Asked by Tiffany
If 14.23 mL of 0.203 M NaOH is required to neutralize .269 g of a monoprotic unknown acid, what is the molecular weight of the unknown acid?
Answers
Answered by
DrBob222
NaOH + HA --> NaA + H2O
mols NaOH = M x L = ?
mols HA = same since the ratio of acid to base is 1:1 (look at the coefficients)
mols HA = grams/molecular weight. You know mols HA and grams HA, solve for molecular weight.
mols NaOH = M x L = ?
mols HA = same since the ratio of acid to base is 1:1 (look at the coefficients)
mols HA = grams/molecular weight. You know mols HA and grams HA, solve for molecular weight.
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