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
i don't know how to find the molar mass
oops..i posted this question twice
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 = 1.3E-4 = (H^+)(A^-)/(HA)
1.3E-4 = 5.62E-5)(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).
mol HA + NaOH --> NaA + H2O
i x 0.030
f x-0.030 0 0.030
pH = pKa + log(0.030/x-0.030)
x = 0.042 mol
PM=5.5g/0.042mol = 129.1 g/mol