Question

At 25 °C, an aqueous solution has an equilibrium concentration of 0.00489 M for a generic...

At 25 °C, an aqueous solution has an equilibrium concentration of 0.00489 M for a generic cation, A (aq), and 0.00245 M for a generic anion, B2–(aq). What is the equilibrium constant, Ksp, of the generic salt A2B(s)?

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Answer #1
Concepts and reason

The concept used in this question is based on the solubility product. The solubility product of the given salt needs to be calculated. That is why this concept is used in this question.

Initially, the expression for the solubility product of the given salt is written. Then, the solubility product of the given salt is calculated by using the expression of the solubility product.

Fundamentals

The solubility product of any salt can be expressed as the product of the concentration of the ions of product that are formed in the solution raised to the power of their stoichiometric coefficients.

For example, consider a salt AX. Its dissociation can be expressed by the following equation.

AX(s) =
A(aq) + X (aq)

The solubility product at the equilibrium of the above reaction can be given as shown below.

[-x][_v]= y
…… (1)

Here, the solubility product is Ksp, the concentration of the products so formed are and .

The balanced chemical equation for the dissociation of at equilibrium is shown below.

A,B(s)
=
2A+ (aq) + B?- (aq)

The solubility product of the above reaction can be expressed as follows.

[-a]][-v]= x

Substitute the given value of [A*] =0.00489 M
and [B2- ]=0.00245 M
in equation (1).

K. =[A+][B]
= (0.00489 M)” x(0.00245 M)
= 5.86x10-8

Therefore, the solubility product of the given salt (A,B)
is 5.86x10 -
.

Ans:

The solubility product of the given salt (A,B)
is 5.8610
.

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