$25.3 ext{ g}$ of Sodium carbonate $ ext{Na}_2 ext{CO}_3$ is dissolved in enough water to make $250 ext{ m

Question

$25.3	ext{ g}$ of Sodium carbonate $	ext{Na}_2	ext{CO}_3$ is dissolved in enough water to make $250	ext{ mL}$ of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, $	ext{Na}^+$ and carbonate ion $	ext{CO}_3^{2-}$ are respectively (Molar mass of $	ext{Na}_2	ext{CO}_3 = 106	ext{ g mol}^{-1}$)

Accepted Answer

Moles of $	ext{Na}_2	ext{CO}_3 = \frac{	ext{Mass}}{	ext{Molar mass}} = \frac{25.3	ext{ g}}{106	ext{ g mol}^{-1}} = 0.2387	ext{ mol}$.
Volume of solution $= 250	ext{ mL} = 0.250	ext{ L}$.
Molarity of $	ext{Na}_2	ext{CO}_3 = \frac{0.2387	ext{ mol}}{0.250	ext{ L}} = 0.9548	ext{ M} \approx 0.955	ext{ M}$.
The dissociation reaction of sodium carbonate is:
$	ext{Na}_2	ext{CO}_3 ightarrow 2	ext{Na}^+ + 	ext{CO}_3^{2-}$
Since $1	ext{ mole}$ of $	ext{Na}_2	ext{CO}_3$ gives $2	ext{ moles}$ of $	ext{Na}^+$ and $1	ext{ mole}$ of $	ext{CO}_3^{2-}$ ,
Concentration of $	ext{Na}^+ = 2 	imes 0.955	ext{ M} = 1.910	ext{ M}$.
Concentration of $	ext{CO}_3^{2-} = 1 	imes 0.955	ext{ M} = 0.955	ext{ M}$.

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