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Posted by on Saturday, November 26, 2011 at 5:45pm.

The pH of 0.15 mol/L of hydrocyanic acid is 5.07. Calculate Ka of hydrocyanic acid.

  • Chemistry - , Saturday, November 26, 2011 at 6:41pm

    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.

  • Chemistry - , Saturday, November 26, 2011 at 7:24pm

    How do you calculate the [H^+] and where did you get the -1E-6 from?

  • Chemistry - , Saturday, November 26, 2011 at 7:33pm

    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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