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Types of Solution

If 1.06 g o...

Question

If $1.06$ g of $N a_{2} C O_{3}$ is dissolved in enough water to make $250$ mL of solution, then the molar concentrations of $N a^{+}$ is $x \times 10^{- 2} M$ . Here, $x$ is :

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Solution

The dissociation equilibrium is $N a_{2} C O_{3} ⇋ 2 N a^{+} + {C O_{3}}^{2 -}$ .
The molar concentration of sodium carbonate is $\frac{1.06 g}{106 g / m o l} \times \frac{1000 m L}{250 m L} \times 1 L = 0.04 M$ .
This is also equal to the molar concentration of carbonate ion.
The molar concentration of sodium ion is equal to $2 \times 0.04 M = 0.08 M = 8 \times 10^{- 2} M$ .

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