Question

Calculate the molar solubility for these substances: (a) \(\mathrm{PbSO}_{4}, K_{\mathrm{sp}}=1.3 \times 10^{-8}\) (b) \(\mathrm{BaCrO}_{4}, K_{\mathrm{sp}}=8.5 \times 10^{-11}\)

Short answer

(a) \( s \approx 1.14 \times 10^{-4} \) M, (b) \( s \approx 9.22 \times 10^{-6} \) M

Step-by-step solution

Understand the Dissociation Reaction

For calculating the molar solubility, first, write down the dissociation reaction for each substance.(a) \(\text{PbSO}_{4}(s) \rightarrow \text{Pb}^{2+}(aq) + \text{SO}_{4}^{2-}(aq)\)(b) \(\text{BaCrO}_{4}(s) \rightarrow \text{Ba}^{2+}(aq) + \text{CrO}_{4}^{2-}(aq)\)

Define the Solubility

Define the molar solubility for each compound as \( s \). This represents the concentration of the ions in solution at equilibrium.(a) For \( \text{PbSO}_{4} \), \( [\text{Pb}^{2+}] = s \) and \( [\text{SO}_{4}^{2-}] = s \).(b) For \( \text{BaCrO}_{4} \), \( [\text{Ba}^{2+}] = s \) and \( [\text{CrO}_{4}^{2-}] = s \).

Write the Solubility Product Expression

Using the solubility product constant (\( K_{\text{sp}} \)), write the expression for each compound.(a) \( K_{\text{sp}} = [\text{Pb}^{2+}][\text{SO}_{4}^{2-}] = s \times s = s^2 = 1.3 \times 10^{-8} \)(b) \( K_{\text{sp}} = [\text{Ba}^{2+}][\text{CrO}_{4}^{2-}] = s \times s = s^2 = 8.5 \times 10^{-11} \)

Solve for Molar Solubility

Solve for \( s \) (molar solubility) by taking the square root of both sides of the equation.(a) \( s = \sqrt{1.3 \times 10^{-8}} \approx 1.14 \times 10^{-4} \) M(b) \( s = \sqrt{8.5 \times 10^{-11}} \approx 9.22 \times 10^{-6} \) M

Key concepts

Solubility Product Constant

The solubility product constant, represented as \(\text{K}_{\text{sp}}\), is a key concept in solubility calculations.
It defines the equilibrium constant for the dissociation of a sparingly soluble compound into its ions in a given solvent.

The value of \(\text{K}_{\text{sp}}\) is specific to each compound and temperature.
Higher \(\text{K}_{\text{sp}}\) generally indicates greater solubility in water.

In our example, the given \(\text{K}_{\text{sp}}\) values for \(\text{PbSO}_{4}\) and \(\text{BaCrO}_{4}\) allow us to calculate their molar solubility by solving related equilibrium expressions.

Dissociation Reaction

To determine the molar solubility, we need to start with the dissociation reaction.
This reaction shows how the solid compound breaks down into its constituent ions in a solution.

For instance:

• \(\text{PbSO}_{4}(s) \rightarrow \text{Pb}^{2+}(aq) + \text{SO}_{4}^{2-}(aq)\)
• \(\text{BaCrO}_{4}(s) \rightarrow \text{Ba}^{2+}(aq) + \text{CrO}_{4}^{2-}(aq)\)

These equations indicate the stoichiometric relationship between the dissolved ions and the original solid compound.
Understanding the dissociation reaction is crucial for setting up the equilibrium concentrations in the next steps.

Equilibrium Concentration

At equilibrium, the concentration of ions in a saturated solution can be represented by the molar solubility (s).
This is when the solution is in a dynamic equilibrium between the undissolved solid and the dissolved ions.

For our substances:

• \([ \text{Pb}^{2+} ] = s \) and \([ \text{SO}_{4}^{2-} ] = s \) for \(\text{PbSO}_{4}\)
• \([ \text{Ba}^{2+} ] = s \) and \([ \text{CrO}_{4}^{2-} ] = s \) for \(\text{BaCrO}_{4}\)

The molar solubility, s, is our unknown variable representing the concentration of each ion in the solution at equilibrium.

Solubility Product Expression

We can now create the solubility product expression using the dissociation reactions and the \(\text{K}_{\text{sp}}\) values.
This expression is derived from the general rule for equilibrium which is the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients.

For example:
\(\text{PbSO}_{4}\): \(\text{K}_{\text{sp}} = [\text{Pb}^{2+}] \times [\text{SO}_{4}^{2-}] = s \times s = s^2\)
\(\text{BaCrO}_{4}\): \(\text{K}_{\text{sp}} = [\text{Ba}^{2+}] \times [\text{CrO}_{4}^{2-}] = s \times s = s^2\)

These equations allow us to link the known \(\text{K}_{\text{sp}}\) values with the unknown solubilities.

Algebraic Solution

Finally, we solve the solubility product expressions algebraically to find the molar solubility, s.
This involves taking the square root of both sides of the expression.

For instance:

• \(\text{PbSO}_{4}\): \(\text{s} = \text{\textbackslash sqrt}(1.3 \times 10^{-8}) \approx 1.14 \times 10^{-4}\) M
• \(\text{BaCrO}_{4}\): \(\text{s} = \text{\textbackslash sqrt}(8.5 \times 10^{-11}) \approx 9.22 \times 10^{-6}\) M

With these calculations, we can determine how much of each compound will dissolve in water, giving us a clear idea of their respective solubilities.