A volume of 0.800 L of a 2 x 10–4 M Ba(NO3)2 solution is added to 0.200 L of 5 x 10–4 M Li2SO4. Does BaSO4 precipitate? Explain your answer

The BaSO₄ precipitates as a solid.

Explanation:

In order to determine the precipitation, we need to look at the reactants, we have: Ba(NO₃)₂ and Li₂SO₄

Salts from nitrate are all soluble and sulfates can make precipitate if they react to elements from group 2. (Solubility rules) So this reaction, has a precipitate but we can predict it, if we look at the reaction and we take acount the Kps:

Ba(NO₃)₂ (aq) + Li₂SO₄ (aq)  →  2LiNO₃ (aq)  +  BaSO₄ (s) ↓

Let's find out the moles of each ion:

Ba(NO₃)₂ (aq) →  Ba²⁺  (aq)  +  2NO₃⁻  (aq)

0.8 L . 0.0002 M = 1.6×10⁻⁴ moles. These are the moles of Ba²⁺ in 1 L of solution, so [Ba²⁺] = 1.6×10⁻⁴ M

Li₂SO₄ (aq) →  2Li⁺ (aq)  +  SO₄⁻² (aq)

0.2 L . 0.0005 M = 1×10⁻⁴ M. These are the moles of sulfate, in 1 L of solution so [SO₄⁻²] =  1×10⁻⁴ M

BaSO₄ (s)  ⇆   Ba²⁺  (aq) + SO₄⁻² (aq)      Kps = 1.08×10⁻¹⁰

We need to compare Kps to [ Ba²⁺] . [SO₄⁻²]

[ Ba²⁺] . [SO₄⁻²] = 1.6 ×10⁻⁸ >  Kps →  Precipitate.

If  [Ba²⁺] . [SO₄⁻²] = Kps  → Solution is saturated

If [ Ba²⁺] . [SO₄⁻²] < Kps → There is no precipitate formed, or if it is formed, it will be quickly dissolved to reach saturation.

im not exactly sure but i think its b!