Answer:
If you poured 1 mole of acetic acid into a pool and 1 mole of acetic acid into a cup, it would not take the same amount of drops of NaOH solution to neutralize both. This is because the concentration of acetic acid in the pool would be much lower than in the cup, and therefore more NaOH would be required to neutralize the cup.
To understand why this is the case, we need to look at the concept of molarity. Molarity is defined as the number of moles of solute per liter of solution. Therefore, if you have 1 mole of acetic acid in a cup with a volume of 0.1 L, the molarity of the solution would be 10 M (1 mole / 0.1 L). If you have 1 mole of acetic acid in a pool with a volume of 1000 L, the molarity of the solution would be 0.001 M (1 mole / 1000 L).
When you add NaOH to an acidic solution like acetic acid, a neutralization reaction occurs. The general equation for this reaction is:
acid + base → salt + water
In the case of acetic acid and NaOH, the specific equation is:
CH3COOH + NaOH → CH3COONa + H2O
This reaction shows that one mole of NaOH reacts with one mole of acetic acid to produce one mole of sodium acetate and one mole of water.
Therefore, if you add drops of NaOH to both the cup and the pool containing acetic acid, it will take more drops to neutralize the cup than it will to neutralize the pool. This is because there are more moles of acetic acid per liter in the cup than in the pool.