Erhatstil

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7+ Year Member
Jul 9, 2008
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Find the Ka of HCN, given that a 0.20 M solution of HCN([FONT=TimesNewRoman,Italic][FONT=TimesNewRoman,Italic]aq..) is 0.002 % ionized at 25 °C.

[H+][CN-] = Ka and .002%= 2.0x10^-5
[HCN]

(2.0x10^-5)2 =Ka doesn't seem to work. why is [H+] not equal to the given ionization ???
(0.2M)

answer key is 8.0x10^-11, my answer was 2.0x10^-9. answer key always says [H+]=(2.0x10^-5)(2.0) instead of the given ionization...

thanks for the help..

 

UndergradGuy7

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Jun 23, 2007
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You have a minor mistake.


HCN + H2O -> H3O+ and CN-

So the amount ionized is 0.002% or 0.002/100=2 x 10^-5
multiplied by the molarity 0.20M * 2 x10^-5 = 4 x 10^-6 M of H3O+ and CN-

The equation is [H3O+][CN-]/HCN = K

Notice that [H3O+] = [CN-]

so (x)(x)/(HCN) = K
we know that the amount ionized = 4x10^-6 = H3O+ = CN-
so (4x10^-6)(4.0x10^-6)/(0.20) = K = 8.0 x 10^-11
 
Apr 7, 2012
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Because that is the concentration of your original solution. You have 0.2M solution that was 0.002% ionized. If it was 0.5M solution 0.002% ionized we would have had a different value. Hope this helps. Good luck.