Calculate the molar solubility of silver chloride (AgCl) in:
a-deionized water and
b-.20M NaCl solution
Given Ksp=1.6*10^-10
.........AgCl ==> Ag^+ + Cl^-
I........solid....0.......0
C........-x.......x.......x
E........solid....x.......x
Ksp = 16E-10 =- (Ag^+)(Cl^-)
Substitute and solve for x = (AgCl) in M.
for part B. You need to calculate (or find in a table) the activity of 0.20M NaCl solution and recalculate the solubility using activities instead of molarity..
To calculate the molar solubility of silver chloride (AgCl), we need to use the solubility product constant (Ksp) and the common ion effect.
a) Molar solubility in deionized water:
In deionized water, we assume there are no common ions present. Therefore, we can directly use the Ksp expression for AgCl:
AgCl ⇌ Ag+ + Cl-
Ksp = [Ag+][Cl-]
Since AgCl is a 1:1 salt, the molar solubility of AgCl is equal to the concentration of Ag+ or Cl-. Let's assume the molar solubility of AgCl is x:
Ksp = x * x
1.6 * 10^-10 = x^2
Take the square root of both sides:
x = √(1.6 * 10^-10) = 1.26 * 10^-5
Therefore, the molar solubility of AgCl in deionized water is 1.26 * 10^-5 M.
b) Molar solubility in 0.20 M NaCl solution:
In this case, NaCl is added to the solution, which provides the common ion Cl-. According to the common ion effect, the presence of a common ion reduces the solubility of a salt. We need to consider the concentration of Cl- ions from NaCl.
Let's assume the molar solubility of AgCl is y. Since AgCl dissociates into Ag+ and Cl-:
AgCl ⇌ Ag+ + Cl-
The concentration of Cl- ions will be the initial concentration from NaCl (0.20 M) plus y (from AgCl dissociation):
[Cl-] = 0.20 + y
Now we can write the Ksp expression considering the concentration of Cl-:
Ksp = [Ag+][Cl-]
1.6 * 10^-10 = y * (0.20 + y)
Simplify the equation:
1.6 * 10^-10 = 0.20y + y^2
Rearrange the equation:
y^2 + 0.20y - 1.6 * 10^-10 = 0
Since the value of y is small compared to 0.20, we can approximate the equation and solve for y:
y^2 + 0.20y ≈ 0.20y
0.20y - 1.6 * 10^-10 ≈ 0
Now solve the quadratic equation using the quadratic formula:
y ≈ (-0.20 ± √(0.04 + 4(1.6 * 10^-10)))/(2)
y ≈ (-0.20 ± √(0.04 + 6.4 * 10^-10))/(2)
y ≈ (-0.20 ± √(6.4 * 10^-10))/(2)
Considering that the value of y is small, we can ignore the negative root:
y ≈ (√(6.4 * 10^-10))/(2)
y ≈ (8 * 10^-6)/(2)
y ≈ 4 * 10^-6
Therefore, the molar solubility of AgCl in a 0.20 M NaCl solution is approximately 4 * 10^-6 M.
To calculate the molar solubility of silver chloride (AgCl), we need to use the concept of the solubility product constant (Ksp). The Ksp is a measure of the extent to which a sparingly soluble compound, like AgCl, will dissolve in a solvent. It represents the product of the concentrations of the ions formed when the compound dissociates in solution.
The balanced chemical equation for the dissociation of AgCl is:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Let's calculate the molar solubility in each case:
a) Deionized Water (pure water)
In pure water, we assume the AgCl completely dissociates into Ag+ and Cl-. Let's assign 's' as the molar solubility of AgCl. Using stoichiometry, the concentration of Ag+ and Cl- ions will also be 's'. Therefore, the expression for the Ksp is:
Ksp = [Ag+][Cl-] = s * s = s^2
Given that Ksp = 1.6 × 10^-10, we can set up the equation:
s^2 = 1.6 × 10^-10
To solve for 's', we take the square root of both sides of the equation:
s = √(1.6 × 10^-10) ≈ 1.26 × 10^-5
Therefore, the molar solubility of AgCl in deionized water is approximately 1.26 × 10^-5 mol/L.
b) 0.20 M NaCl Solution
When AgCl is added to a solution of NaCl, the common ion effect comes into play. NaCl is a soluble ionic compound that dissociates into Na+ and Cl- ions. Since Cl- ions are already present in solution, it will affect the solubility of AgCl.
The additional Cl- ions from NaCl will react with the Ag+ ions formed by the dissociation of AgCl. This will result in the formation of more silver chloride, decreasing the solubility of AgCl.
To calculate the molar solubility in a 0.20 M NaCl solution, we need to consider the initial concentration of Cl- ions as 0.20 M.
Let's assign 'x' as the additional molar solubility beyond what would be expected in pure water. The total concentration of Cl- ions will be the sum of the initial concentration from NaCl (0.20 M) and that provided by the additional AgCl dissolved (x). Therefore, the concentration of Cl- ions is [Cl-] = 0.20 + x.
The solubility of AgCl in the presence of Cl- ions can be expressed as:
s = molar solubility of AgCl in pure water (1.26 × 10^-5 M) + x
The expression for Ksp is:
Ksp = [Ag+][Cl-] = (s - x) * (0.20 + x)
Given that Ksp = 1.6 × 10^-10, we can set up the equation:
(1.26 × 10^-5 + x) * (0.20 + x) = 1.6 × 10^-10
You will need to solve this quadratic equation to find the value of 'x', which represents the additional molar solubility of AgCl in the presence of Cl- ions.
Using a reliable calculator or algebraic solving methods, you can find that 'x' is approximately 2.3 × 10^-8 M.
Therefore, the molar solubility of AgCl in a 0.20 M NaCl solution is approximately 1.26 × 10^-5 M (solubility in pure water) + 2.3 × 10^-8 M (additional solubility due to Cl- ions) = 1.26 × 10^-5 M + 2.3 × 10^-8 M.