$25.0 mL$ of a solution containing $NaOH$ and $Na X_{2} CO X_{3}$ is titrated with $0.100 M$ $HCl$ . With phenolphthalein the end point is at $18.0 mL$ ; adding methyl orange, a further $6.0 mL$ is needed. Taking NaOH = 40, which of the following are correct?

AThe $Na X_{2} CO X_{3}$ in the portion is $0.600$ mmol.
BThe mass of $NaOH$ in the portion is $0.060 g$ .
CThe total volume of acid from the start to the methyl orange end point is $24.0 mL$ .
DThe $NaOH$ in the portion is $1.20$ mmol.

Answer: A, C, D

The second stage (

HCO X_{3} X^{-}

H X_{2} CO X_{3}

) used 6.0 mL, so carbonate = 6.0 x 0.100 = 0.600 mmol. First stage consumed NaOH + carbonate: 1.80 mmol, so NaOH = 1.80 - 0.600 = 1.20 mmol = 0.048 g (not 0.060 g). Total titre = 18.0 + 6.0 = 24.0 mL.

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25.0 mL of a solution containing NaOH and NaX2​COX3​ is titrated with 0.100 M HCl. With phenolphthalein the end point is at 18.0 mL; adding methyl orange, a further 6.0 mL is needed. Taking NaOH = 40, which of the following are correct?

Answer: A, C, D

More questions on Back titration and hardness of water

$25.0 mL$ of a solution containing $NaOH$ and $Na X_{2} CO X_{3}$ is titrated with $0.100 M$ $HCl$ . With phenolphthalein the end point is at $18.0 mL$ ; adding methyl orange, a further $6.0 mL$ is needed. Taking NaOH = 40, which of the following are correct?