Chapter 16: Problem 40
The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{PbI}_{2}\right)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of lead iodide in each of the following. \(\mathbf{a}\). water b. \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)
Short Answer
Expert verified
The solubility of \(\mathrm{PbI_{2}}\) in water is approximately \(1.52 \times 10^{-3}\) M, and in the presence of \(0.10\,\mathrm{M}\, \mathrm{Pb}( \mathrm{NO}_3)_2\), it is approximately \(5.92 \times 10^{-5}\) M.
Step by step solution
01
Write the balanced chemical equation and the Ksp expression
We will begin by writing the balanced chemical equation for the dissolution of lead iodide in water and the corresponding Ksp expression.
For the dissolution of \(\mathrm{PbI_{2}}\) in water, we have
\(\mathrm{PbI}_{2}\left(\mathrm{s}\right) \rightleftharpoons \mathrm{Pb}^{2+}\left(\mathrm{aq}\right)+2 \mathrm{I}^{-}\left(\mathrm{aq}\right)\)
The solubility product constant (Ksp) expression is
\(K_\mathrm{sp} = [\mathrm{Pb}^{2+}][\mathrm{I}^{-}]^2\)
We are given that \(K_\mathrm{sp} = 1.4 \times 10^{-8}\).
02
Calculate the solubility of PbI₂ in water
In water, let the solubility of \(\mathrm{PbI}_{2}\) be S molar. From the stoichiometry of the balanced chemical equation, if S moles of \(\mathrm{PbI_{2}}\) dissolve, S moles of \(\mathrm{Pb}^{2+}\) ions and 2S moles of \(\mathrm{I}^{-}\) ions would be formed.
Hence, using the Ksp expression and substituting the given value:
\(1.4 \times 10^{-8} = [S][2S]^2\)
Now, solve for S:
\(1.4 \times 10^{-8} = 4S^3\)
\(S^3 = \frac{1.4 \times 10^{-8}}{4}\)
\(S^3 = 3.5 \times 10^{-9}\)
\(S = \sqrt[3]{3.5 \times 10^{-9}}\)
\(S = 1.52 \times 10^{-3}\)
Thus, the solubility of \(\mathrm{PbI_{2}}\) in water is approximately \(1.52 \times 10^{-3}\) M.
03
Calculate the solubility of PbI₂ in 0.10M Pb(NO₃)₂
In the presence of \(0.10\,\mathrm{M}\,\mathrm{Pb}(\mathrm{NO}_3)_2\), the added Pb²⁺ ions will affect the solubility of \(\mathrm{PbI_{2}}\). The \(\mathrm{Pb}^{2+}\) ions in solution will come from both the \(\mathrm{Pb}(\mathrm{NO}_3)_{2}\) and the dissolved \(\mathrm{PbI}_{2}\).
Let the solubility of \(\mathrm{PbI_{2}}\) in the presence of \(0.10\,\mathrm{M}\) Pb²⁺ be S' molar. Then, the concentration of \(\mathrm{Pb}^{2+}\) ions from the dissolved \(\mathrm{PbI}_{2}}\) would be S' and from \(\mathrm{Pb}(\mathrm{NO}_3)_2\) would be 0.10 M.
Hence, using the Ksp expression in this case:
\(1.4 \times 10^{-8} = [S' + 0.10][2S']^2\)
Now, solve for S':
\(1.4 \times 10^{-8} = (0.10 + S')[4S'^2]\)
Assuming \(S' \ll 0.10\), we can approximate and simplify the equation as:
\(1.4 \times 10^{-8} \approx 0.10[4S'^2]\)
\(S'^2 \approx 3.5 \times 10^{-8}\)
\(S' \approx \sqrt{3.5 \times 10^{-8}}\)
\(S' \approx 5.92 \times 10^{-5}\)
Thus, the solubility of \(\mathrm{PbI_{2}}\) in the presence of \(0.10\,\mathrm{M}\) Pb(NO₃)₂ is approximately \(5.92 \times 10^{-5}\) M.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product Constant
The solubility product constant, commonly referred to as \(K_{sp}\), is a crucial concept in understanding the solubility of ionic compounds. It is a special equilibrium constant which applies to the dissolution of a sparingly soluble compound into its constituent ions. For example, when lead iodide \((\text{PbI}_2)\) dissolves in water, it dissociates into lead ions \((\text{Pb}^{2+})\) and iodide ions \((\text{I}^-)\). The overall process can be represented as follows:
Knowing \(K_{sp}\) allows us to calculate solubility in moles per liter, as seen in exercises where we find the solubility of lead iodide in water and other solutions.
