Question

If you used 30.5mL of water to wash 0.3333g silver chloride precipitate and this wash liquid only had time to become 75.0% saturated with AgCl how many grams of AgCl would you be washing away? Ksp for AgCl = 1.8x 10^-10.

Answer 1

AgCl (s) + H2O <------> Ag+(aq) + Cl-(aq)

Ksp = [Ag+(aq)][ Cl-(aq)]= 1.8*10^-10

So sollubity of = [Ag+(aq)] = [ Cl-(aq)]= √(1.8*10^-10) = 1.34 *10^-5 moles/L

Thus if water is 100% saturated 1 litrecontain 1.34 *10^-5 moles AgCl.

Here water is only 75% saturated so water contains only (1.34 *10^-5 moles*75)/100 = 1.01*10^-5 moles AgCl/Litre.

Here there's only 30.5 ml water,hence it contains (1.01*10^-5 moles *30.5)/1000 = 3.07*10^-7 moles AgCl.

1 mole AgCl = 143.32 g

So 3.07*10^-7 moles AgCl = (3.07*10^-7*143.32) = 0.000044 g

Grams of AgCl washed away = 4.4*10^-5 g