Answer:
a. 2 H⁺ (aq) + 2 OH⁻ (aq) → 2H₂O(l)
b. Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
Explanation:
First of all, we must determine if the reactants produce precipitates:
A. HNO₃, Ca(OH)₂
B. Pb(NO₃)₂, KI
A. This is a neutralization reaction:
Acid + Base → Water and Salt
The net ionic equation shows, that water is produced
We dissociate the compounds:
HNO₃ → H⁺(aq) + NO₃⁻(aq)
Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq)
Ionic complete equation:
2H⁺(aq) + 2NO₃⁻(aq) + Ca²⁺(aq) + 2OH⁻(aq) → Ca²⁺(aq) + 2NO₃⁻(aq) + H₂O(l)
Net ionic equation: 2 H⁺ (aq) + 2 OH⁻ (aq) → H₂O(l)
B. Pb(NO₃)₂, KI
We dissociate: Pb(NO₃)₂ (aq) → Pb²⁺(aq) + 2NO₃⁻(aq)
KI(aq) → K⁺(aq) + I⁻(aq)
Salts from nitrate are soluble but, when the iodide reacts, it can make a lead iodide precipitate
Ionic complete equation:
Pb²⁺(aq) + 2NO₃⁻(aq) + K⁺(aq) + 2I⁻(aq) → K⁺(aq) + 2NO₃⁻(aq) +PbI₂(s)
Net ionic equation: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s)
We cancel the repeated ions
