Question

At 25 °C, only 0.0240 mol of the generic salt AB3 is soluble in 1.00 L of water. What is the Ksp of the salt at 25 °C? AB,() = A3+ (aq) + 3B (aq) Ksp =

Answer 1

Concepts and reason

The concepts used to solve this problem are solubility and solubility product.

The solubility product, is directly proportional to the solubility of the salt. It is basically the product of the solubility of each ion formed after the dissociation of salt in water. These concentrations are raised to the power that is equal to the stoichiometric coefficients of that ion in a balanced equilibrium reaction.

Fundamentals

Solubility product is a kind of equilibrium constant. It is a product of solubility of dissociated ions of the ionic compound. The higher value of solubility product implies higher solubility of the ionic compound.

The molar solubility is defined as the number of moles of solute in of solution.

The molar solubility is calculated as:

Here, is the number of moles of salt and is the volume of the solution.

When the salt is dissolved in water, it dissociates into corresponding anions and cations. The equation for dissociation of is:

Now, calculate the molar solubility of the salt.

The molar solubility is calculated as:

Since, is dissolved in of water.

Therefore, substitute for and for.

1 mole of dissociates to form 1 mole of and 3 moles of .

Construct an equilibrium table as follows:

The expression for is:

Substitute for and for in the above expression.

Substitute for in the above equation.

Ans:

The of the salt at is .

Answer 2

At equilibrium:

AB3 <----> A3+ + 3 B-

   s 3s

Ksp = [A3+][B-]^3

Ksp = (s)*(3s)^3

Ksp = 27(s)^4

Ksp = 27(2.4*10^-2)^4

Ksp = 8.958*10^-6

Answer: 8.96*10^-6