A $20.0 mL$ sample containing only $Na X_{2} CO X_{3}$ and $NaHCO X_{3}$ is titrated with $0.150 M$ $HCl$ . The phenolphthalein end point is at $9.6 mL$ . After that, a further $24.4 mL$ is needed to reach the methyl orange end point. Find the concentration of $NaHCO X_{3}$ originally present in g per litre (two decimals; $NaHCO X_{3} = 84 g mo l^{- 1}$ ).

Numerical answer type, answer in g/L, tolerance 0.05.

Answer: 9.32 g/L

Phenolphthalein step:

Na X_{2} CO X_{3} = 0.150 \times 9.6 = 1.440

mmol, which forms

1.440

mmol

HCO X_{3} X^{-}

. The second step (

24.4 mL

) titrates all bicarbonate:

0.150 \times 24.4 = 3.660

mmol. Original

NaHCO X_{3} = 3.660 - 1.440 = 2.220

mmol. Concentration

= 2.220 \times 84/20.0 = 9.32 g L^{- 1}