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Stoichiometry

A 0.6234 g ...

Question

A $0.6234$ g sample that might contain $N a O H, N a_{2} C O_{3}, N a H C O_{3}$ or a mixture of $N a O H + N a_{2} C O_{3}$ or $N a_{2} C O_{3} + N a H C O_{3}$ is titrated with $0.1062 M H C l$ by two-indicator method. It is found that $40.38$ mL of the acid is required to reach the phenolphthalein end-point. Methyl orange is then added to the solution and the titration continued using an additional $12.83$ mL of the acid. Thus, the sample contained :

A

N a O H + N a_{2} C O_{3} + N a H C O_{3}

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B

N a O H + N a_{2} C O_{3}

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C

N a_{2} C O_{3} + N a H C O_{3}

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D

none of the above

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Solution

The correct option is A $N a O H + N a_{2} C O_{3}$
The phenolphthalein indicator shows the end point which corresponds to $50 %$ neutralization of sodium carbonate.
Methyl orange indicator gives the end point corresponding to $100 %$ of sodium carbonate.
Since, $40.38$ mL $>$ $12.83$ mL, the sample must contain $N a O H$ and $N a_{2} C O_{3}$ .

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