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.

To solve this problem, you need to use the given information and the equations for pH and Ka of a weak acid. Here is the step-by-step solution:

1. Start with the equation for pH: pH = -log(H+). Since pH is given as 4.49, you can calculate the concentration of H+ using the formula: H+ = 10^(-pH).

2. From the given value, pH = 4.49, you can calculate H+ as follows: H+ = 10^(-4.49) = 3.235 x 10^(-5).

3. The equation for Ka of a weak acid is: Ka = (H+)(A-)/(HA), where HA represents the concentration of acid, H+ represents the concentration of hydrogen ions, and A- represents the concentration of the acid's conjugate base.

4. Substituting the given value of Ka (5.72 x 100) and the calculated value of H+ (3.235 x 10^(-5)) into the equation, you get: (3.235 x 10^(-5))^2/(HA) = 5.72 x 100.

5. Solve this equation for HA to find the concentration of the acid (in mols/L).

6. The obtained value for HA will be a small number, such as 1.83 x 10^(-10) M. Make sure to convert it to the desired unit of mM/L (millimoles per liter) by multiplying by 10^6.

So, the final answer would be 1.83 x 10^(-10) M x 10^6 uM/M = 1.83 x 10^(-4) uM/L.