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

Which one of the following is most soluble?

The solubility of $AgCl$ in its solution is $1.25 \times 10^{-5} \ mol \ dm^{-3}$. What is the solubility product of $AgCl$?

In the reaction $Fe(OH)_{3(s)} \rightleftharpoons Fe^{3+}_{(aq)} + 3OH^{-}_{(aq)}$, if the concentration of $OH^{-}$ ions is decreased by $\frac{1}{4}$ times, then the equilibrium concentration of $Fe^{3+}$ will increase by: (in $\times$)

The molar solubility of $PbI_2$ in $0.2 \ M \ Pb(NO_3)_2$ solution in terms of solubility product of $PbI_2$ is

The solubility in water of a sparingly soluble salt $AB_2$ is $1.0 \times 10^{-5} \ mol \ L^{-1}$. Its solubility product constant will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module