25.0 mL of a solution of Na2CO3 and NaHCO3 is titrated with 0.100 M HCl. The phenolphthalein end point (CO3^2- -> HCO3^-) is reached after 12.0 mL; a further 18.0 mL is needed to reach the methyl orange end point. Which statements are correct?

The molarity of Na2CO3 in the sample is 0.048 M.

Na2CO3 in the sample is 1.20 mmol.

NaHCO3 originally present is 0.60 mmol.

The molarity of Na2CO3 in the sample is 0.048 M.

Na2CO3 in the sample is 1.20 mmol.

The total volume of acid from the start to the methyl orange end point is 24.0 mL.

NaHCO3 originally present is 0.60 mmol.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3eq220

$25.0 mL$ of a solution of $Na X_{2} CO X_{3}$ and $NaHCO X_{3}$ is titrated with $0.100 M$ $HCl$ . The phenolphthalein end point ( $CO X_{3} X^{2 -} \to HCO X_{3} X^{-}$ ) is reached after $12.0 mL$ ; a further $18.0 mL$ is needed to reach the methyl orange end point. Which statements are correct?

AThe molarity of $Na X_{2} CO X_{3}$ in the sample is $0.048 M$ .
B $Na X_{2} CO X_{3}$ in the sample is $1.20$ mmol.
CThe total volume of acid from the start to the methyl orange end point is $24.0 mL$ .
D $NaHCO X_{3}$ originally present is $0.60$ mmol.

Up to phenolphthalein,

12.0 \times 0.100 = 1.20

mmol of acid converts all

CO X_{3} X^{2 -}

HCO X_{3} X^{-}

, so

Na X_{2} CO X_{3}

= 1.20

mmol,

0.048 M

. The second stage uses

18.0 \times 0.100 = 1.80

mmol for all

HCO X_{3} X^{-}

(1.20 from carbonate), so original

NaHCO X_{3}

= 0.60

mmol. Total acid volume is

12.0 + 18.0 = 30.0 mL

, not 24.0.