Solubility product expression of salt MX_4 wh | vedclass

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

Question

(A) The dissociation of the salt $MX_4$ is represented as: $MX_{4(s)} \rightleftharpoons M_{(aq)}^{4+} + 4X_{(aq)}^{-}$ If the solubility of the salt

Vedclass · Accepted Answer

$256S^5$

Vedclass · Answer

(A) The dissociation of the salt $MX_4$ is represented as: $MX_{4(s)} ightleftharpoons M_{(aq)}^{4+} + 4X_{(aq)}^{-}$ If the solubility of the salt is $S$,then the concentration of $M^{4+}$ is $S$ and the concentration of $X^{-}$ is $4S$. The solubility product constant $K_{sp}$ is given by the product of the concentrations of the ions raised to their stoichiometric coefficients: $K_{sp} = [M^{4+}][X^{-}]^4$ Substituting the values: $K_{sp} = (S)(4S)^4$ $K_{sp} = S 	imes 256S^4$ $K_{sp} = 256S^5$

Solubility product expression of salt $MX_4$ which is sparingly soluble with a solubility $(S)$ can be given as

Similar Questions

The solubility product of $AgCl$ is $1.5625 \times 10^{-10}$ at $25\ ^oC$. Its solubility in grams per litre will be

The solubility of a salt $M_2X_3$ is $y \ mol \ m^{-3}$. Its solubility product is:

The solubility product of magnesium hydroxide $Mg(OH)_2$ is $1.2 \times 10^{-11}$. Calculate its solubility in pure water and $0.05 \ M$ $NaOH$.

The $K_{sp}$ of $AgBr$ at $25 \, ^\circ C$ is $4.9 \times 10^{-13}$. If the molar mass of $AgBr$ is $188 \, g/mol$,how much $AgBr$ is required to saturate $20 \, L$ of water?

At $T(K)$,the solubility product of $AgBr$ is $4 \times 10^{-13}$. What is its solubility in $0.1 \ M$ $KBr$ solution?

Vedclass Test Series

Exam Paper Generator

Online Exam Module