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Introduction and Degree of Hydrolysis

25.3 g of sod...

Question

25.3 g of sodium carbonate is dissolved in enough water to make 250 ml of solution. If sodium carbonate dissociates completely, then what is the molar concentration of sodium ions and carbonate ions?

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Solution

Moles of $N a_{2} C O_{3} = \frac{m a s s}{m o l a r m a s s}$
$= \frac{25.3}{106}$
$= 0.239 m o l e s$
Molarity of $N a_{2} C O_{3}$ solution $= \frac{n u m b e r o f m o l e s}{v o l u m e i n l i r e}$
$= \frac{0.239}{0.25} (250 m l = 0.25 L)$
$= 0.955 M$
Now, 1 mole of $N a_{2} C O_{3}$ contains 2 moles of $N a^{+}$ and 1 mole of $C O_{3}^{2 -}$ ions.
Therefore,
Concentration of $N a^{+} = 2 \times 0.955 M = 1.91 M$
Concentration of $C O_{3}^{2 -} = 1 \times 0.955 M = 0.955 M$

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