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?

Jul 6, 2017

• ${\text{NH}}_{4}^{+}$: $0.272 M$

• ${\text{SO}}_{4}^{2 -}$: $0.136 M$

Explanation:

We can solve this problem using some molarity calculations:

$\text{molarity" = "mol solute"/"L soln}$

We should convert the given mass of ${\text{(NH"_4")"_2"SO}}_{4}$ to moles using its molar mass (calculated to be $132.14$ $\text{g/mol}$):

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 $100$ $\text{mL soln}$, 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

$10$ $\text{mL}$ of this solution is added to $50$ $\text{mL H"_2"O}$, which makes a $60$-$\text{mL}$ total solution.

We can now use the dilution equation

${M}_{1} {V}_{1} = {M}_{2} {V}_{2}$

to find the molality of the new, $60$-$\text{mL}$ solution:

$\left(\textcolor{g r e e n}{0.0817 M}\right) \left(10 \textcolor{w h i t e}{l} \text{mL") = (M_2)(60color(white)(l)"mL}\right)$

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

This means that there are $0.136$ moles of ${\text{(NH"_4")"_2"SO}}_{4}$ per liter of solution.

Let's recognize that $1$ ${\text{mol (NH"_4")"_2"SO}}_{4}$ contains

• $2$ ${\text{mol NH}}_{4}^{+}$

• $1$ ${\text{mol SO}}_{4}^{2 -}$

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-)