Problem 70 - Chemistry Matter and Change

Question

Write the \(K_{s p}\) expression for lead chromate \(\left(\mathrm{Pb} \mathrm{CrO}_{4}\right)\) and calculate its solubility in mol/L. \(K_{\mathrm{sp}}=2.3 \times 10^{-13}\)

Step-by-Step Solution

Answer

The \(K_{sp}\) expression for lead chromate (PbCrO\(_4\)) is given by: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{CrO}_4^{2-}] \] Given the provided value of \(K_{sp} = 2.3 \times 10^{-13}\), the solubility (s) of PbCrO\(_4\) in mol/L can be calculated as follows: \[ s = \sqrt{2.3 \times 10^{-13}} \] This results in a solubility of approximately \(1.52 \times 10^{-7}\) mol/L for lead chromate in water.

Step 1: Write the dissolution equation for lead chromate

First, we need to write the balanced dissolution equation for lead chromate: \[ \mathrm{PbCrO}_4(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq) \] This equation represents the dissolution of solid lead chromate into its constituent ions in aqueous solution.

Step 2: Write the \(K_{sp}\) expression for lead chromate

According to the dissolution equation, the solubility product constant (\(K_{sp}\)) expression can be written as: \[ K_{sp} = [\mathrm{Pb}^{2+}] [\mathrm{CrO}_4^{2-}] \] Note that, as concentrations of solids are not included in equilibrium expressions, the concentration of PbCrO\(_4\) is not included in the \(K_{sp}\) expression.

Step 3: Define variables for the solubility

Let the solubility of PbCrO\(_4\) in mol/L be represented by "s". This means that the molar concentration of \(\mathrm{Pb}^{2+}\) and \(\mathrm{CrO}_4^{2-}\) in the solution at equilibrium will also be "s". Now, we can rewrite the \(K_{sp}\) expression using "s": \[ K_{sp} = (s)(s) = s^2 \]

Step 4: Plug in the given value of \(K_{sp}\) and solve for solubility "s"

We have the value of \(K_{sp}\): \[ K_{sp} = 2.3 \times 10^{-13} \] Now, we substitute this value into the rewritten \(K_{sp}\) expression and solve for "s": \[ 2.3 \times 10^{-13} = s^2 \] To find "s", we take the square root of both sides: \[ s = \sqrt{2.3 \times 10^{-13}} \] Calculating the value of "s" gives: \[ s \approx 1.52 \times 10^{-7} \, \text{mol/L} \] Thus, the solubility of lead chromate (PbCrO\(_4\)) in water is approximately \(1.52 \times 10^{-7}\) mol/L.

Key Concepts

Solubility Calculations

Solubility calculations involve determining the amount of a substance that can dissolve in a solvent at a given temperature to form a saturated solution. Understanding solubility is crucial for predicting how substances will behave in various environments, such as biological systems or industrial processes. Solubility is typically expressed in units of moles per liter (mol/L) for chemical solutions.

In the case of salts like lead chromate \(\mathrm{PbCrO}_4\), the solubility calculations become slightly more complex due to the dissociation of the solid into ions when it dissolves in water. The concentration of each ion in a saturated solution is directly related to the solubility product constant \(K_{sp}\). By leveraging the \(K_{sp}\) value, solubility calculations can be performed by setting up an equilibrium expression and solving for the molar solubility, which provides insight into how much of the substance will dissolve under equilibrium conditions.

Chemical Equilibrium

Chemical equilibrium is the state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of reactants and products over time. At equilibrium, a dynamic balance exists where reactants are converted to products and vice versa at the same rate.

For solubility equilibrium involving sparingly soluble ionic compounds like lead chromate, the equilibrium lies far to the left, favoring the solid form. However, a small amount does dissolve and then dissociates into ions. In this context, \(K_{sp}\) is a special equilibrium constant known as the solubility product constant. It quantifies the extent of dissolution and depends only on temperature and the nature of the ionic compound, not on the concentrations of other species in the solution. \(K_{sp}\) values are used to predict whether a precipitate will form in a solution (a concept known as the common ion effect) and to compare the solubilities of different ionic compounds under similar conditions.

Dissolution Equation

The dissolution equation represents the process by which a solid substance dissolves in a solvent to form a homogeneous solution. For ionic compounds, this involves breaking down into constituent ions. The general form of a dissolution equation for a salt is shown by the compound dissolving into its individual cations and anions. In our example of lead chromate \(\mathrm{PbCrO}_4\), the reaction is written as follows:

\[\mathrm{PbCrO}_4(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + \mathrm{CrO}_4^{2-}(aq)\]
From this, we can derive the solubility product expression \(K_{sp}\), which only includes the concentrations of the dissolved ions, as the concentration of the solid is considered constant and thus omitted from the expression. By understanding the dissolution equation, chemists can predict solubility behavior, calculate the \(K_{sp}\), and manipulate the conditions to favor either the dissolving of more solid or the precipitation of dissolved ions.