Determine the pH of 0.100 M aqueous HCN to two decimal places, and determine the concentration of CN- at equlibrium to two singificant figures. Can you show me all the steps to this question? Please and thank you!

Answer 1

...........HCN + H2O ---> H3O^+ + CN^-
initial....0.100...........0.......0
change.....-x..............+x......+x
final...0.100-x.............x.......x

Ka = (H3O^+)(CN^-)/(HCN)
Now substitute into Ka expression the values from the ICE chart.
Ka for HCN is 7.2E-4 from an old book of mine but you need to use the value found in your text/notes.

7.2E-4 = (x)(x)/(0.1-x)
Solve for x = (H3O^+) and convert to pH.
(CN^-) is the same.
NOTE: If you solve the EXACT equation I have written above you must solve a quadratic equation. Most calculators do that with no problem.
Post your work if you get stuck.

Answer 2

So is Ka water H2o? not sure where you got 7.20 in my chart 7.2 goes with H2PO4^-

there is nothing for (H3O+) but I got 4.69 for CN- and 9.31 for HCN
I am so lost...

Answer 3

If you re-read my post I told you where I obtained 7.2E-4. Since you didn't include it in your post I had to look in one of my old texts BUT I told you to look in your text and use the value you found there.
What do you mean there is nothing for H3O^+. That's what you are calculating. And HCN can't possibly be 9.31. It was 0.100 in the problem.