What is the molar mass of an acid if 0.422 g of the acid are neutralized by 17.5 mL of 0.268 M NaOH?
NAOH:
0.268*1000/17.5=15.3
Acid:0.422/0.268=1.57 mr
I think
What is the molar mass of an acid if 0.422 g of the acid are neutralized by 17.5 mL of 0.268 M NaOH?
3 answers
No. To start, you don't have enough information to answer because the problem doesn't tell you if the acid is monoprotic, diprotic, triprotic, etc. If you assume it is monoprotic, it is done this way.
HA + NaOH ==> NaA + H2O
mols NaOH = M x L = 0.268 x 0.0175 = approx 0.0047 but that's just an estimate and you must recalculate for a better answer. Then mols = grams/molar mass or 0.047 = 0.422/molar mass and solve for molar mass of approx 90. But if diprotic, then H2A + 2NaOH ==> Na2A + 2H2O.
mols NaOH = M x L = approx 0.0047
mols H2A = 0.0047 mols NaOH x (1 mol H2A/2 mols NaOH - 0.0047/2 = approx 0.0024 mols H2A. Then molar mass = 0.422/0.0024 = approx 180. Triprotic acid is done similarly.
HA + NaOH ==> NaA + H2O
mols NaOH = M x L = 0.268 x 0.0175 = approx 0.0047 but that's just an estimate and you must recalculate for a better answer. Then mols = grams/molar mass or 0.047 = 0.422/molar mass and solve for molar mass of approx 90. But if diprotic, then H2A + 2NaOH ==> Na2A + 2H2O.
mols NaOH = M x L = approx 0.0047
mols H2A = 0.0047 mols NaOH x (1 mol H2A/2 mols NaOH - 0.0047/2 = approx 0.0024 mols H2A. Then molar mass = 0.422/0.0024 = approx 180. Triprotic acid is done similarly.
"it" or "jan" or whatever name you use, you need to stop posting answers since you don't know the subjects well enough.