Question

a. What is the pH of a 0.040 M solution of HCN?

Find the concentrations of all the major species in the solution?

[HCN], [H3O+], [CN-]

Answer 1

1)

ka of HCN = 6.2*10^-10

Lets write the dissociation equation of HCN

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

4*10^-2 0 0

4*10^-2-x x x

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

Ka = 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

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.2*10^-10)*4*10^-2) = 4.98*10^-6

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

so, x = 4.98*10^-6 M

So, [H+] = x = 4.98*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (4.98*10^-6)

= 5.30

2)

[HCN] = 0.040-x

= 0.040 - 4.98*10^-6 M

= 0.040 M

[H+] = x = 4.98*10^-6 M

[CN-] = x = 4.98*10^-6 M