Posted by Lena on Thursday, July 23, 2009 at 10:54pm.
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 - DrBob222, Thursday, July 23, 2009 at 11:24pm
% 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 - Lena, Thursday, July 23, 2009 at 11:36pm
Do we always have to do an ICE table for weak acids and bases?
chem - DrBob222, Thursday, July 23, 2009 at 11:49pm
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 - DrBob222, Thursday, July 23, 2009 at 11:50pm
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 - Lena, Friday, July 24, 2009 at 5:39pm
Yes I figured out that problem. :) Thanks a lot!
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