1.50 g of a solid containing NaOH, $Na X_{2} CO X_{3}$ and inert matter is dissolved in water and made up to 250 mL. A 25.0 mL aliquot needs 22.0 mL of 0.100 M HCl to reach the phenolphthalein end point, and a further 8.0 mL to the methyl orange end point. Find the mass percentage of $Na X_{2} CO X_{3}$ in the solid ( $Na X_{2} CO X_{3}$ = 106 g/mol; one decimal place).

Numerical answer type, answer in %, tolerance 0.06.

Answer: 56.5 %

The extra 8.0 mL (second stage) corresponds to

Na X_{2} CO X_{3}

: 0.100

\times

8.0 = 0.800 mmol per aliquot. For 250 mL this is 8.00 mmol = 0.848 g. Percentage = 0.848/1.50

\times

100 = 56.5 % (the first-stage volume would additionally give NaOH, which is not asked).