Calculate the molar solubility of silver chloride (AgCl) in:
a-deionized water and
b-.20M NaCl solution
Given Ksp=1.6*10^-10
please show work.
To calculate the molar solubility of silver chloride (AgCl) in deionized water and a 0.20 M NaCl solution, we need to consider the effect of the common ion on the solubility.
a) Deionized water:
In deionized water, there is no common ion present. Therefore, the molar solubility of AgCl will be simply given by the square root of the Ksp value.
Step 1: Write the balanced chemical equation for the dissolution of AgCl:
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)
Step 2: Write the expression for the solubility product constant (Ksp):
Ksp = [Ag+][Cl-]
Step 3: Substitute the known values into the expression:
Ksp = [Ag+][Cl-] = (Ag+) * (Cl-)
Step 4: Use the Ksp value given (Ksp = 1.6 × 10^-10) to find the concentration:
(Ag+)(Cl-) = 1.6 × 10^-10
Since AgCl is a 1:1 electrolyte, the concentration of Ag+ and Cl- will be equal: (Ag+) = (Cl-). Therefore, we can rewrite the equation as:
(Ag+)^2 = 1.6 × 10^-10
Step 5: Take the square root of both sides to solve for (Ag+):
(Ag+) = sqrt(1.6 × 10^-10) = 4 × 10^-6 M
Therefore, the molar solubility of silver chloride (AgCl) in deionized water is 4 × 10^-6 M.
b) 0.20 M NaCl solution:
In a 0.20 M NaCl solution, there is a common ion present (Cl-) from NaCl that can affect the solubility of AgCl. This is known as the common ion effect, where the presence of an ion in solution decreases the solubility of a slightly soluble salt.
To calculate the molar solubility in the presence of the common ion, we need to consider the new concentration of chloride ions.
Step 1: Calculate the concentration of chloride ions in the 0.20 M NaCl solution.
[Cl-] = 0.20 M
Step 2: Write the expression for the solubility product constant (Ksp):
Ksp = [Ag+][Cl-]
Step 3: Substitute the known values into the expression:
Ksp = [Ag+][Cl-] = (Ag+)(0.20 M)
Step 4: Use the Ksp value given (Ksp = 1.6 × 10^-10) and the concentration of chloride ions to find the concentration of Ag+:
(Ag+)(0.20 M) = 1.6 × 10^-10
Step 5: Solve for (Ag+):
(Ag+) = (1.6 × 10^-10)/(0.20 M) = 8 × 10^-10 M
Since (Ag+) = (Cl-), the molar solubility of AgCl in a 0.20 M NaCl solution is 8 × 10^-10 M.