posted by Jenni on .
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!
...........HCN + H2O ---> H3O^+ + CN^-
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.
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...
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.