Importantly, the size of the \(K_{sp}\) value provides insight into the solubility level: smaller \(K_{sp}\) values indicate lower solubility, while larger values suggest higher solubility.
- \(\text{PbI}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)\)
Knowing \(K_{sp}\) allows us to calculate solubility in moles per liter, as seen in exercises where we find the solubility of lead iodide in water and other solutions.
Importantly, the size of the \(K_{sp}\) value provides insight into the solubility level: smaller \(K_{sp}\) values indicate lower solubility, while larger values suggest higher solubility.
Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry describing a state where the concentrations of reactants and products remain constant over time. This balance occurs when the rate of the forward reaction equals the rate of the reverse reaction.
In the case of the dissolution of lead iodide \((\text{PbI}_{2})\), chemical equilibrium is reached when solid \(\text{PbI}_{2}\) is in equilibrium with its ions \(\text{Pb}^{2+}\) and \(\text{I}^{-}\) in solution. This means:
Understanding chemical equilibrium is essential, as it dictates the conditions under which ionic compounds dissolve and the maximum concentration of ions that can be achieved in solution. For instance, adding more lead nitrate \((\text{Pb(NO}_3\text{)}_2)\) shifts the equilibrium by introducing more \(\text{Pb}^{2+}\) ions, affecting the solubility of \(\text{PbI}_{2}\) due to the common ion effect.
In the case of the dissolution of lead iodide \((\text{PbI}_{2})\), chemical equilibrium is reached when solid \(\text{PbI}_{2}\) is in equilibrium with its ions \(\text{Pb}^{2+}\) and \(\text{I}^{-}\) in solution. This means:
- The rate of \(\text{PbI}_{2}\) dissolving to form \(\text{Pb}^{2+}\) and \(\text{I}^{-}\) is equal to the rate of \(\text{Pb}^{2+}\) and \(\text{I}^{-}\) recombining to form solid \(\text{PbI}_{2}\).
Understanding chemical equilibrium is essential, as it dictates the conditions under which ionic compounds dissolve and the maximum concentration of ions that can be achieved in solution. For instance, adding more lead nitrate \((\text{Pb(NO}_3\text{)}_2)\) shifts the equilibrium by introducing more \(\text{Pb}^{2+}\) ions, affecting the solubility of \(\text{PbI}_{2}\) due to the common ion effect.
Ionic Solubility
Ionic solubility pertains to how well an ionic compound dissolves in a solvent, usually water. It is influenced by several factors including temperature, presence of common ions, and the nature of the solute and solvent.
In the context of lead iodide \((\text{PbI}_{2})\), its solubility in water is determined by its \(K_{sp}\). By solving the \(K_{sp}\) expression, we can determine the concentration of ions in a saturated solution.
When additional ions are introduced into the solution, such as \(\text{Pb}^{2+}\) ions from \(\text{Pb(NO}_3\text{)}_2)\), the solubility of \(\text{PbI}_{2}\) is affected. This is due to the "common ion effect," where the presence of shared ions decreases the solubility of an ionic compound.
In the context of lead iodide \((\text{PbI}_{2})\), its solubility in water is determined by its \(K_{sp}\). By solving the \(K_{sp}\) expression, we can determine the concentration of ions in a saturated solution.
When additional ions are introduced into the solution, such as \(\text{Pb}^{2+}\) ions from \(\text{Pb(NO}_3\text{)}_2)\), the solubility of \(\text{PbI}_{2}\) is affected. This is due to the "common ion effect," where the presence of shared ions decreases the solubility of an ionic compound.
- The solution already contains \(\text{Pb}^{2+}\), reducing the amount of \(\text{PbI}_{2}\) that can dissolve without exceeding the saturation point.