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

oops..i posted this question twice

To find the molar mass of the weak acid, we can use the formula weight = mass / moles.

First, we need to find the number of moles of the weak acid in the sample. We can do this by using the concentration of NaOH and the volume of NaOH used in the reaction.

Given:
Mass of weak acid sample = 5.55 g
Volume of NaOH used = 5.00 mL
Concentration of NaOH = 6.00 M

Since the volume of NaOH used is in milliliters, we need to convert it to liters:
Volume of NaOH (in liters) = 5.00 mL / 1000 mL/L = 0.005 L

Now we can calculate the number of moles of NaOH used:
Moles of NaOH = concentration of NaOH × volume of NaOH (in liters)
Moles of NaOH = 6.00 M × 0.005 L = 0.030 moles

Since NaOH reacts with the weak acid in a 1:1 ratio, the number of moles of the weak acid is also 0.030 moles.

Now we can use the number of moles and the mass of the weak acid sample to calculate its molar mass:
Molar mass = mass / moles
Molar mass = 5.55 g / 0.030 moles ≈ 185 g/mol

Therefore, the molar mass of the weak acid is approximately 185 g/mol.

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