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.

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)

Answer Link

Edufirst Edufirst · Answer 2 · 2018-08-10T22:36:58+02:00

Answer:

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

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.