10 ml of $N a H C O_{3}$ and $N a_{2} C O_{3}$ solution requires 2 ml of 0.1 M $H_{2} S O_{4}$ using phenolphthalein as indicator. Weight of $N a_{2} C O_{3}$ in mixture is (in gr)

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Detailed Solution

To neutralize whole $N a_{2} C O_{3}$ , Volume of $H_{2} S O_{4}$ will be twice of given volume. $N a_{2} C O_{3} + H^{+} \to N a H C O_{3} + N a^{+}$

$N_{1} V_{1} = N_{2} V_{2} N_{1} = \frac{N_{2} V_{2}}{V_{1}} N_{1} = \frac{(2 \times 0.1) \times (2 \times 2)}{10} = 0.08 N$

Weight of $N a_{2} C O_{3}$ in 1 $L = 0.08 \times 53$ $= 4.24 g$
Ln 1 L weight of $N a_{2} C O_{3} = 4.24 g$
Ln 10 ml $N a_{2} C O_{3}$ will be $= \frac{4.24}{1000} \times 10$ $= 0.0424 g$