Posted by Brandon on Saturday, November 26, 2011 at 5:45pm.
Convert pH to (H^+). It's approximately, but you need to do it exactly, 9E-6
...............HCN ==> H^+ + CN^-
initial........0.15M...0.......0
change........-1E-6...1E-6...1E-6
equil.......0.15-1E-6..1E-6...1E-6
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 -1E-6 from?
pH = -log(H^+)
5.07 = -log(H^+)
-5.07 = log(H^+)
(H^+) = 8.51E-6
Note that at the top of the page I estimated the (H^+) to be 9E-6M and noted that you should calculate it more accurately. Then I substituted into the ICE chart and I used 1E-6. That was a typo but since it was just an estimate it doesn't make much difference. You should substitute 8.51E-6 for (H^+).
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