Answer:
Sr²⁺ (aq)+ SO₄⁻²(aq) → SrSO₄(s)
Explanation:
The reactants of our reaction are:
SrCl₂ and K₂SO₄
The products of the reactios are:
SrSO₄ and KCl
We write the molecular reaction:
SrCl₂(aq) + K₂SO₄(aq) → 2KCl(aq) + SrSO₄(s) ↓
To state the net ionic equation we split the salts by their ions from the aqueous solutions. The solid product stays the same.
Sr²⁺ (aq) + 2Cl⁻(aq) + 2K⁺(aq) + SO₄⁻²(aq) → 2K⁺(aq) + 2Cl⁻(aq) + SrSO₄(s)
As the potassium cation and the chloride are repeated, they are spectators ions so they won't react in water. Finally our net ionic equation will be:
Sr²⁺ (aq)+ SO₄⁻²(aq) → SrSO₄(s)
Sulfates from elements of group 2, always produces precipitate
Taking into account the definition of net ionic equation, a precipitate of SrSO₄ is formed when When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, and the net ionic equation is:
Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)
The net ionic equation is a chemical equation for a reaction that lists only the species that participate in the reaction.
In other words, an ionic equation is a chemical equation where electrolytes in aqueous solution are written as dissociated ions. Usually this is a salt dissolved in water.
The balanced equation will be:
SrCl₂(aq) + K₂SO₄(aq) → 2KCl(aq) + SrSO₄(s)
where (aq) means aqueous and (s) solid.
Taking into account that:
The total ionic equation in separated aqueous solution will be:
Sr⁺²(aq) + 2 Cl⁻(aq) + 2 K⁺(aq) + SO₄⁻²(aq) → 2 K⁺(aq) + 2 Cl⁻(aq) + SrSO₄(s)
A spectator ion is an ion that appears both as a reactant and as a product in an ionic equation. Spectator ions can be either cations (positively charged ions) or anions (negatively charged ions).
When writing a net ionic equation, spectator ions found in the original equation are ignored. Thus, the total ionic reaction is different from the net chemical reaction.
In this case, K⁺ and Cl⁻ are the spectator ions because they appear unchanged in both the product and the reagent. So these ions cancel out by writing the net ionic equation and you get:
Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)
Since SrSO₄ is an insoluble salt, it will precipitate.
Finally, a precipitate of SrSO₄ is formed when When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, and the net ionic equation is:
Sr⁺²(aq) + SO₄⁻²(aq) → SrSO₄(s)
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