Question

Hydrocyanic acid, HCN is a weak acid with a Ka of 4.9 x 10^-10. What pH would a 0.50M solution of HCN have? Calculate the Kb value for NaCN using information from the previous question.

Answer 1

1)

HCN dissociates as:

HCN -----> H+ + CN-

0.5 0 0

0.5-x x x

Ka = [H+][CN-]/[HCN]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((4.9*10^-10)*0.5) = 1.565*10^-5

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

so, x = 1.565*10^-5 M

So, [H+] = x = 1.565*10^-5 M

use:

pH = -log [H+]

= -log (1.565*10^-5)

= 4.8054

Answer: 4.81

2)

use:

Kb = Kw/Ka

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/4.9*10^-10

Kb = 2.041*10^-5

Answer: 2.04*10^-5