If you had added 50 mL of distilled water to one the vinegar sample instead of 30 mL, would the titration have required more NaOH, less NaOH or the same amount of NaOH? Why?
same amount of acid, so same amount of base.
If you had added 50 mL of distilled water instead of 30 mL to the vinegar sample, the titration would have required the same amount of NaOH. The reason for this is that the concentration of the vinegar sample would not have changed by adding more water. When performing a titration, the concentration of the analyte (in this case, acetic acid in vinegar) is what determines the amount of titrant (in this case, NaOH) needed to reach the endpoint. Adding more water would dilute the vinegar, but the concentration of acetic acid would remain the same because water itself does not affect the concentration. Therefore, the same amount of NaOH would still be required to neutralize the acetic acid in the vinegar sample.
To determine whether adding 50 mL of distilled water instead of 30 mL to the vinegar sample would require more, less, or the same amount of NaOH in a titration, we need to understand the concept of moles and the stoichiometry of the reaction.
In a titration, a known volume and concentration of a solution (in this case, NaOH) are added to react with a known volume and concentration of another solution (in this case, vinegar). The goal is to determine the amount of substance present in the sample being titrated.
In this case, the reaction that occurs between NaOH and vinegar (acetic acid) can be represented by the balanced chemical equation:
CH3COOH + NaOH → CH3COONa + H2O
From the balanced equation, we can see that the stoichiometry of the reaction is 1:1 for acetic acid (CH3COOH) and NaOH. This means that one mole of acetic acid reacts with one mole of NaOH.
When performing a titration, the volume and concentration of the NaOH solution (solute) and the volume of the vinegar sample (solvent containing acetic acid) are known. By measuring the volume of NaOH solution required to reach the equivalence point (where all the acetic acid is neutralized), we can determine the amount of acetic acid present.
Now, let's compare two scenarios: adding 30 mL of distilled water and adding 50 mL of distilled water to the vinegar sample.
Scenario 1: Adding 30 mL of distilled water
In this case, if you initially add 30 mL of distilled water to the vinegar sample, the total volume would be increased, but the concentration of acetic acid remains the same. Therefore, during the titration, the same amount of NaOH solution would be required to reach the equivalence point since the stoichiometry of the reaction remains unchanged.
Scenario 2: Adding 50 mL of distilled water
On the other hand, if you add 50 mL of distilled water to the vinegar sample, the total volume would increase even further. In this scenario, the concentration of acetic acid in the solution would be diluted. As a result, more NaOH solution would be required to neutralize all the acetic acid present because there is more solvent (water) but the same amount of acetic acid molecules.
Therefore, if you added 50 mL of distilled water instead of 30 mL to the vinegar sample, the titration would require more NaOH solution because the acetic acid concentration has been further diluted.
In summary, adding more distilled water to the vinegar sample would result in a more diluted acetic acid solution, requiring more NaOH solution during the titration to neutralize the same amount of acetic acid.