Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag+ ion in the solution after equilibrium has been reestablished. For AgCl, Ksp = 1.8 × 10¯10
To calculate the number of moles of Ag+ ion in the solution after equilibrium has been reestablished, we need to determine the change in concentration of Ag+ ion due to the addition of KCl.
First, let's calculate the initial concentration of Ag+ ion in the saturated aqueous solution of AgCl.
Given:
Mass of KCl added = 0.200 g
Ksp of AgCl = 1.8 × 10¯10
Since AgCl is a sparingly soluble salt, we can assume that it fully dissociates in water.
Step 1: Calculate the number of moles of KCl added.
Molar mass of KCl (K = 39.10 g/mol, Cl = 35.45 g/mol) = 39.10 + 35.45 = 74.55 g/mol
Number of moles of KCl = mass / molar mass = 0.200 g / 74.55 g/mol = 0.00268 mol
Step 2: Calculate the increase in the concentration of Cl- ion due to the addition of KCl.
Since AgCl fully dissociates, the increase in Cl- ion concentration is equal to the increase in KCl concentration.
Step 3: Calculate the solubility of AgCl in the saturated solution before the addition of KCl.
The solubility of AgCl can be calculated using the Ksp expression:
Ksp = [Ag+][Cl-]
Given Ksp = 1.8 × 10¯10, and assuming x represents the concentration of Ag+ and Cl- ions (since they are in a 1:1 ratio), we have:
Ksp = x * x = x^2
Solving for x:
x = sqrt(Ksp) = sqrt(1.8 × 10¯10) = 1.34 × 10¯5 M
Thus, the concentration of Ag+ and Cl- ions before adding KCl is 1.34 × 10¯5 M.
Step 4: Calculate the concentration of Cl- ions after adding KCl.
Since the total volume does not change after adding KCl, the increase in Cl- ion concentration is equal to the concentration of KCl added.
Concentration of Cl- ions after adding KCl = 0.00268 mol / 1 L = 0.00268 M
Step 5: Calculate the new concentration of Ag+ ions.
Since the concentration of Cl- ions increased from 1.34 × 10¯5 M to 0.00268 M, it means that the concentration of Ag+ ions decreased by the same amount.
Change in concentration of Ag+ ions = -0.00268 M
New concentration of Ag+ ions = Initial concentration of Ag+ ions - Change in concentration
New concentration of Ag+ ions = 1.34 × 10¯5 M - 0.00268 M = 1.34 × 10¯5 M - 2.68 × 10¯3 M = -2.666 × 10¯3 M
However, a negative concentration is not physically meaningful since concentration represents the amount of substance per unit volume. Therefore, we can conclude that the concentration of Ag+ ions after equilibrium has been reestablished is zero.
Thus, the number of moles of Ag+ ion in the solution after equilibrium has been reestablished is zero.
To calculate the number of moles of Ag+ ion in the solution after equilibrium has been reestablished, we need to consider the solubility equilibrium of AgCl.
The solubility product constant (Ksp) for AgCl is given as 1.8 × 10^(-10). The balanced equation for the dissociation of AgCl in water is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Let's assume x moles of AgCl dissolve in the saturated aqueous solution. As AgCl dissociates completely, the concentration of Ag+ ions and Cl- ions formed would also be x moles/L.
According to the problem statement, 0.200 g of KCl is added to the solution, which doesn't change the total volume.
To calculate the moles of Ag+ ions at equilibrium, we'll follow these steps:
Step 1: Convert 0.200 g of KCl to moles.
The molar mass of KCl is approximately 74.55 g/mol. Dividing the mass by the molar mass gives:
0.200 g / 74.55 g/mol ≈ 0.00268 mol KCl
Step 2: Since KCl completely dissociates, the moles of Cl- ions produced would be equal to the moles of KCl added:
0.00268 mol Cl- ions
Step 3: As the solution is saturated with AgCl, the equilibrium concentration of Cl- ions should match the concentration of Cl- ions formed from AgCl dissociation.
Therefore, the concentration of Cl- ions at equilibrium is also 0.00268 mol/L.
Step 4: As AgCl dissociates completely, the moles of Ag+ ions formed would also be equal to the concentration of Cl- ions, which is 0.00268 mol.
Hence, the number of moles of Ag+ ion in the solution at equilibrium is 0.00268 mol.
........AgCl(s) ==> Ag^+(aq) + Cl^-(aq)
..........x...........x.........x
How many moles KCl were added. That is 0.200g/molar mass KCl = about 0.003 but you need to be more accurate. In 1L that will be 0.003 M.
..........KCl ==> K^+ + Cl^-
initial..0.003....0......0
equil......0....0.003....0.003
Now substitute into the Ksp expression for AgCl.
1.8E-10 = (Ag^+)(Cl^-)
(Ag^+) = x from AgCl.
(Cl^-) = x from AgCl + 0.003 from KCl. You have one unknown, solve for that which is the solubility of AgCl under these conditions.