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_{a}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}= x

^{2}/(0.0253 - x)

0 = -x

^{2}- 1.79x10

^{-2}x + 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 -log

_{10}(4.10x10

^{-2}), but apparently that's wrong.

Any help? Thank you!