Jan 24, 2015

You would need $\text{0.078 g}$, or $\text{78 mg}$ of ${\text{Na"_2"SO}}_{4}$.

The key factor in solving this problem is the balanced chemical equation for the dissociation of ${\text{Na"_2"SO}}_{4}$ in water:

$N {a}_{2} S {O}_{4 \left(a q\right)} \to 2 N {a}_{\left(a q\right)}^{+} + S {O}_{4 \left(a q\right)}^{2 -}$

Notice that 1 mole of ${\text{Na"_2"SO}}_{4}$ produces 2 moles of ${\text{Na}}^{+}$ ions.

What a $\text{25}$ ppm solution of ${\text{Na}}^{+}$ ions means is that you must have $\text{25 mg}$, or $\text{25" * 10^(-3)"g}$, of sodium ions in $\text{1 L}$, or $\text{1000 mL}$, of water. You can determine the moles of sodium ions your solution must contain by using

${\text{25" * 10^(-3)"g" * ("1 mole Na"^(+))/("23.0 g") = "1.09" * 10^(-3)"moles Na}}^{+}$

Since the mole ratio between ${\text{Na"_2"SO}}_{4}$ and ${\text{Na}}^{+}$ is $1 : 2$, the number of moles of the former you'll need is

$1.09 \cdot {10}^{_ 3} {\text{moles Na"^(+) * ("1 mole Na"_2"SO"_4)/("2 moles Na"^(+)) = 0.55 * 10^(-3) "moles Na"_2"SO}}_{4}$

Therefore, the mass of ${\text{Na"_2"SO}}_{4}$ needed is

$\text{0.55" * 10^(-3) "moles Na"_2"SO"_4 * ("142.0 g")/("1 mole") = 78 * 10^(-3)"g" = "78 mg}$