Question

The solubility of the ionic compound MX3, having a molar mass of 288 g/mol, is 3.60 x 10-2 g/L. Calculate the KSP of the compound.

Answer 1

$K_{sp}$ of the compound is found to be  5.04 ×$10^{-10}$.

Solubility :Solubility can be define as the amount of a substance that dissolves or mixes in a given amount of solvent at specific conditions.

Solubility equilibrium

Ksp = $[A^{+} ]^{a}$ $[B^{-} ]^{b}$

Ksp = solubility product constant

A+ = cation in an aquious solution

B- = anion in an aqueous solution

a, b = relative concentrations of a and b

Given,

Solubility = s = 3.60 × $10^{-2}$ g/L

molar mass = 288 g/ mol

∴ s= 3.60 × $10^{-2}$ g/L ÷ 288 g/ mol = 1.25 ×$10^{-4}$ mol/ L

Reaction:

MX3 ⇄ M + 3X

           s       3s

$K_{sp}$ =[ $M^{+3}$] [ $X^{-1}$$]^{3}$ = solubility product

∴ $K_{sp}$ =$[s]^{} [3s]^{3}$

∴ $K_{sp}$ = 3 $s^{4}$

∴ $K_{sp}$ = 3 × (3.60 × $10^{-2}$ $)^{4}$

∴ $K_{sp}$ = 503.8848 ×$10^{-8}$  = 5.04 ×$10^{-10}$

Learn more about solubility here .....

brainly.com/question/23946616

#SPJ1