Question

What is the pH of a 0.84 M solution of NaCN (Ka of HCN=6.2X10-10)?

Answer 1

we have below equation to be used:

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)/6.2*10^-10

Kb = 1.613*10^-5

CN- dissociates as

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

0.84 0 0

0.84-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.613*10^-5)*0.84) = 3.681*10^-3

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

so, x = 3.681*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (3.681*10^-3)

= 2.43

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.43

= 11.57

Answer: 11.57