A 0.5-L saturated solution of silver chloride, AgCl, is evaporated to dryness. The residue is equivalent to 1.4 x 10^-3 mole. Calculate the experimental value of the Ksp of silver chloride. Show all of your work.
....AgCl ==> Ag^+ + Cl^-
......x.......x......x
mols AgCl = 1.4E-3 mols in 1/2 L; therefore, you will have 1.4E-3 mol x 2 = 2.8E-3 mols in a L.
Ksp = (Ag^+)(Cl^-)
Ag^+ = Cl^- = 2.8E-6
Substitute and solve for Ksp. Not good experimental result.
To calculate the experimental value of the solubility product constant (Ksp) of silver chloride (AgCl), we need to use the given information.
Step 1: Write the balanced chemical equation:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)
Step 2: Determine the molar concentration of AgCl in the solution:
Given that the solution is saturated, it means that the concentration of Ag⁺ and Cl⁻ ions should be equal. Hence, the concentration of Ag⁺ and Cl⁻ is equal to the concentration of AgCl.
The given solution volume is 0.5 L, and the number of moles of AgCl in the residue is 1.4 × 10⁻³ mol. Using these values, we can calculate the molar concentration (C) of AgCl as follows:
C = moles of solute / volume of solution
C = 1.4 × 10⁻³ mol / 0.5 L
C = 2.8 × 10⁻³ M
Step 3: Calculate the concentration of Ag⁺ and Cl⁻ ions:
Since AgCl dissociates into one Ag⁺ ion and one Cl⁻ ion, the concentration of Ag⁺ and Cl⁻ is also 2.8 × 10⁻³ M.
Step 4: Write the expression for the solubility product constant (Ksp):
Ksp = [Ag⁺][Cl⁻]
Step 5: Substitute the concentration values into the Ksp expression:
Ksp = (2.8 × 10⁻³ M) × (2.8 × 10⁻³ M)
Ksp = 7.84 × 10⁻⁶
Therefore, the experimental value of the Ksp of silver chloride (AgCl) is 7.84 × 10⁻⁶.