At 25^{circ}C,the solubility product (K_{sp}) | vedclass

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

Question

(A) The dissociation of $M(OH)_2$ is given by: $M(OH)_2(s) \rightleftharpoons M^{2+}(aq) + 2OH^-(aq)$.Let the solubility be $s$. Then $[M^{2+}] = s$ a

Vedclass · Accepted Answer

$4 - \log 4$

Vedclass · Answer

(A) The dissociation of $M(OH)_2$ is given by: $M(OH)_2(s) ightleftharpoons M^{2+}(aq) + 2OH^-(aq)$.Let the solubility be $s$. Then $[M^{2+}] = s$ and $[OH^-] = 2s$.The solubility product expression is $K_{sp} = [M^{2+}][OH^-]^2 = (s)(2s)^2 = 4s^3$.Given $K_{sp} = 32 	imes 10^{-12}$,we have $4s^3 = 32 	imes 10^{-12}$,which simplifies to $s^3 = 8 	imes 10^{-12}$.Taking the cube root,$s = 2 	imes 10^{-4} \ M$.The concentration of hydroxide ions is $[OH^-] = 2s = 2 	imes (2 	imes 10^{-4}) = 4 	imes 10^{-4} \ M$.The $pOH$ is calculated as $pOH = -\log[OH^-] = -\log(4 	imes 10^{-4}) = -(\log 4 + \log 10^{-4}) = -(\log 4 - 4) = 4 - \log 4$.

At $25^{\circ}C$,the solubility product $(K_{sp})$ of $M(OH)_2$ is $32 \times 10^{-12}$. What is the $pOH$ of a saturated solution of $M(OH)_2$?

Similar Questions

The solubility of a sparingly soluble salt $AX_2$ is $1 \times 10^{-4} \ mol \ dm^{-3}$ at $298 \ K$. Calculate its solubility product $(K_{sp})$?

If the solubility product of $CaSO_4$ is $2.5 \times 10^{-5}$,then its solubility will be .......

Which statement is incorrect for the solubility product $(K_{sp})$?

The best explanation for the solubility of $MnS$ in dil. $HCl$ is that

How many grams of $CaC_2O_4$ will dissolve in distilled water to make one litre of saturated solution (in $g$)? (Solubility product of $CaC_2O_4$ is $2.5 \times 10^{-9} \ mol^2 \ L^{-2}$ and its molecular weight is $128$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module