What is the concentration of $C O_{3}^{2 -}$ in a $0.1 M N a_{2} C O_{3}$ solution after the hydrolysis reactions have come to an equilibrium? ( $K_{a} (H C O_{3}^{-}) = 4.7 \times 10^{- 11}$ )

2 \times 10^{4}

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6.3 \times 10^{- 4}

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2 \times 10^{- 4}

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9.54 \times 10^{- 2}

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Solution

The correct option is D $9.54 \times 10^{- 2}$
The first step of hydrolysis of $C O_{3}^{2 -}$ ions is predominant.
$C O_{3}^{2 -} (a q) 0.1 (1 - h) + H_{2} O (l) \to H C O_{3}^{-} (a q) 0.1 h + O H^{-} (a q) 0.1 h$
$K_{h} = \frac{0.1 h^{2}}{(1 - h)}$
$K_{h} = \frac{K_{w}}{K_{a}} = \frac{1 \times 10^{- 14}}{4.7 \times 10^{- 11}} K_{h} = 2.12 \times 10^{- 4}$
Finding $\frac{C}{K_{h}}$

$\frac{C}{K_{h}} = \frac{0.1}{2.12 \times 10^{- 4}} = 471.7$
Since $\frac{C}{K_{h}} > 100$ , we can use $1 - h \approx 1$
$K_{h} = 0.1 \times h^{2} h = \sqrt{\frac{2.12 \times 10^{- 4}}{0.1}} = 0.046 [C O_{3}^{2 -}]_{e q b} = 0.1 \times (1 - h) = 0.1 \times (1 - 0.046) [C O_{3}^{2 -}]_{e q b} = 9.54 \times 10^{- 2}$

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