A 1.0 L solution has 10.00 grams of CaCl2 dissolved in it. What is the Molar solubility of AgCl in this same solution? The Ksp for AgCl is 1.6 x 10-10 at this temperature. Hint: watch out for the common ion again.
i figured it out
To calculate the molar solubility of AgCl in the given solution, we need to consider the common ion effect. The presence of CaCl2 already contributes chloride ions (Cl-) in the solution, which can affect the solubility of AgCl.
To find the molar solubility of AgCl, we can use the Ksp expression for AgCl, which is:
Ksp = [Ag+][Cl-]
Given that the Ksp for AgCl is 1.6 x 10^-10 and the concentration of chloride ions ([Cl-]) is determined by the dissolving of CaCl2, we need to calculate the concentration of chloride ions in the solution.
First, we need to find the molar concentration (molarity) of CaCl2. This can be calculated using the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
Given that we have 10.00 grams of CaCl2 and a total solution volume of 1.0 L, we need to convert grams to moles using the molar mass of CaCl2. The molar mass of CaCl2 is 40.08 g/mol for calcium (Ca) and 35.45 g/mol for chlorine (Cl).
moles of CaCl2 = 10.00 g / (40.08 g/mol + 2(35.45 g/mol))
≈ 10.00 g / 111.98 g/mol
≈ 0.0892 mol
Now we can calculate the molarity of CaCl2:
Molarity (CaCl2) = 0.0892 mol / 1.0 L
= 0.0892 M
Since CaCl2 dissociates into three ions in solution (one Ca2+ ion and two Cl- ions), the concentration of chloride ions contributed by CaCl2 is 2 times the molar concentration of CaCl2. Thus:
[Cl-] = 2 * Molarity (CaCl2)
= 2 * 0.0892 M
= 0.1784 M
Now, we can substitute the value of [Cl-] into the Ksp expression for AgCl and solve for [Ag+]:
Ksp = [Ag+][Cl-]
1.6 x 10^-10 = [Ag+][0.1784 M]
Solving for [Ag+], we get:
[Ag+] = (1.6 x 10^-10) / (0.1784 M)
= 8.98 x 10^-10 M
Therefore, the molar solubility of AgCl in the given solution is approximately 8.98 x 10^-10 M.