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Common Ion Effect

25 mL solutio...

Question

$25$ mL solution containing a mixture of $N a_{2} C O_{3}$ and $N a O H$ required $19.5$ mL of $0.995 N$ $H C l$ for the phenolpthalein end-point and in a separate titration, $25$ mL of the same mixture required $25.9$ mL of same $H C l$ for methyl orange end-point. The concentration (in g/L) of $N a_{2} C O_{3}$ in the mixture is $9 X$ . Find the value of $X$ .

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Solution

$25.9 - 19.5 = 6.4 m L$ of $0.995 N H C l$ corresponds to the half neutralization of $N a_{2} C O_{3}$ .
The number of moles of $N a_{2} C O_{3}$ is $0.995 N \times 6.4 \times 10^{- 3} = 6.368 \times 10^{- 3} m o l e s$
The concentration of $N a_{2} C O_{3}$ is $\frac{6.368 \times 10^{- 3} m o l e s \times 106 g / m o l}{25 \times 10^{- 3} L} = 27 g / L$ .
value of $X$ = $3$
$19.5 - 6.4 = 13.1 m L$ of $0.995 N H C l$ corresponds to the neutralization of $N a O H$ .
The number of moles of $N a O H$ is $0.995 N \times 13.1 \times 10^{- 3} = 13.03 \times 10^{- 3} m o l e s$
Thus, the concentration of $N a O H$ is $\frac{13.03 \times 10^{- 2} m o l e s \times 40 g / m o l}{25 \times 10^{- 3} L} = 20.86 g / L$

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