Chemistry
posted by Brandon .
The pH of 0.15 mol/L of hydrocyanic acid is 5.07. Calculate Ka of hydrocyanic acid.

Convert pH to (H^+). It's approximately, but you need to do it exactly, 9E6
...............HCN ==> H^+ + CN^
initial........0.15M...0.......0
change........1E6...1E6...1E6
equil.......0.151E6..1E6...1E6
Ka = (H^+)(CN^)/(HCN)
Substitute from the ICE chart and solve for Ka. 
How do you calculate the [H^+] and where did you get the 1E6 from?

pH = log(H^+)
5.07 = log(H^+)
5.07 = log(H^+)
(H^+) = 8.51E6
Note that at the top of the page I estimated the (H^+) to be 9E6M and noted that you should calculate it more accurately. Then I substituted into the ICE chart and I used 1E6. That was a typo but since it was just an estimate it doesn't make much difference. You should substitute 8.51E6 for (H^+).