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A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alcohol and 25.0 mL of DI water. The titration of this solution with 0.1052 M NaOH required 15.62 mL to reach the phenolphthalein endpoint. Determine the moles of NaOH that reacted with the acetylsalicylic acid.

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Answer:

Moles of NaOH that reacted:
n_(NaOH)=1.64*10^(-3)mol

Step-by-step explanation:

During the titration, all moles of NaOH added in the solution, react with the acetylsalicylic acid neutralizing each other.

The titration required 15.62 ml (0.01562 L) of the 0.1052 M NaOH solution. The moles of NaOH in that volume are:


n_(NaOH)=0.01562 L * (0.1052mol)/(L)


n_(NaOH)=1.64*10^(-3)mol

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