Solveeit Logo
Login

Question

Question: Which of the following will have lowest value of \(K_{sp}\) at room temperature?...

Which of the following will have lowest value of KspK_{sp} at room temperature?

A

Be(OH)2Be(OH)_{2}

B

Mg(OH)2Mg(OH)_{2}

C

Ca(OH)2Ca(OH)_{2}

D

Ba(OH)2Ba(OH)_{2}

Answer

Be(OH)2Be(OH)_{2}

Explanation

Solution

: Be(OH)2Be(OH)_{2} is least soluble in water hence it will have lowest value of KspK_{sp}

Be(OH)2Be(OH)_{2} Be2++2OH−Be^{2 +} + 2OH^{-}

Ksp=[Be2+][OH−]2K_{sp} = \lbrack Be^{2 +}\rbrack\lbrack OH^{-}\rbrack^{2}