# Finding pH using ICE chart/quadratic formula on the last step!

Calculate the pH of a 0.0253 M aq solution of a monoprotic acid. The Ka for the acid is 1.79 x 10-2

So a monoprotic acid will have the form HA --> H + A, correct?

I 0.0253 0 0
C -x x x
E (0.0253-x) x x

EDIT: I just realized the formatting is all messed up on my chart, but it shouldn't be too difficult to discern what I'm doing.

So we have:

1.79 x 10-2 = x2/(0.0253 - x)
0 = -x2 - 1.79x10-2x + 4.53x10-4

After using the quadratic formula, we find the acceptable value of x (the one that is positive) to be:

4.10 x 10-2 M

Now my question is: how do I find the pH? I thought I could use -log10(4.10x10-2), but apparently that's wrong.

Any help? Thank you!

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epenguin
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Calculate the pH of a 0.0253 M aq solution of a monoprotic acid. The Ka for the acid is 1.79 x 10-2

So a monoprotic acid will have the form HA --> H + A, correct?

I 0.0253 0 0
C -x x x
E (0.0253-x) x x

EDIT: I just realized the formatting is all messed up on my chart, but it shouldn't be too difficult to discern what I'm doing.

So we have:

1.79 x 10-2 = x2/(0.0253 - x)
0 = -x2 - 1.79x10-2x + 4.53x10-4

After using the quadratic formula, we find the acceptable value of x (the one that is positive) to be:

4.10 x 10-2 M

Now my question is: how do I find the pH? I thought I could use -log10(4.10x10-2), but apparently that's wrong.

Any help? Thank you!
Your method and your quadratic equation I think are OK so there must be some mistake somewhere in arithmetic.

I don't know how you know it is wrong. But I think a way to know without anyone telling you (in other words better than 'apparently', rather the sort of checkup thinking you should habitually do) is that your [H+] result is higher than the total concentration of acid that is there, which is impossible.

I went though a calc. and if you get pH 1.85 same as me it is probably not coincidence, if you get different check a bit but I also am error-prone.

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