Question #742ad

1 Answer
Jan 24, 2015

You would need "0.078 g", or "78 mg" of "Na"_2"SO"_4.

The key factor in solving this problem is the balanced chemical equation for the dissociation of "Na"_2"SO"_4 in water:

Na_2SO_(4(aq)) -> 2Na_((aq))^(+) + SO_(4(aq))^(2-)

Notice that 1 mole of "Na"_2"SO"_4 produces 2 moles of "Na"^(+) ions.

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

"25" * 10^(-3)"g" * ("1 mole Na"^(+))/("23.0 g") = "1.09" * 10^(-3)"moles Na"^(+)

Since the mole ratio between "Na"_2"SO"_4 and "Na"^(+) is 1:2, the number of moles of the former you'll need is

1.09 * 10^(_3)"moles Na"^(+) * ("1 mole Na"_2"SO"_4)/("2 moles Na"^(+)) = 0.55 * 10^(-3) "moles Na"_2"SO"_4

Therefore, the mass of "Na"_2"SO"_4 needed is

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