Question

What is the pH of a 1.24 mol/L solution of HCN(aq) if its Ka = 6.2 x 10^-10?

I am not sure how to this question... I did 6.2x10^-10/ 1.24?? I am not sure if it is right OR are we suppose to find the Kb value which i already did Kb= kw/ka = 1.0x10^-14/6.2x10^-10 = 1.61x10^-5 then divide it by 1.24 = 1.30X10^-5???

Answer 1

HCN is an acid. Treat it as an acid.
HCN==> H^+ + CN^-

Ka = (H^+)(CN^-)/(HCN) = 6.2 x 10^-10

Now do the ICE table.
I = initial concns:
C = change in concns:
E = equilibrium concns:

Initial (before any ionization takes place):
(HCN) = 1.24
(H^+) = 0
(CN^-) = 0

change:
(H^+) = +y
(CN^-) = +y
(HCN) = -y

equilibrium:
(HCN) = 1.24 - y
(H^+) = 0 + y = y
(CN^-) = 0 + y = y

Substitute the equilibrium concns into the Ka expression and solve for y = (H^+), then pH = -log(H^+).