Question

The solubility product of $A{g_2}Cr{O_4}$ is $2.6 \times {10^{ - 2}}$ at $25^\circ C$. Calculate the solubility of the compound.

Answer 1

Hint: Solubility product constants are used to describe saturated solutions of ionic compounds of relatively low solubility.

Complete step by step solution:
When two electrolytic solutions are combined, a precipitate may, or may not form. In order to determine whether or not a precipitate will form or not, two factors are examined. First, determine the possible combinations of ions that could result when the two solutions are combined if the concentrations of the ions are great enough so that the reaction quotient Q exceeds the ${K_{sp}}$ value.
$A{g_2}Cr{O_4}(s) \to 2A{g^ + }(aq) + Cr{O_4}^{2 - }(aq)$
${K_{sp}} = [A{g^{2 + }}][Cr{O_4}^{2 - }]$
Solubility $({K_{sp}})$ is the ability of a substance to dissolve. The solubility product constant $({K_{sq}})$ describes the equilibrium between a solid and its constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. For example, the higher the $({K_{sq}})$ the more soluble the compound is. $({K_{sq}})$ is defined in terms of activity rather than concentration because it is a measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. $({K_{sq}})$ is used to describe the saturated solution of ionic compounds.
Let "x" be the number of moles of silver chromate that dissolves in every litre of solution (its solubility)
Considering the rate of equation:
$2.6\times{10^{ - 2}}$ = ${[2x]^2}[x]$
$2.6\times{10^{ - 2}}$=$4{x^3}$
$x = 0.1866\,moles$

Note:
The solubility of a substance in a solvent is the total amount of the solute that can be dissolved in the solvent at equilibrium. On the other hand, the solubility product constant is an equilibrium constant that provides insight into the equilibrium between the solid solute and its constituent ions that are dissociated across the solution. The product solubility is dependent on the factor like The common-ion effect, The diverse-ion effect, the presence of ion-pairs