Question

At 25 °C only 0.0630 mol of the generic salt AB2 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C?

Answer 1

Concepts and reason

The concept used to solve this problem is to determine the product of total ion in the solutions because the solubility product is the product of the ion concentrations.

Fundamentals

The solubility product is denoted by ${K_{{\rm{sp}}}}$ , it is the product of the ion concentrations.

The molar solubility is denoted by $S$ . It is the number of moles of solute, which is dissolve in the 1.00 liter of the solution.

The balanced chemical reaction for dissolution of generic salt; is as follows:

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The solubility product ${K_{{\rm{sp}}}}$ , for the generic salt; is as follows:

Here, the solubility of is $S$ .

According to the problem; 0.0630 mol of the generic salt is soluble in 1.00 L of water therefore the molarity of salt is .

The solubility product is denoted by ${K_{{\rm{sp}}}}$ , it is the product of the ion concentrations.

The solubility product ${K_{{\rm{sp}}}}$ , for the generic salt; is as follows:

Herenow calculate the solubility from solubility product as follows:

Ans:

Thus the solubility product of is ..