Which hydroxide will have lowest value of solubility product at normal temperature $(25^{o} C)$ ?

M g {(O H)}_{2}

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C a {(O H)}_{2}

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B a {(O H)}_{2}

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B e {(O H)}_{2}

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Solution

The correct option is D $B e {(O H)}_{2}$
As we move down the group of alkaline earth metals lattice enthalpy decreases much faster than hydration enthalpy so solubility of hydroxides increases down the group. So $B e (O H)_{2}$ is least soluble and lowest value of solubility product.

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{M g (O H)}_{2}

5 \times 10^{- 19}

25^{o} C

{M g (O H)}_{2}

H_{2} O ?

(1) B e S O_{4} > M g S O_{4} > C a S O_{4} > B a S O_{4}

(2) B e (O H)_{2} > M g (O H)_{2} > C a (O H)_{2} > B a (O H)_{2}

(3) L i O H < N a O H < K O H < R b O H

(4) L i_{2} C O_{3} < N a_{2} C O_{3} < K_{2} C O_{3} < R b_{2} C O_{3}

p H

M g (O H)_{2}

10.45

25^{\circ} C

B a (O H)_{2} > S r (O H)_{2} > C a (O H)_{2} > B e (O H)_{2}

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