Posted by mel on .
A solution is prepared by dissolving 16.5 g ammonium sulfate in enough water to make 145.0 mL of stock solution. A 10.50 mL sample of this stock solution is added to 57.70 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
16.5 g (NH4)2SO4 in 145.0 mL solution =
16.5 g/145 mL = xx g/mL and use this below.
Concn of diluted solution is
xx g/mL x 10.50 mL/(57.70+10.50) = yy g (NH4)2SO4/mL.
How is the concentration to be measured? g/mL is what the yy is above.
(SO4^=) = same as (NH4)2SO4 = zz g/mL.
(NH4^+) = 2 x [(NH4)2SO4] = ww g/mL.
If you want these numbers in molarity, just multiply zz or ww g/mL x 1000 and divide by molar mass (NH4(2SO4.
Check my thinking. Check my work.
Suppose 15.o mL of 0.315 M NH4CL is combine with 10.0mL OF 0.430 M (NH4)3PO4. What is the total concentration of ammonium ions in this mixture?