Question

A 44.9 mL sample of a 0.350 M solution of NaCN is titrated by 0.260 M HCl. K_b for CN^- is 2.0×10^-5. Calculate the pH of the solution:

(a) prior to the start of the titration	pH =
(b) after the addition of 48.4 mL of 0.260 M HCl	pH =
(c) at the equivalence point	pH =
(d) after the addition of 76.2 mL of 0.260 M HCl.	pH =

Answer 1

a ) Kb = 2 x 10⁵

[OH^-] = underroot Kb x c

= underroot 2 x 10⁵ x 0.350

= 2.603 x 10 ^-3M

p^OH = - log [OH^-]

= - log 2.603 x 10 ^-3

   ^{= 2.58}

pH = 14 - 2.58 =11.42

b)Vol after mixing = 48.4 + 44.9 = 93.3

molarity of HCL = 48.4 / 93.3 M

Molarity of NaCN = 44.9 / 93.3 M

Net molarity = 48.4 / 93.3 - 44.9 / 93.3

= 0.037

[H⁺] = 0.037 = 3.7 x 10^-2

pH = -log  3.7 x 10^-2

   ^{= 1.43}

^{c )} At equivqlence point pH = kb = 2

d)

Vol after mixing = 76.2 + 44.9 = 121.1

molarity of HCL = 76.2 / 121.1 M

Molarity of NaCN = 44.9 / 121.1 M

Net molarity = 76.2 / 121.1 - 44.9 / 121.1

= 0.25

[H⁺] = 0.25 = 2.5 x 10^-1

pH = -log 2.5 x 10^-1

= 0.6021