Question

Review I Constants Periodic Table You may want to reference (Pages 336-337) Section 8.6 while completing this problem Part A Lead ions can be removed from solution by precipitation with sulfate ions. Suppose that a solution contains lead() nitrate. Enter the complete ionic equation to show the reaction of aqueaus lead(l) nitrate with aqueous potassium sulfate to form solid lead(Il) sulfate and aqueous potassium nitrate Express your answer as a chemical equation. Identify all of the phases your answer. | ΑΣΦ ? Pb (aq) 2NO(aq) +2K* (aq) SO42 (aq)>2K* (aq) +2NO (aq)PbSO, (s) Previous Answers Request Answer Submit Part B Enter the net ionic equation to show the reaction of aqueous lead(ll) nitrate with aqueous potassium sulfate to form solid lead(ll) sulfate and aqueous potassium nitrate Express your answer as a chemlcal equatlon. ldentlfy all of the phases In your answer. ΑΣφ ? w Request Answer Submit

Answer 1

Aqueous lead (II) nitrate (Pb(NO₃)₂) reacts with aqueous potassium sulphate (K₂SO₄) to form solid lead (II) sulphate (PbSO₄) and aqueous potassium nitrate (KNO₃) as per the following equation

$Pb(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow PbSO_4(s)+KNO_3(aq)$

There are 2 K on left and one on right so to balance it out write 2 as coefficient of KNO₃

$Pb(NO_3)_2(aq)+K_2SO_4(aq)\rightarrow PbSO_4(s)+2KNO_3(aq)$

Part A: Now split the aqueous compounds into ions as these essentially exist as ions in solution

$Pb^{2+}(aq)+2NO_3^-(aq)+2K^+(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2K^+(aq)+2NO_3^-(aq)$

This is the total or complete ionic equation.

Part B: Now cancel the common ions on both sides (2K⁺ and SO₄^2-) as these don't take part in the reaction and are spectator ions.

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

This is the net ionic equation.