A $25.0 mL$ aliquot of a solution containing only $NaOH$ and $Na X_{2} CO X_{3}$ is titrated with $0.100 M$ $HCl$ . The phenolphthalein end point needs $20.0 mL$ and a further $5.0 mL$ takes it to the methyl orange end point. The mass of $NaOH$ in the aliquot ( $NaOH = 40$ ) is:

Answer: A

The second step converts

HCO X_{3} X^{-}

(formed from carbonate) to

H X_{2} CO X_{3}

Na X_{2} CO X_{3} = 0.100 \times 5.0 = 0.50

mmol. At phenolphthalein,

NaOH + Na X_{2} CO X_{3} = 0.100 \times 20.0 = 2.00

mmol, so

NaOH = 1.50

mmol. Mass

= 1.50 \times 40 = 60

mg.