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Introduction and Degree of Hydrolysis

25.3g of sodi...

Question

$25.3 g$ of sodium carbonate, ${N a}_{2} {C O}_{3}$ is dissolved in enough water to make $250 m L$ of solution. If sodium cabonate dissociates completely, molar concentration of sodium ion, ${N a}^{+}$ and carbonate ion ${C O}_{3}^{2 -}$ are respectively: (Molar mass of ${N a}_{2} {C O}_{3} = 106 g {m o l}^{- 1}$ )

0.955 M

and

1.910 M

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1.910 M

and

0.955 M

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1.90 M

and

1.910 M

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0.477 M

and

0.477 M

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Solution

The correct option is B $1.910 M$ and $0.955 M$
Number of moles of $N a_{2} C O_{3} = \frac{25.3}{105.96} = 0.238 m o l$

No. of moles of $N a_{2} C O_{3} / v o l u m e i n l i t e r s$ $= \frac{0.238}{0.25}$ $= 0.952 \frac{m o l e}{L}$ $(1000 m l = 1 L)$

Finding the no. of ions:
$N a_{2} C O_{3} ⟶ 2 N a^{+} + C O_{3}^{- -}$
$(2 m o l e s o f N a^{+} a n d 1 m o l e o f C O_{3}^{- -})$

Finding molarity
$M o l a r i t y o f N a^{+} = n o . o f m o l e s o f i o n s i n s o l u t i o n$
$= 2 \times 0.952 = 1.904 m o l e s o f N a^{+} / l i t e r$
$= 1.904 M$
Molarity of $C O_{3}^{- -} = 1 \times 0.952$
$= 0.952 m o l e s o f c a r b o n a t e i o n / L i t e r$ .
So $B$ option is correct.

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