Asked by John

A chemist takes a 647 gram sample of this hydrate. She removes all the water possible from the sample and collects it. To this water, she adds 7.6 grams of lead(II) nitrate. Afterwards she wants to precipitate all the lead. She wants to use a solution of aluminum sulfate to precipitate the lead. The aluminum sulfate solution is 0.07 M and she has 1.0 L of it.
a. What is the Molarity of the lead(II) nitrate solution ?
b. How many ml of aluminum sulfate solution will the chemist need?
c. What will be the resulting Molarity of the aluminum cation left in the resulting solution mixture once she combines her volumes ?
Answered by DrBob222
John, I don't believe you have all of the information listed. You can do b but not a or c with the information listed.
Answered by John
Oops, the hydrate is Copper Sulfate Pentahydrate
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