To find the concentration of the acetate ion (CH₃COO⁻) in the solution after adding 0.1 moles of CH₃COOAg to 1 L of 0.1 M HCl, we need to consider the dissociation of CH₃COOAg and the effect of HCl on the equilibrium. Let's break this down step by step.
Understanding the Components
We have two main components in this scenario:
- CH₃COOAg: This is silver acetate, which will dissociate in solution to produce acetate ions (CH₃COO⁻) and silver ions (Ag⁺).
- HCl: This is a strong acid that will dissociate completely in water, providing H⁺ ions that can affect the equilibrium of the acetate ion in solution.
Dissociation Reactions
When CH₃COOAg dissolves in water, it dissociates as follows:
CH₃COOAg (s) ⇌ CH₃COO⁻ (aq) + Ag⁺ (aq)
Since we are adding 0.1 moles of CH₃COOAg to 1 L of solution, it will produce 0.1 moles of CH₃COO⁻ and 0.1 moles of Ag⁺ in the solution.
Effect of HCl on the Equilibrium
Next, we need to consider the presence of HCl. The H⁺ ions from HCl can react with the acetate ions to form acetic acid (CH₃COOH):
CH₃COO⁻ (aq) + H⁺ (aq) ⇌ CH₃COOH (aq)
This reaction will shift the equilibrium to the left, consuming some of the acetate ions. To find the final concentration of CH₃COO⁻, we need to determine how much acetate is converted to acetic acid.
Calculating the Concentration
Initially, we have:
- Concentration of CH₃COO⁻ = 0.1 M (from CH₃COOAg)
- Concentration of H⁺ = 0.1 M (from HCl)
Now, we can use the equilibrium constant for acetic acid (Kₐ) to find out how much acetate will remain in solution. The dissociation constant for acetic acid is given as:
Kₐ = [H⁺][CH₃COOH] / [CH₃COO⁻]
Assuming that x moles of CH₃COO⁻ react with H⁺ to form CH₃COOH, we can express the concentrations at equilibrium:
- [CH₃COO⁻] = 0.1 - x
- [CH₃COOH] = x
- [H⁺] = 0.1 (since HCl is a strong acid, it remains essentially unchanged)
Substituting into the Kₐ expression:
10⁻⁵ = (0.1)(x) / (0.1 - x)
Solving for x
Assuming x is small compared to 0.1, we can simplify the equation:
10⁻⁵ = (0.1)(x) / 0.1
Thus, we have:
10⁻⁵ = x
Now, substituting back to find the concentration of CH₃COO⁻:
[CH₃COO⁻] = 0.1 - 10⁻⁵ ≈ 0.1 M (since 10⁻⁵ is negligible compared to 0.1)
Final Concentration of CH₃COO⁻
Therefore, the concentration of the acetate ion (CH₃COO⁻) in the solution is approximately:
0.1 M
However, we must consider the solubility product of AgCl (Kₛₚ = 10⁻¹⁰). The presence of Ag⁺ ions can lead to the precipitation of AgCl if the concentration of Cl⁻ ions is sufficient. Since we have 0.1 M HCl, the concentration of Cl⁻ is also 0.1 M, which is greater than the solubility product of AgCl. Therefore, Ag⁺ will precipitate out as AgCl, effectively removing Ag⁺ from the solution.
In conclusion, the concentration of CH₃COO⁻ in the solution remains approximately 0.1 M, as the precipitation of AgCl does not significantly affect the acetate concentration under these conditions. The final answer is:
- (1) 10⁻²
- (2) 10⁻³
- (3) 10⁻⁴
- (4) 10⁻⁵
The correct option is (1) 10⁻², as the concentration of CH₃COO⁻ is effectively 0.1 M in this scenario.
