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

In which of the following solutions will the solubility of $Ag_2CO_3$ $(K_{sp} = 8 \times 10^{-12})$ be maximum at $30 \, ^\circ C$?

The expression for the solubility product of $Al_2(SO_4)_3$ is

Calculate the molar solubility of $Ni(OH)_{2}$ in $0.10 \, M$ $NaOH$. The solubility product constant $(K_{sp})$ of $Ni(OH)_{2}$ is $2.0 \times 10^{-15}$.

At $T(K)$ in a saturated solution of $MgCO_3$ and $Ag_2CO_3$, if the concentration of $Mg^{2+}$ ion is $3.2 \times 10^{-5} \ M$, then the concentration of $Ag^{+}$ ion in the solution will be [Given: $K_{sp}(MgCO_3) = 1.6 \times 10^{-6}$ and $K_{sp}(Ag_2CO_3) = 8.0 \times 10^{-12}$ at $T(K)$]

Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4} \ M \ Na_2CO_3$ solution. At which concentration of $Ba^{2+}$,precipitate of $BaCO_3$ begins to form? ($K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module