A solution contains only $Na X_{2} CO X_{3}$ and $NaHCO X_{3}$ . It is titrated with $0.100 M$ $HCl$ : the phenolphthalein end point is reached after $12.0 mL$ and the methyl orange end point (total volume from the start) after $32.0 mL$ . Calculate the mass of $NaHCO X_{3}$ in the sample in mg (nearest integer; $NaHCO X_{3} = 84 g mo l^{- 1}$ ).

Numerical answer type, answer in mg, tolerance 0.5.

Answer: 67 mg

Phenolphthalein step converts

Na X_{2} CO X_{3} NaHCO X_{3}

Na X_{2} CO X_{3} = 0.100 \times 12.0 = 1.20

mmol. The remaining

32.0 - 12.0 = 20.0 mL

neutralises all bicarbonate (original plus formed), i.e.

2.00

mmol, of which

1.20

mmol came from carbonate. Original

NaHCO X_{3} = 2.00 - 1.20 = 0.80

mmol, mass

= 0.80 \times 84 = 67.2

mg.