Question

Determine the molar solubility of PbSO4 in pure water. Ksp(PbSO4)= 1.82 x 10^-8.

Determine the molar solubility of PbSO4 in pure water. Ksp(PbSO4)= 1.82 x 10^-8.
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Answer #1
Concepts and reason

The concept of solubility product (K)
sp
is used in this problem.

The solubility of a given compound in pure water can be determined with the help of the expression of the solubility product. First, write the expression of solubility product for a given molecule and then calculate its solubility.

Fundamentals

Solubility:

Solubility is explained in terms of the maximum quantity of a solute that can be dissolved in the solvent and forms a solution. It is a chemical property which is defined as the ability of a solute to dissolve in a solvent.

Solubility product:

Solubility product basically defines the equilibrium between solids and its respective ions in the solution. The value of the solubility product represents the degree to which a chemical compound dissociates in water.

Example –

The dissociation reaction of ionic solid (АB)
is given as shown below.

АB(s)
A* (aq)+B (aq)

The expression corresponding to the solubility product is shown below.

FA в]
K
sp

The balanced chemical equation for the dissociation reaction of PbSO4
is as follows:

PbSO4Pb2 +SO%

The expression corresponding to the solubility product is given as shown below.

,-[Pb* [so]
K
sp
…… (1)

Consider the S
is the molar solubility of the given compound.

The molar solubility of in pure water is calculated as shown below.

Substitute 1.82x108
as the value of К.
sp
and as the molar solubility of in the equation .

1.82x10(S)(s)
V1.82x10
=1.35x10 M

Ans:

The molar solubility of PbSO
4
in pure water is 1.35x10 M
4
.

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