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During boiling of water containing temporary hardness $M g {(H C O_{3})}_{2}$ converted in to $M g {(O H)}_{2}$ but $C a {(H C O_{3})}_{2}$ converts into $C a C O_{3}$ . the reason is

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$K_{s p} o f M g C O_{3} > K_{s p} o f M g {(O H)}_{2}$

$K_{s p} o f C a {(O H)}_{2} > K_{s p} o f C a C O_{3}$

$K_{s p} o f M g {(O H)}_{2} > K_{s p} o f M g {(O H)}_{3}$

$K_{s p} o f M g {(O H)}_{2} = K_{s p} o f M g C O_{3}$

answer is B.

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Detailed Solution

$K_{s p} o f M g C O_{3} = 3.5 \times 10^{- 8} K_{s p} o f M g {(O H)}_{2} = 1.8 \times 10^{- 11}$

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