Question

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?

Answer 1

• `"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-)`