Question

Write complete ionic equation to show the reaction of aqueous lead(II) nitrate with aqueous potassium sulfate to form solid lead(II) sulfate and aqueous potassium nitrate.

Answer 1

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + SO₄²⁻ (aq) → PbSO₄(s) + 2K⁺ + 2 NO₃⁻ (aq)

Step-by-step explanation:

1) Start understanding the nature of the reactants, the kind of reaction that occurs, and the nature of the products:

i) word equation (given):

aqueous lead(II) nitrate + aqueous potassium sulfate → solid lead(II) sulfate + aqueous potassium nitrate

ii) molecular equation, including the phases:

Pb(NO₃)₂ (aq) + K₂SO₄ (aq) → PbSO₄(s) + 2KNO₃ (aq)

iii) It is a combination of two salts to form two new salts, in a double replacement reaction.

2) Separate the ionic compounds in solution into its ions

i) Reactant side:

Pb(NO₃)₂ (aq) = Pb²⁺ (aq) + 2NO₃⁻ (aq)

K₂SO₄ (aq) = 2K⁺ (aq) + SO₄²⁻ (aq)

ii) Product side

PbSO₄(s) remains unchanged since it did not ionize

2KNO₃(aq) = 2K⁺ + 2 NO₃⁻ (aq)

3) Write both sides, i.e. the whole reaction in the ionic form:

Pb²⁺ (aq) + 2NO₃⁻ (aq) + 2K⁺ (aq) + SO₄²⁻ (aq) → PbSO₄(s) + 2K⁺ + 2 NO₃⁻ (aq)

That is the total ionic equation.

Answer 2

The correct answer to this question is this one:

Pb(NO3)2(aq)+K2SO4(aq)→PbSO4(s)+2KNO3(aq)
Everything except the lead sulfate is soluble, so they all dissociate.

Pb2++2NO−3+2K++SO2−4→PbSO4(s)+2K++2NO−3
Cancel out stuff that appears on both sides, and your net ionic equation is

Pb2+(aq)+SO2−4(aq)→PbSO4(s)