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?

1 Answer
Jul 6, 2017

Answer:

  • #"NH"_4^+#: #0.272M#

  • #"SO"_4^(2-)#: #0.136M#

Explanation:

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-)#