Asked by harley
.256g unknown acid is titrated with .12M NaOH. It requires 34.8ml of base to reach the equivalence point. What is the molar mass of the acid?
Answers
Answered by
DrBob222
I don't know and we can't find out.
mols NaOH = M x L = approx 0.0042
IF the unknown acid is a monoprotic acid, then mols acid = mols base and
mols acid = grams/molar mass
You know mols acid and grams acid, calculate molar mass acid. However, the problem doesn't say it is a monoprotic acid. If it is diprotic then
H2A + 2NaOH = 2H2O + Na2A
and mols NaOH above x 1/2 = mols H2A
Then mols acid = grams acid/molar mass and this molar mass will be 1/2 the one above. Of course it could be triprotic and
H3A + 3NaOH ==> 3H2O + Na3A
etc.
mols NaOH = M x L = approx 0.0042
IF the unknown acid is a monoprotic acid, then mols acid = mols base and
mols acid = grams/molar mass
You know mols acid and grams acid, calculate molar mass acid. However, the problem doesn't say it is a monoprotic acid. If it is diprotic then
H2A + 2NaOH = 2H2O + Na2A
and mols NaOH above x 1/2 = mols H2A
Then mols acid = grams acid/molar mass and this molar mass will be 1/2 the one above. Of course it could be triprotic and
H3A + 3NaOH ==> 3H2O + Na3A
etc.
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