Asked by sea lively
If 0.314g sample of a solid monoprotic acid were dissolved in water and titrated to an end point that required 44.2 mL of 0.115M NaOH, what is the molecular mass of the unknown acid?
Answers
Answered by
DrBob222
HA + NaOH ==> NaA + H2O
mols NaOH = M x L = ?
1 mol HA = 1 mol NaOH; therefore, mols HA = mols NaOH.
Then mol = g/molar mass. You know mpols and grams, solve for molar mass.
mols NaOH = M x L = ?
1 mol HA = 1 mol NaOH; therefore, mols HA = mols NaOH.
Then mol = g/molar mass. You know mpols and grams, solve for molar mass.
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