# 2nd Year Chem, pH problem Part 2!?

Calculate the pH of a soln of acetic acid that is 3% ionized, Ka = 1.8 x 10^-5.

Ok, thanks to the help from a very kind person, I realized what I needed to. But now, Im getting the wrong answer over and over again. I get 3.13. But thats not one of the options I have for the answers. If anyone could help me out I would appricate it.

That is the answer look for an option thats rounded up or down?

You're right. Working backwards from your answer:

pH = -log [H+]

3.13 = -log [H+]

[H+] = .00074

[H+] / [HAc] = .03

.00074 / [HAc] = .03

[HAc] = .025

Ka = [H+][Ac-] / [HAc]

1.8 x 10^-5 = .00074^2 / .025

1.8 x 10^-5 is close to 2.2 x 10^-5

The denominator should have been 0.97 instead of 0.03. It is 97% not ionized.

Try the calculation again

Does pH of 2.38 fit better?

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Ok, thanks to the help from a very kind person, I realized what I needed to. But now, Im getting the wrong answer over and over again. I get 3.13. But thats not one of the options I have for the answers. If anyone could help me out I would appricate it.

**Answer:**That is the answer look for an option thats rounded up or down?

You're right. Working backwards from your answer:

pH = -log [H+]

3.13 = -log [H+]

[H+] = .00074

[H+] / [HAc] = .03

.00074 / [HAc] = .03

[HAc] = .025

Ka = [H+][Ac-] / [HAc]

1.8 x 10^-5 = .00074^2 / .025

1.8 x 10^-5 is close to 2.2 x 10^-5

The denominator should have been 0.97 instead of 0.03. It is 97% not ionized.

Try the calculation again

Does pH of 2.38 fit better?

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