Find the molar solubility of BaCrO4(Ksp=2.1 ×10^-10) in (a) pure water and (b) 1.5 ×10^-3 M Na2

Find the molar solubility of BaCrO4(Ksp=2.1 ×10^-10) in (a)...

Key Concepts

Solubility Product (Ksp)

The solubility product is the equilibrium constant for the dissolution of a sparingly soluble ionic compound into its constituent ions in a saturated solution. It quantitatively expresses the product of the ion concentrations, each raised to the power of its stoichiometric coefficient, and is a crucial factor in determining the solubility of the compound under various conditions.

Molar Solubility

Molar solubility refers to the number of moles of a solute that can dissolve per liter of solution to form a saturated solution. It is a direct measure of how much of a sparingly soluble salt can dissolve in a given solvent, finding its importance in understanding and calculating saturation in chemical systems.

Common Ion Effect

The common ion effect describes the reduction in solubility of a sparingly soluble salt when a solution already contains one of the ions present in the salt. This phenomenon arises because the additional ion shifts the equilibrium position according to Le Châtelier's principle, leading to a decrease in the amount of salt that can dissolve.

Equilibrium Expressions for Sparingly Soluble Salts

This concept involves writing the algebraic expression of the solubility product for a sparingly soluble salt, where the product of the dissolved ion concentrations (each raised to the appropriate power) is equated to its Ksp value. It is essential for solving problems related to solubility in both pure water and solutions containing common ions.

Transcript

00:01 So here we're given information about specific chemical reaction where barium chromate solid may dissociate to a small amount to barium ion and our chromate anion.

00:11 And we're given that the ksp is equivalent to 2 .1 times 10 to the negative 10.

00:18 And essentially the ksp is equivalent to the concentrations of the products of the reactants and basically concentration of species that changed significantly, so we're only interested in our aqueous species, since the concentration of our solid species don't change significantly.

00:36 So we can write this expression.

00:39 And we know that when, essentially, when one of these compounds dissociate, the other compound dissociates in the more than one of these, when barium chromate dissociates, the barium ion and the chromate ion are produced with equal quantities.

00:55 So the concentration of barium is equivalent to the concentration of chromate and we can just write this as s or the solubility is equivalent to the concentration of barium times the concentration of chromate.

01:07 So the solubility squared is equivalent to 2 .1 times 10 to the negative 10.

01:14 And we can evaluate this to be that the solubility is equivalent to 1 .4 times 10 to the negative 5 molars.

01:25 So this is for the first part.

01:29 And the second part, the question is essentially addressing the common ion effect, where we already have a significant concentration of chromate, and we're going to dissolve essentially more barium chromate.

01:42 So we're given that we have a significant concentration of chromates of 1 .5 times 10 to the negative 3 molars.

01:53 In this case, the dissociation is going to be much less due to the significant concentration of the chromate anion...

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