What is the equilibrium concentration of silver ions at 25 degrees C in a 1.0-L saturation solution of silver carbonate to which .20 mole of NaHCO3 has been added? The Ksp of Ag2CO3 is 8.1 x 10-12 at 25 degrees C
To find the equilibrium concentration of silver ions in the solution, we need to consider the equilibrium reaction that occurs between silver carbonate (Ag2CO3) and water (H2O):
Ag2CO3 ⇌ 2 Ag+ + CO3 2-
From the equation, we see that for every one molecule of silver carbonate that dissolves, it forms two silver ions (Ag+). The equilibrium concentration of silver ions can be determined using the solubility product constant (Ksp) and the stoichiometry of the reaction.
The Ksp expression for Ag2CO3 is given by:
Ksp = [Ag+]^2 [CO3 2-]
We are given the Ksp value of Ag2CO3 as 8.1 x 10^-12, which indicates the product of the silver ion concentration and the carbonate ion concentration at equilibrium.
We also know that 0.20 moles of NaHCO3 has been added to a 1.0-L saturation solution of silver carbonate. When NaHCO3 dissolves, it dissociates to produce carbonate ions (CO3 2-). Since the carbonate ions come from the added NaHCO3, the initial concentration of carbonate ions is 0.20 M.
Since the stoichiometry of the reaction states that for every two silver ions formed, one carbonate ion is also formed, the initial concentration of the silver ions (Ag+) is 2x, where x is the equilibrium concentration.
Substituting these values into the Ksp expression, we have:
8.1 x 10^-12 = (2x)^2 (0.20)
Simplifying, we get:
8.1 x 10^-12 = 4x^2 (0.20)
Solving for x, we can rearrange the equation as follows:
x^2 = (8.1 x 10^-12) / (4 * 0.20)
x^2 = 2.025 x 10^-12
x = √(2.025 x 10^-12)
x ≈ 1.424 x 10^-6
Therefore, the equilibrium concentration of silver ions (Ag+) in the solution at 25 degrees Celsius is approximately 1.424 x 10^-6 M.