Question

When an aqueous solution of strontium chloride is added to an aqueous solution of potassium sulfate, a precipitation reaction occurs. Write the balanced net ionic equation of the reaction.

Answer 1

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

Answer 2

Answer:

SO4^2-(aq) + Sr^2+(aq) → SrSO4(s)

Explanation:

Step 1: Data given

an aqueous solution of strontium chloride = SrCl2(aq)

an aqueous solution of potassium sulfate = K2SO4(aq)

Step 2: The unbalanced equation

K2SO4(aq) + SrCl2(aq) → KCl(aq) + SrSO4(s)

Step 3: Balancing the equation

K2SO4(aq) + SrCl2(aq) → KCl(aq) + SrSO4(s)

On the left side we have 2x K (in K2SO4), on the right side we have 1x K (in KCl). To balance the amount of K on both sides, We have to multiply KCl (on the right side) by 2.

K2SO4(aq) + SrCl2(aq) → 2KCl(aq) + SrSO4(s)

Step 4: The net ionic equation:

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will.

2K+(aq) + SO4^2-(aq) + Sr^2+(aq) + 2Cl-(aq) → 2K+(aq) + 2Cl-(aq) + SrSO4(s)

After canceling those spectator ions in both side, look like this:

SO4^2-(aq) + Sr^2+(aq) → SrSO4(s)