Question

What does Ksp in chemistry stand for?

Answer 1

We can say that solubility is the property of a solid, fluid or vaporous compound substance called solute to break up in a solid, fluid or vaporous dissolvable. We have to know that the dissolvability of a substance generally relies upon the physical and compound properties of the solute and dissolvable just as on temperature, and presence of different compounds (counting changes to the pH) of the solution.

Complete step by step answer:
In terms of equilibrium, solubility is characterized as the most extreme measure of solute that can be broken up in a dissolvable at balance.
Let us now define Ksp.
We can say Ksp is nothing but the solubility product constant.
We can say that the solubility product constant, Ksp, is the balance steady for a strong substance dissolving in an aqueous solution. It addresses the level at which a solute breaks up in solution. The more dissolvable a substance is, the higher the Ksp value it has.
When we take a binary salt $M{X_2}$, we can write the dissociation reaction as,
$M{X_2}\left( s \right)\overset {} \leftrightarrows {M^ + }\left( {aq} \right) + 2{X^ - }\left( {aq} \right)$
We can write the expression for the solubility product constant as,
${K_{sp}} = \dfrac{{\left[ {{M^ + }} \right]{{\left[ {{X^ - }} \right]}^2}}}{{\left[ {M{X_2}\left( s \right)} \right]}}$
We know that while we are writing equilibrium expressions, the compounds present in solid state are ignored. So, the expression for solubility product constant becomes,
${K_{sp}} = \left[ {{M^ + }} \right]{\left[ {{X^ - }} \right]^2}$
We have to know that Ksp indicated the maximum extent to which solid could be dissolved in water.

Note: We have to know that solubility of reaction could be decreased by common ions. The value of Ksp would lower for reaction which has common ion at given equilibrium and the value of Ksp would higher for reaction which does not have common ion at given equilibrium
We have to know that when the value of Ksp, the solubility of the compound would be more.