[orchid]

Hello again! I have the question and answer but I do not how they get some parts of the answer. Thanks!

Calculate the ionization, Ka for H3O and the pH of 0.5mol/L hydrocyanic acid at 25degreesC.

HCN + H2O <-> H3O + CN

Equilibrium:  (0.50-x)   x     x

Ka = [H3O][CN]/[HCN]
= 4.9 x 10^-10

x2 = (0.50)(4.9 x 10^-10)
x = 1.6 x 10^-5

Therefore, H3O = 1.6 X 10^-5 mol/L

I did not understand how they got 4.9 x 10^-10. Please *delete me* 

[Albert]

I did not understand how they got 4.9 x 10^-10. Please *delete me* 

Me neither. 

It seems they just took it for a table of ionization constants.
Mind you, did you post the complete text of the problem?

[oraklu]

You can either be asked for Ka OR for pH not for both in this type of question. The question has a missing value, either that of Ka or that of pH.

[orchid]

Well, the exact question is:

Calculate (a) the [H3O+] and (b) the pH of 0.50 mol/L hydrocyanic acid at 25C.

And the answers are:

a) 1.6 x10^-5 mol/L
b) 4.79

[Albert]

So, as I expected, you weren't supposed to calculate any ionization constant, but just to find it out in a table of Kas.

[orchid]

LOL. i see, thanks!