posted by Anonymous on .
A solution is prepared by dissolving 15.6 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL sample of this stock is added to 52.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
I know concentration is Molarity = moles/L I need help starting the problem.
15.6 g (NH4)2SO4 = ?? moles.
moles = 15.6/molar mass (NH4)2SO4.
M (NH4)2SO4 = moles/0.100 L for the stock solution = xx M
Technically you can't go from this solution to the final solution BECAUSE 10 mL of the stock is ADDED to 52 mL water and the final volume may or may not be 62 mL. We will assume it is 62 mL.
final concn (NH4)2SO4 = xxM x (10/62) = yy M. The sulfate will be yy M (1 mole sulfate/1 mole ammonium sulfate) and the NH4 concn will be twice that (2 NH4^+ in 1 mole (NH4)2SO4).