Question

A 44.7 mL sample of a 0.280 M solution of NaCN is titrated by 0.220 M HCl. Kb for CN- is 2.0×10-5. Calculate the pH of the solution:

(a) prior to the start of the titration

(b) after the addition of 28.4 mL of 0.220 M HCl

(c) at the equivalence point

(d) after the addition of 83.6 mL of 0.220 M HCl.

PLEASE SHOW ALL WORK

Answer 1

(a) prior to the start of the titration

molarity of NaCN= 0.280 M

CN- + H2O ---------------> HCN + OH-

0.280 0 0 ------------> initial

0.280 -x x x

Kb = x^2 / 0.280 -x

2.0 x 10^-5 = x^2 / 0.280 -x

x = 2.36 x 10^-3

x = [OH-]=  2.36 x 10^-3

pOH = -log [OH-]

pOH = -log ( 2.36 x 10^-3 )

pOH = 2.63

pH + pOH = 14

pH = 11.37

(b) after the addition of 28.4 mL of 0.220 M HCl

millimoles of NaCN = 44.7 x 0.280 = 12.52

millimoles of HCl = 28.4 x 0.220 = 6.248

NaCN + HCl -------------------> HCN + NaCl

12.52 6.25 0 0

6.27 0 6.25 -

pH = pKa + log [NaCN/HCN]

pH = 9.30 + log (6.27/6.25)

pH = 9.30

(c) at the equivalence point

here millimoles of acid = millimoles of base

44.7 x 0.280 = 0.220 x V

V = 56.89 mL

at equivalence point base volume = 56.89 mL

at equivalence point only HCN is remained.

NaCN + HCl -----------------> HCN

[HCN] concentration = 44.7 x 0.280 / (44.7 +56.89)

                                = 0.123 M

C = 0.123 M

pH = 1/2 [Pka - log C]

      = 1/2 [9.3 -l og 0.123]

      = 5.1

(d) after the addition of 83.6 mL of 0.220 M HCl.

millimoles of NaCN = 44.7 x 0.280 = 12.52

millimoles of HCl = 83.6 x 0.220 = 18.39

NaCN + HCl -------------------> HCN + NaCl

12.52 18.39 0 0

0 5.87 -

strong acid remains

HCl concentration = 5.87 / (44.7+ 83.6) = 0.046 M

pH = -log [H+]

pH = 1.34