[mafagafo]

From http://www.chemicalforums.com/index.php?topic=74746.0

Q: If 25.00mL of 0.20mol/L HCO2H is titrated with 0.20mol/L NaOH, determine the pH after 10.00mL of NaOH has been added.
the correct answer is 3.57.


I am not a good chemistry student, but I want to show my solution to this problem and get some feedback from you. I want to know if it is valid or not. Borek mentioned Henderson-Hasselbalch, I think it is "solvable" without it.

By adding 0.20 * 0.01 = 2e-3 mol NaOH, this same amount of COOH^- will be formed.

So [COOH^-] now is 2e-3 / 0.035 = 5.714e-2.

And [HCOOH] is 3e-3 / 0.035 = 8.5714e-2.

As Ka(HCOOH) = 1.8e-4,

1.8e-4 = [H⁺](5.714e-2)(8.5714e-2)^-1

and

[H⁺] = 2.7001e-4.

pH = -log10([H⁺]) = 3.6 (2 significant figures as I used a Ka with only 2 s.f.)

mafagafo.

[Xenonman]

I can't answer the problem right now, but, the Henderson-Hasselbalch happens to be a derivation of the Ka equation, as seen here:
http://en.wikipedia.org/wiki/Henderson_hasselbach#Derivation.
It's a more convenient way of calculating pH's of buffers.  Your approach is also valid, and it looks fine from here, as you considered the added volume of both solutions. 

It is good to see you try to find new ways of solving problems. It may not always work, but it's a good exercise anyway.

[mafagafo]

Thanks a lot. I just discovered this forum now. I think I can learn a bit here and there.

I did not know https://en.wikipedia.org/wiki/Henderson_hasselbach#Derivation.

Thanks for the link. It is so obvious now.

[mafagafo]

pH = pKa + log10([A^-]/[AH])
pH = -log10(1.8e-4) + log10((2e-3 / 0.035)/(3e-3 / 0.035))
pH = 3.6

It is easier.

[Borek]

pH = -log10(1.8e-4) + log10((2e-3 / 0.035)/(3e-3 / 0.035))

[tex]pH = -\log_{10}(1.8\times10^{-4}) + \log_{10}\left(\frac{\frac{2\times10^{-3}}{0.035}}{\frac{3\times10^{-3}}{0.035}}\right)[/tex]

Have you noticed how the volume (0.035) cancels out, leaving you with just a ratio of amounts of HCOO^- and HCOOH?

[mafagafo]

Yes, but I have a question about this:

If [H] is the mol/L concentration of hydrogen in a solution. (Right?) How would I denote the amount of hydrogen (e.g. 2 moles)?

H is just hydrogen, right?

Thank you all for answering.

[Borek]

H is just hydrogen, right?

H is hydrogen, but we are talking about calculating pH, which is related to the concentration of H⁺.

[mafagafo]

I was not clear enough. Read that H as any element you want. My question is more about notation than about this problem.

• = mol (x) per liter

?? = mol (x)

Is there a way to indicate amount of substance?

[Xenonman]

The mole is an SI unit, with clear instructions on its symbol, the magnitude it measures, and the symbol of said magnitude.
Same for concentration of stuff.

On another note, something fun happens when you neutralize the half of the starting moles.

[mafagafo]

I noticed it. Why can't I edit that post anymore?

- EDIT -

What I wanted is:
[tex]n_{H_2}[/tex]

[Xenonman]

Users have 30ish minutes to edit their posts.

n_H is somewhat ambiguous. For hydrogen, you should use either n_{H_2} or n_H^±, depending on circumstances.