Calculate the solubility of AgCl in the following solution:
100mL of .004 M CaCl2
let x = solubility of AgCl in moles/L, then
AgCl(s) ==> Ag^+ + Cl^-
...x.........x......x
............CaCl2(aq) ==> Ca^2+ + 2Cl^-
initial......0.004..........0.......0
change......-0.004.........0.004..0.008
equil.........0............0.004..0.008
Ksp = (Ag^+)(Cl^-)
Now substitute from the ICE charts (both of them).
(Ag^+) = x
(Cl^-) = x from the AgCl and 0.008 from the CaCl2 for a total of x+0.008
Solve for x = solubility AgCl in moles/L.
Convert from 1 L soln to 100 mL soln.
To calculate the solubility of AgCl in a solution of CaCl2, we need to consider the common ion effect. The addition of CaCl2 adds chloride ions (Cl-) to the solution, which can affect the solubility of AgCl.
The solubility product constant (Ksp) for AgCl is given by:
Ksp = [Ag+][Cl-]
Let's assume that x represents the solubility of AgCl in moles per liter (M).
Since AgCl dissociates to form one Ag+ ion and one Cl- ion, the concentration of Ag+ ions and Cl- ions in the solution will both be equal to x M.
However, the concentration of Cl- ions in the solution is not only from AgCl, but also from CaCl2.
Given that the concentration of CaCl2 is 0.004 M, the concentration of Cl- ions in the solution (from CaCl2) is also 0.004 M.
Therefore, the total concentration of Cl- ions in the solution is:
[Cl-] = 0.004 M (from CaCl2) + x M (from AgCl)
Now we can substitute the values into the solubility product constant equation:
Ksp = [Ag+][Cl-] = (x)(0.004 M + x)
The value of Ksp for AgCl is 1.77 x 10^-10.
To solve for x, we can rearrange the equation:
Ksp = (x)(0.004 M + x)
1.77 x 10^-10 = x(0.004 M + x)
Now we solve for x using algebraic methods.
1. Multiply both sides of the equation by (0.004 M + x):
1.77 x 10^-10 (0.004 M + x) = x(0.004 M + x)
2. Expand the equation:
7.08 x 10^-13 M + 1.77 x 10^-10 x = 0.004 Mx + x^2
3. Rearrange the equation to bring all terms to one side:
x^2 + (0.004 M - 1.77 x 10^-10) x + 7.08 x 10^-13 M = 0
4. Solve the quadratic equation using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
Using a = 1, b = (0.004 M - 1.77 x 10^-10), and c = 7.08 x 10^-13 M, we can calculate the roots of the quadratic equation to find the solubility of AgCl.
Note: The calculations for the roots of the quadratic equation can get quite complex. If you provide specific values for the concentration of CaCl2, we can proceed with numeric calculations to determine the solubility of AgCl in the given solution.
To calculate the solubility of AgCl in a solution, we need to determine the common ion effect. In this case, CaCl2 is the common ion.
The common ion effect states that the solubility of a slightly soluble salt is reduced when a common ion is added to the solution. This means that the presence of Ca2+ ions from CaCl2 will decrease the solubility of AgCl.
To calculate the solubility of AgCl, we need to determine the concentration of Ag+ ions, as AgCl dissociates into Ag+ and Cl- ions. Let's use the solubility product constant (Ksp) of AgCl to find the concentration of Ag+ ions.
The balanced equation for the dissociation of AgCl is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
The Ksp expression for AgCl is:
Ksp = [Ag+][Cl-]
Since AgCl is a slightly soluble salt, we can assume that the solubility of AgCl is "x" moles/L. Therefore, the concentration of Ag+ ions will also be "x" moles/L.
Since AgCl has a 1:1 stoichiometry, we can substitute "x" into the Ksp expression:
Ksp = (x)(x)
Given that the value of Ksp for AgCl is 1.77 x 10^-10, we can solve for "x" using the equation:
1.77 x 10^-10 = x^2
Taking the square root of both sides:
sqrt(1.77 x 10^-10) = x
Calculating the square root:
x ≈ 1.33 x 10^-5 mol/L
Therefore, the solubility of AgCl in the given solution is approximately 1.33 x 10^-5 mol/L.