mr.pyro - 7-10-2006 at 15:18
I have this question that I'm really stuck on:
A chemist wanted to determine the concentration of a solution of lactic acid, HC3H5O3. She found that the pH of the solution was 2.27. What was the
concentration of the solution? The Ka of lactic acid is 1.4 x 10−4.
I think an ICE table is involved but I'm not sure how to set it up. I think I'm suppose to det it up backwards?
Any help would be greatly appreciated.
Magpie - 7-10-2006 at 17:34
I don't know what an ICE table is but can guess it means "initial concentration" and "equilibrium."
Solving this involves Ka (for the acid) and Kw (for water). Set these two parameters up as mathematical definitions and see if that gives you any
ideas.
The_Davster - 7-10-2006 at 18:09
Not sure what is meant by 'set it up backwards' but you use the pH to get the [H+], which equells the [lactate] if lactic acid is monoprotic, and from
there you use the ka value and ka equation to get the acid concentration.
mr.pyro - 8-10-2006 at 11:12
Thanks rogue, that helped.
Nerro - 8-10-2006 at 11:19
Just curious, is the answer someting like 0,206M?
mr.pyro - 8-10-2006 at 17:36
Yea. Chem dept says .21 is the right one. looks good nerro
Nerro - 9-10-2006 at 02:05
OK, I was just cautious not to make myself look like a retard 
the theory is quite straight forward,
10^(-pH) = [H3O+] so 10^(-2.27) = [H3O+]
Ka = [H3O+][A-]/[HA] so [H3O+]² = 1.4*10^-4 * [HA] = [H3O+]²
that gives you 0.206M acid, because the pH is so low the Kw shouldn't really affect the outcome anymore. That's only a problem at pH's quite close to
7.