Question #63681

1 Answer
Ernest Z.
Apr 9, 2017

In Experiment I, you are titrating only the "Na"_2"CO"_3, while in Experiment II you are titrating the "NaHCO"_3 formed in Experiment I plus what was present in the original mixture.

Experiment I

Experiment IExperiment I

In Experiment I, you are titrating the mixture of "Na"_2"CO"_3 and "NaHCO"_3 to a phenolphthalein end-point.

Phenolphthalein changes colour around pH 9.

The titration curve for "Na"_2"CO"_3 looks like this:

www.chemguide.co.ukwww.chemguide.co.uk

At the endpoint, you will have neutralized only the "Na"_2"CO"_3.

You will not yet have neutralized any "NaHCO"_3, so the milliequivalents of "NaHCO"_3 neutralized are zero.

Experiment II

Experiment IIExperiment II

In Experiment II, you are titrating both the "NaHCO"_3 that was formed in Experiment I and the "NaHCO"_3 that was present in the beginning.

Thus, the milliequivalents of base ("NaHCO"_3) are those you formed in Experiment I plus those that were initially present.

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