Posted by Meche on Wednesday, February 28, 2007 at 9:03pm.
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).
Right. 1.83 x 10^-10 M x 10^6 (uM/M) = 1.83 x 10^-?? uM.
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