So what would the explanation for this be? Its really confusing me.
The Ksp is independent of [NaOH] while the solubility of Ca(OH)2 decreases as the [NaOH] increases. Explain why this is the case.
( in the data table, the naoh concentration is 0 M, and the solubility is 2.30 x 10^-2, and ksp is around 5.5x10^-5; next table the concentration of naoh increases and the solubility and ksp go down)
Ca(OH)2 ==> Ca^2+ + 2OH^-
and Ksp = (Ca^2+)(OH^-)^2
Ksp is a constant. NaOH is a strong base and ionizes 100%; therefore, NaOH ==> Na^+ + OH^-. Le Chatelier's Principle tells you that when the OH^- increases, due to the NaOH, the Ksp equilibrium of Ca(OH)2 is forced to the left thereby decreasing the solubility of Ca(OH)2 in the solution.
The given information suggests that the solubility of Ca(OH)2 decreases as the concentration of NaOH increases, while the solubility product constant (Ksp) remains independent of NaOH concentration.
Let's break this down step-by-step:
1. Solubility Product Constant (Ksp):
- Ksp is a constant that represents the equilibrium constant for the dissolution of an insoluble compound.
- For Ca(OH)2, the balanced equation for its dissolution is: Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq).
- The Ksp expression for this reaction is: Ksp = [Ca2+][OH-]².
- The value of Ksp for Ca(OH)2 is approximately 5.5x10^-5.
2. Effect of [NaOH] on Solubility:
- When NaOH is added to a solution that contains Ca(OH)2, it reacts with Ca2+ ions to form a precipitate of Ca(OH)2.
- The reaction equation is: Ca2+(aq) + 2OH-(aq) + Na+(aq) + Cl-(aq) ⇌ Ca(OH)2(s) + Na+(aq) + Cl-(aq).
- The concentration of NaOH is represented by [NaOH].
- As [NaOH] increases, more OH- ions are present in the solution, resulting in a shift in equilibrium towards the formation of Ca(OH)2 precipitate rather than calcium ions (Ca2+) and hydroxide ions (OH-).
- Therefore, the solubility of Ca(OH)2 decreases as the concentration of NaOH increases.
3. Independent Ksp:
- Ksp represents the equilibrium constant for the dissolution of Ca(OH)2, which is only influenced by the products (Ca2+ and OH-) and not by the presence of NaOH.
- The stoichiometry of the reaction remains the same, and the concentrations of Ca2+ and OH- ions are what determine the value of Ksp.
- Since Ksp remains constant, it implies that the ratio of [Ca2+][OH-]² remains unchanged, and the solubility of Ca(OH)2 cannot be affected by the concentration of NaOH.
In summary, the solubility of Ca(OH)2 decreases as the concentration of NaOH increases due to the reaction between NaOH and Ca2+ ions, leading to the formation of Ca(OH)2 precipitate. However, the solubility product constant (Ksp) remains independent of NaOH concentration, as it is determined by the concentration of Ca2+ and OH- ions only.
To understand why the Ksp is independent of [NaOH], but the solubility of Ca(OH)2 decreases as the [NaOH] increases, we need to examine the concept of equilibrium and the reactions involved.
First, let's define some terms:
- Ksp (Solubility Product Constant) is the equilibrium constant for the dissolution of a slightly soluble compound in water. It represents the product of the concentrations of the ions in the solution, each raised to the power of its stoichiometric coefficient.
- Solubility refers to the maximum amount of a solute that can dissolve in a given solvent at a specific temperature, usually expressed in moles per liter (M).
Ca(OH)2 (calcium hydroxide) dissociates in water according to the following reaction:
Ca(OH)2(s) ⇌ Ca2+(aq) + 2OH-(aq)
Now, let's consider the effect of [NaOH] on Ca(OH)2 solubility:
1. When [NaOH] is 0 M:
At this point, no NaOH is present to react with Ca(OH)2. The solubility and Ksp are determined solely by the dissolution of Ca(OH)2. The given solubility of 2.30 x 10^-2 (in mol/L) indicates that this is the maximum amount of Ca(OH)2 that can dissolve at this temperature. The Ksp, which is around 5.5 x 10^-5, is calculated from the concentrations of Ca2+ and OH- ions in the solution.
2. When [NaOH] increases:
As the concentration of NaOH increases, more OH- ions are introduced into the solution. These OH- ions can react with Ca2+ ions from the dissociation of Ca(OH)2. The reaction between Ca2+ and OH- ions forms Ca(OH)2(s) again, decreasing the solubility. This decrease in solubility results in a decrease in the concentration of Ca2+ and OH- ions in the solution, which in turn decreases the Ksp.
In summary, the Ksp of Ca(OH)2 remains constant because it is determined solely by the concentration of Ca2+ and OH- ions in the solution, which are in turn determined by the dissolution of Ca(OH)2. However, as the concentration of NaOH increases, more OH- ions become available to react with Ca2+ ions, leading to the precipitation of Ca(OH)2 and a decrease in solubility.
To further analyze and interpret the data table provided, compare the values of [NaOH], solubility, and Ksp for different concentrations of NaOH. Note how the solubility decreases as the NaOH concentration increases and observe the corresponding changes in the Ksp.