Question

Calculate the PH of an aqueous solution at 298 K that is 0.095 M in (HCN). (Ka for HCN = 4.9 x 10^-10)

Answer 1

Lets write the dissociation equation of HCN

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

9.5*10^-2 0 0

9.5*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 ((4.9*10^-10)*9.5*10^-2) = 6.823*10^-6

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

so, x = 6.823*10^-6 M

So, [H+] = x = 6.823*10^-6 M

we have below equation to be used:

pH = -log [H+]

= -log (6.823*10^-6)

= 5.17

Answer: 5.17