Which of the following will have lowest value of $K_{s p}$ at room temperature?

B e (O H)_{2}

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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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Solution

The correct option is A $B e (O H)_{2}$
${B e (O H)}_{2}$ is least soluble in water due to low hydration energy of ${B e}^{2 +}$ and covalent nature. It has lowest solubility product.
${B e (O H)}_{2} ⇌ {B e}^{2 +} + 2 O H^{-}$
$K_{s p} = {[B e}^{2 +}] {[O H^{-}]}^{2}$

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Similar questions

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}

K_{S P}

K_{s p}

K_{s p}

25^{o} 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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