11. A solution is prepared by dissolving 25.0 g of ammonium sulfate in 100 mL of solution. Then, 10.0 mL of this stock was added to 50.0 mL of water. Calculate the concentrations of ammonium and sulfate in the final solution. Assume the volumes are additive. Imol Oo189 13a.149 moles Concentration of original ammonium sulfate solution: 1.89 M NHY2 Concentration of final solution (total volume = 60 mL) after dilution: MIV1 = M2V2 M2 = 0.315 M (NH4)2SO4 M (SO,2-) = 0.315 M SO M (NH4) = 2 x (0.315) = 0.630 M NH4
11. A solution is prepared by dissolving 25.0 g of ammonium sulfate in 100 mL of solution. Then, 10.0 mL of this stock was added to 50.0 mL of water. Calculate the concentrations of ammonium and sulfate in the final solution. Assume the volumes are additive. Imol Oo189 13a.149 moles Concentration of original ammonium sulfate solution: 1.89 M NHY2 Concentration of final solution (total volume = 60 mL) after dilution: MIV1 = M2V2 M2 = 0.315 M (NH4)2SO4 M (SO,2-) = 0.315 M SO M (NH4) = 2 x (0.315) = 0.630 M NH4
Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter4: Chemical Reactions In Solution
Section: Chapter Questions
Problem 4.110QE
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Why do I keep getting 3.15 M for the part undelined in blue?
Transcribed Image Text:11. A solution is prepared by dissolving 25.0g of ammonium sulfate in 100 mL of solution.
Then, 10.0 mL of this stock was added to 50.0 mL of water. Calculate the concentrations of
ammonium and sulfate in the final solution. Assume the volumes are additive.
25g Imol 0.189
132.149 moles
Concentration of original ammonium sulfate solution: 1.89 M
NHY2 Concentration of final solution (total volume = 60 mL) after dilution:
M¡V1 = M2V2
M2 = 0.315 M (NH4)2SO4
M (SO4²-) = 0.315 M SO, M (NH4*) = 2 x (0.315) = 0.630 M NH4*
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