# Homework Help: Aspirin + NaOH => ?

1. Dec 18, 2008

### tgt

1. The problem statement, all variables and given/known data
Does this reaction occur?

Aspirin = C9H8O4 (numbers are meant to be subscripts)

Aspirin + NaOH => what?

3. The attempt at a solution

Aspirin + NaOH => 9 carbon dioxide + 4 water molecules + 9 hydrogens (i.e H2)

2. Dec 18, 2008

### Staff: Mentor

No. You have to look at the structure of aspirin, using just overall formula will lead you nowhere.

3. Dec 18, 2008

### symbolipoint

The empirical formula for Aspirin is not exact enough; you want to know that Aspirin is acetyl salicylic acid. (or is it three separate words, "acetylsalicylic acid ?")

4. Dec 18, 2008

### tgt

5. Dec 18, 2008

### Staff: Mentor

Read the wikipedia article, it clearly states what should happen when you put aspirin in NaOH solution.

6. Dec 19, 2008

### lesieux

Aspirin has got a carboxyl group -COOH and an ester group -CO-OCH3 (acetyl); the first function reacts as an acid and becomes -COO-, the second function overcomes saponification RCOOR' + OH– -> RCOO– + R'OH

7. Dec 19, 2008

### symbolipoint

Would the saponification reaction be a much slower reaction, possibly relying on more extreme conditions other than moderate concentrations at room temperature?

8. Dec 19, 2008

### Staff: Mentor

Aspirin in basic solution hydrolises quite fast. Fast enough that it will at least partially hydrolise during standard titration with base, giving unreproducible results.

9. Dec 19, 2008

### lesieux

That is the reason for you need to use an excess of sodium hydroxide (at high temperature to ensure that all the aspirin reacts) to titrate aspirine and then, in a second step, titrate the excess of sodium hydroxide. But you will have to consider that one mol of aspirin reacted with two moles of base in your calculation.

10. Dec 19, 2008

### Staff: Mentor

That's right, backtitration is the usual approach.