Chapter 17: Problem 54
A \(1.00-\mathrm{L}\) solution saturated at \(25^{\circ} \mathrm{C}\) with lead(II) iodide contains \(0.54 \mathrm{~g}\) of \(\mathrm{PbI}_{2} .\) Calculate the solubility-product constant for this salt at \(25^{\circ} \mathrm{C}\).
Short Answer
Expert verified
The solubility-product constant (Ksp) for PbI2 at 25掳C is \( 6.41 \times 10^{-9} \).
Step by step solution
01
Write the balanced dissociation equation for PbI2
The balanced dissociation equation for PbI2 in water is:
\[ \mathrm{PbI_2(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)} \]
02
Write the expression for the solubility-product constant, Ksp
Ksp is the product of the equilibrium concentrations of the ions in the saturated solution, each raised to the power of their stoichiometric coefficients. In this case, the expression becomes:
\[ \mathrm{K_{sp} = [Pb^{2+}][I^-]^2} \]
03
Calculate the molar concentrations of the ions in the saturated solution
Given that the mass of dissolved PbI2 is 0.54 g in 1 L of the saturated solution, we can find the moles of \(\mathrm{PbI}_2\) by dividing the mass by its molar mass:
\[ \frac{0.54 \,\mathrm{g}}{461.01 \,\mathrm{g/mol}} = 0.00117 \,\mathrm{mol} \]
Since 1 mol of PbI2 dissociates into 1 mol of Pb虏鈦 and 2 mol of I鈦 ions, we can calculate the molar concentration of each ion in the saturated solution:
\[ [\mathrm{Pb^{2+}}] = 0.00117 \,\mathrm{M} \]
\[ [\mathrm{I^-}] = 2 \times 0.00117 \,\mathrm{M} = 0.00234 \,\mathrm{M} \]
04
Substitute the molar concentrations into the Ksp expression and solve for Ksp
Now, we can substitute the calculated molar concentrations of Pb虏鈦 and I鈦 ions into the Ksp expression:
\[ \mathrm{K_{sp} = (0.00117)(0.00234)^2 =}\, 6.41 \times 10^{-9} \]
The solubility-product constant for PbI2 at 25掳C is \( 6.41 \times 10^{-9} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Lead(II) Iodide
Lead(II) Iodide, denoted as PbI鈧, is a chemical compound that prominently features in discussions about solubility because of its sparingly soluble nature. It is known for its striking bright yellow color, which can serve as an excellent visual indicator in laboratory settings. The compound is composed of one lead (Pb) ion paired with two iodide (I鈦) ions. At room temperature, Lead(II) Iodide exists in a solid state when it is not dissolved in a solvent.
To understand PbI鈧 more thoroughly, one must consider its molecular structure and the interactions between ions. The lead ion (Pb虏鈦) forms a stable coordination with the iodide ions, which is why under normal conditions, PbI鈧 remains largely undissolved. However, when dissolved in water, it dissociates into its constituent ions. The degree to which this dissociation happens is governed by its Solubility Product Constant (Ksp). Understanding this constant helps predict the solubility of PbI鈧 when environmental conditions change. Such insights are important in various fields, ranging from environmental science to industrial applications.
To understand PbI鈧 more thoroughly, one must consider its molecular structure and the interactions between ions. The lead ion (Pb虏鈦) forms a stable coordination with the iodide ions, which is why under normal conditions, PbI鈧 remains largely undissolved. However, when dissolved in water, it dissociates into its constituent ions. The degree to which this dissociation happens is governed by its Solubility Product Constant (Ksp). Understanding this constant helps predict the solubility of PbI鈧 when environmental conditions change. Such insights are important in various fields, ranging from environmental science to industrial applications.
Chemical Equilibrium
Chemical equilibrium refers to a state in which the forward and backward reactions occur at the same rates, leading to constant concentrations of reactants and products. When applied to the case of Lead(II) Iodide, chemical equilibrium is achieved in a saturated solution where the rate of dissolution of PbI鈧 equals the rate of its precipitation.
In a chemical context, this balance is expressed through the equilibrium constant expression, in this case, the solubility product constant or Ksp. For lead(II) iodide in solution, the dissociation can be represented as:
In a chemical context, this balance is expressed through the equilibrium constant expression, in this case, the solubility product constant or Ksp. For lead(II) iodide in solution, the dissociation can be represented as:
- PbI鈧 (s) 鈬 Pb虏鈦 (aq) + 2 I鈦 (aq)
Dissolution Reaction
A dissolution reaction is a process where a solid substance dissolves in a solvent, forming a solution. In the case of Lead(II) Iodide, the dissolution reaction is represented by the separation of PbI鈧 into the ions Pb虏鈦 and I鈦 when it is mixed with water. The reaction can be summarized by the equation:
During the dissolution of PbI鈧, one mole of the solid dissolves to form one mole of Pb虏鈦 ions and two moles of I鈦 ions. The Ksp expression,
- PbI鈧 (s) 鈬 Pb虏鈦 (aq) + 2 I鈦 (aq)
During the dissolution of PbI鈧, one mole of the solid dissolves to form one mole of Pb虏鈦 ions and two moles of I鈦 ions. The Ksp expression,
- Ksp = [Pb虏鈦篯[I鈦籡虏