Posted by help on .
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 formula
pH=kpa+log(base/acid)
Ph=3.89+log(6.00/x)
(x)=2.62
i don't know how to find the molar mass

chemistry 
help,
oops..i posted this question twice

chemistry 
DrBob222,
I assume the unknown acid is a monoprotic one.
..HA...+...NaOH ==> NaA....+ ... H2O
5.55 g..0.005*6.00..0.03 moles..0.03 mols.
0.03 moles/0.750 L = 0.04 M
Ka = 1.3E4 = (H^+)(A^)/(HA)
1.3E4 = 5.62E5)(0.04)/(X)
Solve for X = molarity HA.
Then M x L = M x 0.750 = moles HA
moles HA = g/molar mass.
You have moles HA and g HA, solve for molar mass HA.
Post your work if you get stuck.
There is nothing wrong with the equation you used but the log value skews the answer a little. I can go through that with you if you wish but post your work if you want that done so I can use your values.
I obtained an answer for molar mass of about 430 (that isn't an exact number). 
chemistry 
lili,
mol HA + NaOH > NaA + H2O
i x 0.030
f x0.030 0 0.030
pH = pKa + log(0.030/x0.030)
x = 0.042 mol
PM=5.5g/0.042mol = 129.1 g/mol