Titration of a mixture of $Na_{2}CO_{3}$ and $NaHCO_{3}$

Hi shama, let's solve this titration question step-by-step. First, let's look at the given tables and find the concordant values of the volume of hydrochloric acid used.

For the first table with phenolphthalein, we see that the concordant volumes are ten point zero zero cubic centimeters for trials A and C. Let's find their average.

For the second table with methyl orange, the concordant volumes are nineteen point zero zero cubic centimeters for trials A and C. Let's find their average as well.

Since the volume of the mixture used for titration is typically twenty-five cubic centimeters in standard chemistry practicals, we will use twenty-five cubic centimeters as our base volume, but we will also express our final formulas in terms of a general volume V.

Now, let's address part a, which asks for the balanced chemical equations. First, let's look at the phenolphthalein endpoint.

At the phenolphthalein endpoint, only sodium carbonate reacts with hydrochloric acid to form sodium hydrogencarbonate and sodium chloride.

Now, let's look at the methyl orange endpoint. At this stage, the titration continues, and the sodium hydrogencarbonate is neutralized by the acid.

Let's move on to part b, starting with finding the concentration of sodium carbonate in the mixture.

The volume of hydrochloric acid used to reach the phenolphthalein endpoint is ten cubic centimeters. This volume is only used to convert sodium carbonate to sodium hydrogencarbonate.

Since the concentration of the acid is zero point one molar, the moles of acid used is zero point one times ten divided by one thousand, which is zero point zero zero one zero moles.

From our balanced equation, one mole of sodium carbonate reacts with one mole of hydrochloric acid. Therefore, the moles of sodium carbonate is also zero point zero zero one zero moles.