What volume of 0.15M NaOH would be required to neutralize 0.65g of pure aspirin?

Question

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Answer 1

Aspirin is O-acetylsalicylic acid, CH3COO-C6H4-COOH, or C9H8O4. The molecular weight is 180. Let aspirin be called A.

0.65gA x 1molA/180gA x 1molNaOH/1molA x 1000mLNaOH/0.15molNaOH = (0.65)(1000)/(180)(0.15) = 24mL

The 0.65gA is given. The first factor is from the molecular weight. gA cancel, leaving molA. The second factor is from the balanced equation. molA cancel, leaving molNaOH. The third factor is from the molarity. molNaOH cancel, leaving mLNaOH.