Chapter 16: Problem 25
One gram of \(\mathrm{PbCl}_{2}\) is dissolved in \(1.0 \mathrm{~L}\) of hot water. When the solution is cooled to \(25^{\circ} \mathrm{C}\), will some of the \(\mathrm{PbCl}_{2}\) crystallize out? If so, how much?
Short Answer
Expert verified
Answer: Yes, some of the PbCl2 will crystallize out when the solution cools to 25°C. Approximately 0.72 g of PbCl2 will crystallize out.
Step by step solution
01
Find the solubility product constant (Ksp) for PbCl2 at 25 °C
From reference tables or literature, the solubility product constant (Ksp) of PbCl2 at 25°C is 1.7 x 10^-5.
02
Calculate the initial concentration of ions
According to the balanced reaction:
PbCl2(s) -> Pb^2+ (aq) + 2Cl^- (aq)
1.0 g of PbCl2 was dissolved in 1.0 L of water, so let's find the initial concentrations of Pb^2+ and Cl^- ions. The molar mass of PbCl2 is approximately 278.1 g/mol.
1 g / 278.1 g/mol ≈ 0.0036 mol
So the initial concentration:
Pb^2+: 0.0036 mol / 1.0 L = 0.0036 M
Cl^-: 2 * 0.0036 mol / 1.0 L = 0.0072 M (because there are two 2 Cl^- ions produced for every Pb^2+ ion)
03
Compare the initial ion concentration product to the Ksp value
Let's find the initial ion concentration product: [Pb^2+] * [Cl^-]^2 = 0.0036 * (0.0072)^2 ≈ 1.86 × 10^-4
Since this value (1.86 × 10^-4) is greater than the Ksp (1.7 × 10^-5), precipitation will occur, i.e., some of the PbCl2 will crystallize out when the solution cools to 25°C.
04
Calculate the amount of PbCl2 that will crystallize out
To find the amount of PbCl2 that will crystallize out, we first need to calculate the equilibrium concentrations of both ions at 25°C.
Let x be the concentration of Pb^2+ that will precipitate:
Ksp = [Pb^2+]*(Cl^-)^2
1.7 x 10^-5 = (0.0036 - x) * (0.0072 - 2x)^2
Solve this equation for x ≈ 0.0026 M
Now, let's find the amount of PbCl2 that corresponds to this concentration:
Amount of PbCl2 precipitated = (0.0026 mol/L) * (1 L) * (278.1 g/mol) ≈ 0.72 g
05
Final Answer
When the solution cools to 25°C, some of the PbCl2 will indeed crystallize out of the solution. Approximately 0.72 g of PbCl2 will crystallize out.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chemical Equilibrium
Understanding chemical equilibrium is crucial to interpreting what happens when a substance like PbCl2 dissolves and then starts to crystallize upon cooling. At equilibrium, the rate at which PbCl2 dissolves into its ions is equal to the rate at which the ions come together to form solid PbCl2. However, this state of balance doesn't mean the concentrations of ions and solid are the same; rather, it indicates a constant ratio given by the solubility product constant, Ksp, at a specific temperature.
Chemical equilibrium can be disturbed by changes in conditions, such as temperature. When hot PbCl2 solution is cooled, the solubility of PbCl2 decreases, which shifts the equilibrium to favor the formation of solid PbCl2—an illustration of Le Châtelier's principle. This shift causes precipitation or crystallization of PbCl2 until a new equilibrium is established at the lower temperature.
Chemical equilibrium can be disturbed by changes in conditions, such as temperature. When hot PbCl2 solution is cooled, the solubility of PbCl2 decreases, which shifts the equilibrium to favor the formation of solid PbCl2—an illustration of Le Châtelier's principle. This shift causes precipitation or crystallization of PbCl2 until a new equilibrium is established at the lower temperature.
Saturation and Crystallization
Saturation refers to the point at which a solution contains the maximum concentration of a solute that can be dissolved at a given temperature. Exceeding this point leads to crystallization, where the excess solute forms solid crystals. In our example, the solution initially prepared with hot water was unsaturated because the temperature allowed more PbCl2 to dissolve. But as the solution cools, the solubility limit decreases, leading to a supersaturated solution.
Here, crystallization occurs because the ions in the supersaturated solution start to bind together to form a solid in order to reach a new saturated state. The step-by-step calculation in the problem shows us that the initial ion concentration is higher than what the new solubility limit allows at 25°C, hence indicating that crystallization will happen until the solution is just saturated with respect to PbCl2.
Here, crystallization occurs because the ions in the supersaturated solution start to bind together to form a solid in order to reach a new saturated state. The step-by-step calculation in the problem shows us that the initial ion concentration is higher than what the new solubility limit allows at 25°C, hence indicating that crystallization will happen until the solution is just saturated with respect to PbCl2.
Molar Solubility
Molar solubility is a term that represents the number of moles of a solute that can be dissolved per liter of solution before the solution becomes saturated. It directly relates to the solubility product constant and varies with temperature. The high-temperature molar solubility of PbCl2 is greater than its molar solubility at 25°C, which is why the substance starts to crystallize as the solution cools.
The exercise uses molar solubility to determine how much PbCl2 will crystallize out. By calculating 'x', we find out the equilibrium concentration of Pb^2+ after crystallization, which is essential for calculating how many grams of PbCl2 will precipitate. The precipitation continues until the product of the ion concentrations equals the Ksp, at which point the solution is just saturated, and molar solubility is achieved at the lower temperature.
The exercise uses molar solubility to determine how much PbCl2 will crystallize out. By calculating 'x', we find out the equilibrium concentration of Pb^2+ after crystallization, which is essential for calculating how many grams of PbCl2 will precipitate. The precipitation continues until the product of the ion concentrations equals the Ksp, at which point the solution is just saturated, and molar solubility is achieved at the lower temperature.