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Topic: Help to Understand the value of ionization, Ka  (Read 4918 times)

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Offline orchid

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Help to Understand the value of ionization, Ka
« on: June 13, 2006, 03:44:56 AM »
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*  :)

Offline Albert

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Re: Help to Understand the value of ionization, Ka
« Reply #1 on: June 13, 2006, 05:52:55 AM »
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?

oraklu

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Re: Help to Understand the value of ionization, Ka
« Reply #2 on: June 13, 2006, 07:43:34 AM »
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.

Offline orchid

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Re: Help to Understand the value of ionization, Ka
« Reply #3 on: June 13, 2006, 12:38:18 PM »
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

Offline Albert

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Re: Help to Understand the value of ionization, Ka
« Reply #4 on: June 13, 2006, 12:42:46 PM »
So, as I expected, you weren't supposed to calculate any ionization constant, but just to find it out in a table of Kas.

Offline orchid

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Re: Help to Understand the value of ionization, Ka
« Reply #5 on: June 13, 2006, 01:30:38 PM »
LOL. i see, thanks!

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