1. A solution is prepared by dissolving 10.8g ammonium sulfate(NH4)2SO4 IN enough water to make 100 ml of stock solution. A 10 sample of this stock soluiton is added to 50 ml of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
2.what volume of 0.1 M Na3PO4 is required to precipitate all the lead ions from 150 ml of 0.25M Pb(NO3)2?
- ?Lv 71 decade agoFavorite Answer
The molecular weight of (NH4)2SO4 is 132 g/mol. The molarity of a 100 mL solution containing 10.8 g ammonium sulfate is calculated to be 10.8/132/(1000/100)=0.0818 M. Dilution of this stock solution with water results in a lower molarity which can be calculated using the equation M1V1=M2V2. Since M1=0.0818, V1=10 mL (assumed) and V2=50 mL, M2 is calculated to be 0.01636 M. Ammonium sulfate dissociates in water to form two ammonium ions and one sulfate ion. Assuming complete dissociation, [NH4+]=0.03272 M, [SO42-]=0.01636 M.3Pb2++2PO43-=Pb3(PO4)2, 150 mL of 0.25 M Pb(NO3)2 contains 0.25=n*1000/150 or n=0.0375 mol of lead which would need 0.025 mol phosphate to precipitate completely. For a 0.1 M solution, each 100 mL contains 0.01 mol. Therefore, 250 mL are needed.