What is the correct increasing order of solubility of the following salt?

$N a H C O_{3}, K H C O_{3}, M g (H C O_{3})_{2}, C a (H C O_{3})_{2}$

K H C O_{3} < M g (H C O_{3})_{2} < C a (H C O_{3})_{2} < N a H C O_{3}

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M g (H C O_{3})_{2} < C a (H C O_{3})_{2} < N a H C O_{3} < K H C O_{3}

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N a H C O_{3} < K H C O_{3} < M g (H C O_{3})_{2} < C a (H C O_{3})_{2}

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C a (H C O_{3})_{2} < N a H C O_{3} < K H C O_{3} < M g (H C O_{3})_{2}

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Solution

The correct option is D $N a H C O_{3} < K H C O_{3} < M g (H C O_{3})_{2} < C a (H C O_{3})_{2}$
In the give metal bicarbonates, the anion (bicarbonate ion) is the same. Hence the solubility will depend on the size of cation. The size of $N a^{+}$ ion is smaller than that of $K^{+}$ ion. Hence sodium bicarbonate has high lattice energy and lower solubility than potassium bicarbonate. Similarly, the size of $M g^{+}$ ion is smaller than that of $C a^{+}$ ion. Hence magnesium bicarbonate has high lattice energy and lower solubility than calcium bicarbonate.
Hence, the correct option is $(C)$

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