# chem

posted by on .

Calculate the [h3o+] and [oh-], pH and the percent dissociation for both 0.25 mol/L HCl and 0.25 mol/L HCN.

0.25 mol/L HCL
[H+] = 0.25 mol/L
[OH-] = 1 x 10^-14 / 0.25 = 4 x 10^-14
pH = 0.60
% dissociation = ?
( I don't know how to figure out percent dissociation)

For the HCN I thought I would do the same thing since there is also 0.25 mol/L of it but in the solution there is the number 1.2 x 10^ -5 and I don't understand where it came from. Could someone please help? thanks :)

• chem - ,

% diss = 100% for HCl because there is no undissociated HCl present.
My guess is that 1.2 x 10^-5 is (H3O^+) but you can work it out.
Ka for HCN in my tables (but you use your tables) 6.2 x 10^-10. So set up an ICE chart and calculate (H3O^+) (which I find to be 1.2 x 10^-5), then
you can do pH, pOH, H^+, OH^-, and
% diss = [(H3O^+)/(HCN)]*100

• chem - ,

Do we always have to do an ICE table for weak acids and bases?

• chem - ,

Unless you have a lot of experience and can do some of it in your head, leave out some steps, and keep things straight. The ICE chart is too easy to use not to use it.

• chem - ,

Did you get your KHP/Ca(OH)2 problem solved? I figures the problem to be you had not listed the mass KHP and you were calculating M Ca(OH)2.

• chem - ,

Yes I figured out that problem. :) Thanks a lot!