Question

A solution of 0.170 mol/l ion strength is prepared using NaNO3 (solid) and Na2SO4 (solid). Calculate the mass of each salt that is required to prepare 250 mL of the said solution, knowing that the final Na+ concentration was 0.130 mol/L.

Answer 1

Answer:

1,06 g of NaNO₃ and 1,42 g of Na₂SO₄

Explanation:

In water, NaNO₃ and Na₂SO₄ dissociates as:

NaNO₃ → Na⁺ + NO₃⁻

Na₂SO₄ → 2 Na⁺ SO₄²⁻

The ionic strength is difined as:

∑ $\frac{C_{i} Z_{i}^2}{2}$ Where Ci and Zi are concentration and charge of each ion in solution. Thus:

$\frac{C_{Na+}+C_{NO3-}+4C_{SO4-}}{2}$ = 0,170 mol/L

Knowing $C_{Na^+}$ = 0,130 mol/L you can obtain:

0,210 mol/L = $C_{NO_{3}^-}+4C_{SO_{4}^{2-}}$ (1)

The 0,130 mol/L of Na⁺ comes from  $C_{NaNO_{3}}+2C_{Na_{2}SO_{4}}$, thus:

0,130 mol/L = $C_{NO_{3}^-}+2C_{SO_{4}^{2-}}$ (2)

Replacing (2) in (1)

$C_{NO_{3}^-} = 0,050 M$

And:

$C_{SO_{4}^{2-}} = 0,040 M$

The 0,050 M of NO₃⁻ comes from:

0,050M×0,250L×$\frac{84,99 g}{1molNaNO_{3}}$ = 1,06 g of NaNO₃

The 0,040 M of SO₄⁻ comes from:

0,040M×0,250L×$\frac{142,04 g}{1molNa_{2}SO_{4}}$ = 1,42 g of Na₂SO₄

I hope it helps!