Question

Calculate the OH- concentration and pH of a 3.7×10-3M aqueous solution of sodium cyanide, NaCN. Finally, calculate the CN- concentration. Ka (HCN) = 4.9×10-10.

Answer 1

Ka of HCN = 4.9*10^-10

Find Kb of CN-

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

CN- dissociates as

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

0.0037 0 0

0.0037-x x x

Kb = [HCN][OH-]/[CN-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((2.041*10^-5)*3.7*10^-3) = 2.748*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

2.041*10^-5 = x^2/(3.7*10^-3-x)

7.551*10^-8 - 2.041*10^-5 *x = x^2

x^2 + 2.041*10^-5 *x-7.551*10^-8 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 2.041*10^-5

c = -7.551*10^-8

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.025*10^-7

roots are :

x = 2.648*10^-4 and x = -2.852*10^-4

since x can't be negative, the possible value of x is

x = 2.648*10^-4

So,

[OH-] = x = 2.648*10^-4

use:

pOH = -log [OH-]

= -log (2.648*10^-4)

= 3.5771

use:

PH = 14 - pOH

= 14 - 3.5771

= 10.4229

[HCN] = 0.0037-x

= 0.0037- (2.648*10^-4)

= 3.44*10^-3 M

[OH-] = 2.65*10^-4

pH = 10.42

[HCN] = 3.44*10^-3 M