What is the solubility of silver chloride in water at 25C? (Ksp = 1.6 x 10^-10)
I don't know how to do the work
AgCl(s) ==> Ag^+ + Cl^-
Let x = molar solutility of AgCl.
Then x = molar concentration of Ag^+.
and x = molar concentraton of Cl^-
Then Ksp = (Ag^+)(Cl^-) = 1.6E-10
Substitute from above.
x*x=1.6E-10
Solve for x. That is molar solubility in mols/L. If you want grams AgCl, then M x molar mass = grams.
To find the solubility of silver chloride in water at 25°C, you need to use the given value of Ksp (the solubility product constant).
First, set up an equation using the dissociation of silver chloride:
AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Let x represent the molar solubility of AgCl (in moles per liter). Since silver chloride completely dissociates, the concentration of both Ag+ and Cl- will also be x (in moles per liter).
The solubility product constant expression for AgCl is given by:
Ksp = [Ag+][Cl-]
Substitute the molar solubility (x) into the Ksp expression:
Ksp = x * x
This can be simplified to:
Ksp = x^2 = 1.6 x 10^-10
Solve for x by taking the square root of both sides:
x = √(1.6 x 10^-10)
Now you have the molar solubility of AgCl in moles per liter.
If you want to find the solubility in grams per liter, you can multiply the molar solubility (x) by the molar mass of AgCl. The molar mass of AgCl is the sum of the atomic masses of silver (Ag) and chlorine (Cl), which can be found on the periodic table.
Multiply the molar solubility (x) by the molar mass of AgCl to get the solubility in grams per liter.