Posted by **Meche** on Wednesday, February 28, 2007 at 9:03pm.

My work:

pH = -log(x) = 4.49

x = 3.2359e-5

3.2359e-5 = 5.72/M

When I solved for M, I got that large number... 176764.8987

Ka for a given weak acid is 5.72 x 100. What concentration of acid (mM/L) will give a pH of 4.49?

I got 176764.8987 M (not converted to uM). Though I am assuming I did something wrong since that number is unrealistic.

How would I solve this problem?

I wonder if the Ka actually is 572. Seems large for a weak acid. However, assuming that is correct:

HA ==> H^+ + A^-

Ka=(H^+)(A^-)/(HA) = 572

pH = 4.49 = -log(H^+).

(H^+)= 3.235 x 10^-5

Plug this number in for (H^+) AND for

(A^-) and solve for (HA) and the answer will be in mols/L. There are 1000 millimols in a mol. (HA)= somthing like 10^-12 M.

Actually, it's 10^0, meaning 1. It didn't show up before.

OK, I don't think I'm doing this right.

I got (H^+)= 3.235 x 10^-5

So then would the equation be...

(3.235 x 10^-5)^2/(HA)=5.72

If so, I get a super small number.

Also, the answer is in macromols, not millimols. So then, I would multiply by 1000000, though the answer still appears to be wrong. I'm sorry, can you please help me out? Thanks.

The (H^+) is correct.

You DO (REALLY) get a super small number for (HA)=acid concentration.

Finally, the answer is in moles/liter.

So 1.83 x 10

^{-10} = (acid). The reason it is such a small number is because the acid (for a weak acid) is not all that weak. I only wrote the conversion from mols to millimoles because your original post stated you wanted the answer in mM/L (which means to me you want millimoles/L).

Sorry again about that, when I copied and pasted the problem, some symbols were changed.. I should have looked over it.

So I did it all, and also got 1.83 x 10-10. I multiply by 1000000 to get in uM... and then I think I am done (so I hope).

Thanks

Right. 1.83 x 10^-10 M x 10^6 (uM/M) = 1.83 x 10^-?? uM.

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