log[1.9 x 10^-5] = 4.72

pH = 4.72 + log [0.031] / [0.047]

pH = 4.54

c) millimoles of HCl added = 15 x 0.1 = 1.5

so it is equivalent point

at equivalent point

means all NaN3 becomes HN3

as HN3 is weak acid

pH = 1/2 [pKa - logC]

C = 1.5 / 25 = 0.06 M

pH = 1/2 [4.72 - log 0.06]

pH = 2.97

d) millimoles of HCl added = 27 x 0.1 = 2.7

2.7 - 1.5 = 1.2 mnillimoles HCl left

[HCl] = 1.2 / 37 = 0.032 M

as HCl is strong acid

pH = -log [H⁺]

pH = -log [0.032]

pH = 1.49