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:
∑ [tex]\frac{C_{i} Z_{i}^2}{2}[/tex] Where Ci and Zi are concentration and charge of each ion in solution. Thus:
[tex]\frac{C_{Na+}+C_{NO3-}+4C_{SO4-}}{2}[/tex] = 0,170 mol/L
Knowing [tex]C_{Na^+}[/tex] = 0,130 mol/L you can obtain:
0,210 mol/L = [tex]C_{NO_{3}^-}+4C_{SO_{4}^{2-}}[/tex] (1)
The 0,130 mol/L of Na⁺ comes from [tex] C_{NaNO_{3}}+2C_{Na_{2}SO_{4}}[/tex], thus:
0,130 mol/L = [tex]C_{NO_{3}^-}+2C_{SO_{4}^{2-}}[/tex] (2)
Replacing (2) in (1)
[tex]C_{NO_{3}^-} = 0,050 M[/tex]
And:
[tex]C_{SO_{4}^{2-}} = 0,040 M[/tex]
The 0,050 M of NO₃⁻ comes from:
0,050M×0,250L×[tex]\frac{84,99 g}{1molNaNO_{3}}[/tex] = 1,06 g of NaNO₃
The 0,040 M of SO₄⁻ comes from:
0,040M×0,250L×[tex]\frac{142,04 g}{1molNa_{2}SO_{4}}[/tex] = 1,42 g of Na₂SO₄
I hope it helps!
