chemistry
posted by Anonymous on .
What is the solubility of AgCl when it is in a solution of 0.15 M NaCl? (AgCl has Ksp = 1.8 x 1010.) Make a reaction table. Include rows for initial concentration, change in concentration, and equilibrium concentration. Write down the equation for the Ksp of the reaction. What is the concentration of Cl? Plug that number into the Ksp equation to determine the solubility of AgCl when it is in a solution of 0.15 M NaCl

...........AgCl ==> Ag^+ + Cl^
I............x........0........0
C...........x.........x........x
E...........x..........x......x
........NaCl ==> Na^+ + Cl^
I.......0.15M.....0......0
C.......0.15M....0.15....0.15
E........0.........0.15...0.15
Ksp + (Ag^+)(Cl^)
1.8E10 = (x)(x+0.15) note:(Cl^) = x+ 0.15 or x from the AgCl + 0.15 from the NaCl. This a problem illustrating the common ion effect. The solubility of AgCl is DECREASED significantly by the addition of a common ion.
Solve for x.