# Question #63681

Apr 9, 2017

In Experiment I, you are titrating only the ${\text{Na"_2"CO}}_{3}$, while in Experiment II you are titrating the ${\text{NaHCO}}_{3}$ formed in Experiment I plus what was present in the original mixture.

Experiment I

In Experiment I, you are titrating the mixture of ${\text{Na"_2"CO}}_{3}$ and ${\text{NaHCO}}_{3}$ to a phenolphthalein end-point.

Phenolphthalein changes colour around pH 9.

The titration curve for ${\text{Na"_2"CO}}_{3}$ looks like this:

At the endpoint, you will have neutralized only the ${\text{Na"_2"CO}}_{3}$.

You will not yet have neutralized any ${\text{NaHCO}}_{3}$, so the milliequivalents of ${\text{NaHCO}}_{3}$ neutralized are zero.

Experiment II

In Experiment II, you are titrating both the ${\text{NaHCO}}_{3}$ that was formed in Experiment I and the ${\text{NaHCO}}_{3}$ that was present in the beginning.

Thus, the milliequivalents of base (${\text{NaHCO}}_{3}$) are those you formed in Experiment I plus those that were initially present.