[tkle28]

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?

[Borek]

Tell us how you got 2.26.

(Actually 2.18 doesn't look exactly correct either).

[tkle28]

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.

[Borek]

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?

[tkle28]

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?

[Borek]

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?

[Aldebaran]

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.

[Borek]

2.20 is what I got as well.

(lol, you haven't seen really tedious calculations yet )

[Aldebaran]

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.

[Borek]

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).

[Vidya]

In many exams and books also, approximation is assumed for these type of questions.

[Borek]

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.