A solution is prepared by dissolving 10.8g ammonium sulphate in enough water to make 100ml of stock solution. A 10ml sample of this stock solution is added to 50ml of water. Calculate concentration of ammonium ions and sulphate ions in the final solution?

We can solve this problem using some molarity calculations:

#"molarity" = "mol solute"/"L soln"#

We should convert the given mass of #"(NH"_4")"_2"SO"_4# to moles using its molar mass (calculated to be #132.14# #"g/mol"#):

#10.8cancel("g (NH"_4")"_2"SO"_4)((1color(white)(l)"mol (NH"_4")"_2"SO"_4)/(132.14cancel("g (NH"_4")"_2"SO"_4))) = color(red)(0.0817# #color(red)("mol (NH"_4")"_2"SO"_4#

This is the quantity present in #100# #"mL soln"#, so let's calculate the molarity of the solution (converting volume to liters):

#"molarity" = (color(red)(0.0817)color(white)(l)color(red)("mol (NH"_4")"_2"SO"_4))/(0.100color(white)(l)"L soln") = color(green)(0.817M#

#10# #"mL"# of this solution is added to #50# #"mL H"_2"O"#, which makes a #60#-#"mL"# total solution.

We can now use the dilution equation

#M_1V_1 = M_2V_2#

to find the molality of the new, #60#-#"mL"# solution:

#(color(green)(0.0817M))(10color(white)(l)"mL") = (M_2)(60color(white)(l)"mL")#

#M_2 = ((color(green)(0.817M))(10cancel("mL")))/(60cancel("mL")) = 0.136M#

This means that there are #0.136# moles of #"(NH"_4")"_2"SO"_4# per liter of solution.

Let's recognize that #1# #"mol (NH"_4")"_2"SO"_4# contains

• #2# #"mol NH"_4^+#

• #1# #"mol SO"_4^(2-)#

The concentrations of each ion is thus

#(2)(0.136M) = color(blue)(0.272M# #color(blue)("NH"_4^+#

#(1)(0.136M) = color(purple)(0.136M# #color(purple)("SO"_4^(2-)#