Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: tkle28 on May 11, 2023, 03:22:43 AM
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Hello, I ran across this problem and was wondering if my answer is correct.
The question is: What is the pH of a 0.065 M solution of HF? Ka(HF) = 6.8 x 10^-4. For this one I got 2.26 and apparently the answer was 2.18. Can someone tell me what I did wrong?
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Tell us how you got 2.26.
(Actually 2.18 doesn't look exactly correct either).
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Tell us how you got 2.26.
(Actually 2.18 doesn't look exactly correct either).
To be quite Frank with you, my calculations itself were all over the place. I somehow ended with 2.77 but since 2.26 was the closest answer I picked that. I found H+ by doing root symbol(0.00068) x (0.065) then got my answer of 0.00169, then I just put in -log(0.00169) in got an answer of 2.77. I’m trying to figure out what exactly I’m doing wrong.
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Putting numbers into random formulas is never going to be a valid strategy :/
[itex][H^+] = \sqrt{K_a\times C_a}[/itex] is a formula that is valid only under a specific assumption about the acid dissociation. Do you recall what this assumption is?
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Putting numbers into random formulas is never going to be a valid strategy :/
[itex][H^+] = \sqrt{K_a\times C_a}[/itex] is a formula that is valid only under a specific assumption about the acid dissociation. Do you recall what this assumption is?
Is it whenever it’s weak?
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Is it whenever it’s weak?
No. Just because it is weak is not enough.
In general this formula works reasonably well when the acid is dissociated less than 5%. Is this the case?
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Well solving the quadratic is pretty tedious but I persevered and get 2.199. However it’s rather late at night here and I may well have made a slip or rounding error. I might check it in the morning when I’m more awake.
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2.20 is what I got as well.
(lol, you haven't seen really tedious calculations yet :D )
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To help the OP. Looking at your calculations there’s an error other than the approximation. Root of 0.00068 x 0.065 is 6.6483 x 10^-3 and the log of this is - 2.177 giving pH 2.18 as per the given answer. So it seems the question assumes the approximation method was expected.
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So it seems the question assumes the approximation method was expected.
At nearly 10% dissociation that's a bad idea (I don't blame you for checking, I blame whoever asked the question and expected the answer).
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In many exams and books also, approximation is assumed for these type of questions.
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In many exams and books also, approximation is assumed for these type of questions.
Problem is not with using the approximation, it is with _misusing_ it outside of its applicability.