posted by John on .
Solid Na2SO4 is added to a solution which is 0.020 M in Pb(NO3)2 and 0.045 M in AgNO3. Assume the volume remains constant. Ksp = 2.0 10-8 for PbSO4 and Ksp = 1.2 10-5 for Ag2SO4.
What is the concentration of the first ion precipitated when the second ion starts to precipitate?
Ksp = 2E-8 = (Pb^2+)(SO4^2-)
(SO4^2-) = 2E-8/0.02 = 1E=6M
Ksp = 1.2E-5 = (Ag^+)^2(SO4^2-)
(SO4^2-) = 1.2E-5/(0.045)^2 = 5.9E-3
When Na2SO4 is added incrementally, the first Ksp exceeded will begin to ppt. Ksp for PbSO4 is smaller; therefore, it will be the first ppt. It will continue to ppt as Na2SO4 is added until the Ksp for Ag2SO4 is exceeded. When will that be? When the (SO4^2-) becomes 5.9E-3M (and not before). So what will (Pb^2+) be when SO4^2- is 5.9E-3. Go back to the PbSO4 Ksp, plug in SO4^2- = 5.9E-3 and calculate the (Pb^2+).
Thank you so much Dr.Bob222. You have my eternal gratitude.