# chemistry

posted by .

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.

## Similar Questions

1. ### Conversion

After finishing a Chem problem, I arrived at a answer of 176764.8987 M. How would I convert that to uM/L (micromolar) If M is molar, your number is quite unrealistic. Multipy Molarity by one million to get micromoles per liter.
2. ### Chemistry

If an equal number of moles of the weak acid HOCN and the strong base KOH are added to water, the resulting solution will be acidic, basic or neutral?
3. ### Chemistry/pH- Weak Acid

in response to Chemistry/pH- Weak Acid. but the rule of 5%?
4. ### General Chem

A weak acid has Ka= 1*10^-3. If [HA]= 1.00 M what must be [A-] for the pH to be pH 2.7?
5. ### chemistry

hello, I can't seem to solve this question. i tried to look up someone else who may have asked this question. I found this girl named Sara,but her answer didn't really help me much because it did not answer the question. What is the …
6. ### chemistry

hello, I can't seem to solve this question. i tried to look up someone else who may have asked this question. I found this girl named Sara,but her answer didn't really help me much because it did not answer the question. What is the …
7. ### chemistry

a 5.55g sample of a weak acid with ka=1.3*10^-4 was combined with 5.00ml of 6.00 M NAOH and the resulting solution was diluted to 750mL. The measured pH of the solution was 4.25. what is the molor mass of the weak acid. if used the …
8. ### chemistry

a 5.55g sample of a weak acid with ka=1.3*10^-4 was combined with 5.00ml of 6.00 M NAOH and the resulting solution was diluted to 750mL. The measured pH of the solution was 4.25. what is the molor mass of the weak acid. if used the …
9. ### Chemistry

Strong base is dissolved in 535 mL of 0.200 M weak acid (Ka = 3.16 × 10-5) to make a buffer with a pH of 4.04. Assume that the volume remains constant when the base is added. HA (aq)+ OH^-(aq) -> H2O(l) + A^-(aq) Calculate the …
10. ### Biochemistry

Show Calculations used in mkaing 250mL of 0.5M sodium acetate ph 4.7. You are given: 1M acetic acid (weak acid) 1M sodium hydroxide (strong base) Sodium acetate buffer: ph 4.7 acetic acid pKa: 4.7 So far this is what I've gotten: pH …

More Similar Questions