Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: orchid on June 13, 2006, 03:44:56 AM
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Hello again! I have the question and answer but I do not how they get some parts of the answer. Thanks!
Calculate the ionization, Ka for H3O and the pH of 0.5mol/L hydrocyanic acid at 25degreesC.
HCN + H2O <-> H3O + CN
Equilibrium: (0.50-x) x x
Ka = [H3O][CN]/[HCN]
= 4.9 x 10^-10
x2 = (0.50)(4.9 x 10^-10)
x = 1.6 x 10^-5
Therefore, H3O = 1.6 X 10^-5 mol/L
I did not understand how they got 4.9 x 10^-10. Please *delete me* :)
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I did not understand how they got 4.9 x 10^-10. Please *delete me* :)
Me neither. ???
It seems they just took it for a table of ionization constants.
Mind you, did you post the complete text of the problem?
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You can either be asked for Ka OR for pH not for both in this type of question. The question has a missing value, either that of Ka or that of pH.
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Well, the exact question is:
Calculate (a) the [H3O+] and (b) the pH of 0.50 mol/L hydrocyanic acid at 25C.
And the answers are:
a) 1.6 x10^-5 mol/L
b) 4.79
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So, as I expected, you weren't supposed to calculate any ionization constant, but just to find it out in a table of Kas.
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LOL. i see, thanks!