Question

CN Finally, calculate the CN concentration. (K. (HCN) - 1.9 x 10") Calculate the Ol concentration and pll of a 1.5 x 10-M aqueous solution of sodium cyanide, N [OH- M CN- OM

Answer 1

1)

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-

0.0015 0 0

0.0015-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)*1.5*10^-3) = 1.75*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/(1.5*10^-3-x)

3.061*10^-8 - 2.041*10^-5 *x = x^2

x^2 + 2.041*10^-5 *x-3.061*10^-8 = 0

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

a = 1

b = 2.041*10^-5

c = -3.061*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 = 1.229*10^-7

roots are :

x = 1.651*10^-4 and x = -1.855*10^-4

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

x = 1.651*10^-4

[OH-]= x = 1.651*10^-4

Answer: x = 1.7*10^-4

2)

use:

pOH = -log [OH-]

= -log (1.651*10^-4)

= 3.7824

use:

PH = 14 - pOH

= 14 - 3.7824

= 10.2176

Answer: 10.22

3)

[CN-] = 0.0015-x

= 0.0015 - (1.651*10^-4)

= 1.3*10^-3 M

Answer: M