The solubility product of Mg(OH)_2 is 1.2 tim | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The dissociation of $Mg(OH)_2$ is: $Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^-$Let $S$ be the solubility in $mol/L$. Then $K_{sp} = [Mg^{2+}][OH^

Vedclass · Accepted Answer

$8.16 	imes 10^{-4}$

Vedclass · Answer

(B) The dissociation of $Mg(OH)_2$ is: $Mg(OH)_2 ightleftharpoons Mg^{2+} + 2OH^-$Let $S$ be the solubility in $mol/L$. Then $K_{sp} = [Mg^{2+}][OH^-]^2 = (S)(2S)^2 = 4S^3$.Given $K_{sp} = 1.2 	imes 10^{-11}$,we have $4S^3 = 1.2 	imes 10^{-11}$,so $S^3 = 0.3 	imes 10^{-11} = 3 	imes 10^{-12}$.$S = \sqrt[3]{3 	imes 10^{-12}} = 1.44 	imes 10^{-4} \ mol/L$.The molar mass of $Mg(OH)_2 = 24 + 2(16 + 1) = 58 \ g/mol$.Solubility in $g/L = S 	imes 	ext{Molar mass} = 1.44 	imes 10^{-4} 	imes 58 \approx 8.35 	imes 10^{-3} \ g/L$.Solubility in $g/100 \ cm^3 = \frac{8.35 	imes 10^{-3}}{10} = 8.35 	imes 10^{-4} \ g/100 \ cm^3$. The closest option is $B$.

The solubility product of $Mg(OH)_2$ is $1.2 \times 10^{-11}$. The solubility of this compound in gram per $100 \ cm^3$ of solution is

Similar Questions

What will be the solubility of $AlCl_3$ in a solution of $CaCl_2$ with concentration $C$?

On the addition of a solution containing $CrO_4^{2-}$ ions to a solution containing $Ba^{2+}$,$Sr^{2+}$,and $Ca^{2+}$ ions,the precipitate obtained first will be of

Sodium sulfate is soluble in water,whereas barium sulfate is sparingly soluble. Why?

The solubility product of a sparingly soluble salt $AX$ is $4.9 \times 10^{-13}$. What is its solubility in $mol \ dm^{-3}$?

Solubility product $(K_{sp})$ of the salts $MX$,$MX_2$,and $M_3X$ at temperature $T$ are $4 \times 10^{-8}$,$3.2 \times 10^{-14}$,and $2.7 \times 10^{-15}$ respectively. The solubilities of these salts at temperature $T$ are in the order:

Vedclass Test Series

Exam Paper Generator

Online Exam Module