Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: kraver0 on November 29, 2012, 10:16:02 PM
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A solution that contains 1.48 M hydrofluoric acid, HF (Ka = 7.2E-4), and 3.00 M hydrocyanic acid, HCN (Ka = 6.2E-10).
What is the pH of this mixture of weak acids?
a) 1.49
b) 2.97
c) 4.52
d) 9.04
e)12.51
The correct answer is a. But I don't know how to get this answer. Can someone help me? Thanks!
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Check this:
its in german but the formulas you can read
http://www.chemieunterricht.de/dc2/mwg/ph-saeuremischung.htm
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Could you calculate pH based on the HF present in this solution? This is quite sufficient result in this case.
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use the formula [H+] = square root (Ka x C)
to find out the concentrations of H+ of both the acids. Total the concentrations and use the formula pH = -log[H+]
You get the answer....
step - 1
H+ ions from HCN = 4.31 x 105-
step - 2
H+ ions from HF = 3268.66 x 105-
Add the above concentrations and find negative log... you get 1.48
Hope this helped you.... good luck
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use the formula [H+] = square root (Ka x C)
to find out the concentrations of H+ of both the acids. Total the concentrations and use the formula pH = -log[H+]
You get the answer....
step - 1
H+ ions from HCN = 4.31 x 105-
step - 2
H+ ions from HF = 3268.66 x 105-
Add the above concentrations and find negative log... you get 1.48
Hope this helped you.... good luck
And it is completely wrong.
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can you tell how it is wrong?
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First of all - you can't simply sum concentrations, as H+ produced in one reaction shifts the equilibrium in the other reaction.
Second - you are using an approximate method not checking if it can be used.