Combining aqueous solutions of bai2 and na2so4 affords a precipitate of baso4. which ions are spectator ions in the reaction?

When combining aqueous solutions of BaI₂ and Na₂SO₄ to form a precipitate of BaSO₄, the spectator ions are Na⁺ and I⁻.

Let's consider the molecular equation for the reaction that takes place when combining aqueous solutions of BaI₂ and Na₂SO₄. This is a double displacement reaction.

BaI₂(aq) + Na₂SO₄(aq) ⇒ BaSO₄(s) + 2 NaI(aq)

Then, let's consider the net ionic equation, which includes the ions that participate in the reaction (not spectator ions) and the insoluble species.

Ba²⁺(aq) + SO₄²⁻(aq) ⇒ BaSO₄(s)

As we can see, the ions missing in the net ionic equation because they are spectator ions are Na⁺ and I⁻.

When combining aqueous solutions of BaI₂ and Na₂SO₄ to form a precipitate of BaSO₄, the spectator ions are Na⁺ and I⁻.

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I⁻ (aq) and Na⁺ (aq)

Explanation:

We have the chemical reaction:

BaI₂ + Na₂SO₄ → BaSO₄ + 2 NaI

However if you want to determine the spectator ions you need  to write the states of compounds:

(aq) - ions dissolved in water

(s) - solid

Ba²⁺ (aq) + 2 I⁻ (aq) + 2 Na⁺ (aq) + SO₄²⁻ (aq) → BaSO₄ (s) + 2 Na⁺ (aq) + 2 I⁻ (aq)

The ions which does not change the state and remains dissolved in the solution are spectator ions. For our chemical reaction we have the following spectator ions:

I⁻ (aq) and Na⁺ (aq)

the answer is   c.) 1575 j.

explanation: