posted by adam on .
How many mL of 0.109 M Na2CO3 would be needed to precipitate all of the copper ions in 26.7 mL of 0.159 M CuSO4?
Na2CO3 + CuSO4 --> CuCO3 + Na2SO4
CuCO3 is insoluble and so form ppt.
Na2CO3 : CuSO4
1 : 1
actual mole of CuSO4 = Mv = 0.159M x 26.7e-3L =_______moles = moles of Na2CO3.
volume of Na2CO3 = n/M where n is the mole of Na2CO3 and M is the molarity of Na2CO3. The volume should be in L so you need to convert to mL.
hope that helps.