Silver ions may be recovered from used photographic fixing solution by precipitating it is as silver chloride. The solubility of silver chloride is 1.6 x10-3 g/L. Calculate the Ksp in __(answer) * 10-10 for silver chloride. Give the answer to one decimal place.
Molar Masses: Ag = 107.87 g/mol, Cl = 35.45 g/mol
...............AgCl(s) ==> Ag^+ + Cl^-
I...............solid.............0...........0
C.............solid...............x..........x
E..............solid...............x..........x
Ksp = (Ag^+)(Cl^-)
Ksp = x*x
You know x is 1.6E-3 g/L. Convert that to mols/L
1.5E-3/143.32 = ? Substitute and solve for Ksp.
To calculate the solubility product constant (Ksp), you need to know the concentrations of the silver ions (Ag+) and chloride ions (Cl-) in a saturated solution of silver chloride (AgCl).
First, let's convert the solubility of silver chloride to moles per liter (mol/L). The solubility is given as 1.6 x 10^(-3) g/L.
The molar mass of AgCl is Ag (107.87 g/mol) + Cl (35.45 g/mol) = 143.32 g/mol.
To convert grams to moles, we divide the given mass by the molar mass:
1.6 x 10^(-3) g / (143.32 g/mol) = 1.1 x 10^(-5) mol/L
So, the concentration of Ag+ and Cl- ions in a saturated solution of AgCl is 1.1 x 10^(-5) mol/L.
The balanced chemical equation for the dissociation of silver chloride is:
AgCl(s) ↔ Ag+(aq) + Cl-(aq)
Since the stoichiometry of the reaction is 1:1, the concentration of Ag+ and Cl- ions in the saturated solution is the same.
Therefore, the concentration of Ag+ and Cl- ions is 1.1 x 10^(-5) mol/L each.
Now, to calculate the Ksp, we multiply the concentrations of Ag+ and Cl- ions:
Ksp = [Ag+][Cl-] = (1.1 x 10^(-5) mol/L) x (1.1 x 10^(-5) mol/L)
Ksp = 1.21 x 10^(-10) mol^2/L^2
Finally, we express the Ksp in the requested format, which is __.(answer) x 10^(-10) mol^2/L^2. Rounding to one decimal place:
Ksp = 1.2 x 10^(-10) mol^2/L^2
Therefore, the Ksp of silver chloride is 1.2 x 10^(-10) mol^2/L^2.