Question

What is the solubility of silver chloride in g/l? Kpa=[ Ag+][ Cl-]

Answer 1

Solution

For silver chloride, Ksp = [Ag⁺][Cl^-], where the brackets mean concentration in mol/L. For every mole of silver chloride dissolved, one mole of silver ion and one mole of chloride ion are found in solution. Therefore, the first step in solving the problem is to convert the solutility from g/L to mol/L. The formula weight of silver chloride is 107.8 + 35.4 = 143.3. The solubility of AgCl is

6.56x10^-4g  /1 L x  1 mol /143.3 g= 4.58 x 10^-6 mol/L

Therefore, [Ag⁺] = [Cl^-] = 4.58 x 10^-6 mol/L. Substituting this value into the solubility product expression gives:

K_sp= [Ag⁺][Cl^-] = (4.58x10^-6)(4.58x10^-6)

      = 20.9x10^-12 = 2.09x10^-11

The solubility of silver chloride is 6.56x10^-4 g/L.