A $250 mL$ flask holds a solution of $NaOH$ and $Na X_{2} CO X_{3}$ . A $25.0 mL$ aliquot is titrated with $0.1000 M$ $HCl$ : the phenolphthalein end point is at $21.4 mL$ and the methyl orange end point is at a total of $28.6 mL$ . Find the mass of $NaOH$ in the whole $250 mL$ flask, in grams (three decimals; $NaOH = 40$ ).

Numerical answer type, answer in g, tolerance 0.003.

Answer: 0.568 g

Let

x

and

y

be mmol of

NaOH

and

Na X_{2} CO X_{3}

in the aliquot. Phenolphthalein:

x + y = 0.1000 \times 21.4 = 2.14

. Methyl orange (total):

x + 2 y = 0.1000 \times 28.6 = 2.86

. So

y = 0.72

and

x = 1.42

mmol. The flask holds

10 \times

this:

14.2

mmol

\times 40 mg = 0.568 g