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
3 answers
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).
..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
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
1 answer