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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:
∑ Where Ci and Zi are concentration and charge of each ion in solution. Thus:
= 0,170 mol/L
Knowing = 0,130 mol/L you can obtain:
0,210 mol/L = <em>(1)</em>
The 0,130 mol/L of Na⁺ comes from , thus:
0,130 mol/L = <em>(2)</em>
Replacing (2) in (1)
And:
The 0,050 M of NO₃⁻ comes from:
0,050M×0,250L× = <em>1,06 g of NaNO₃</em>
The 0,040 M of SO₄⁻ comes from:
0,040M×0,250L× = <em>1,42 g of Na₂SO₄</em>
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