Asked by John
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?
What is the concentration of the first ion precipitated when the second ion starts to precipitate?
Answers
Answered by
DrBob222
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+).
(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+).
Answered by
John
Thank you so much Dr.Bob222. You have my eternal gratitude.
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