Write the balanced net ionic equation for the reactions that occur when the given aqueous solutions are mixed. Include the physical states.
A. nitric acid, HNO3 , and calcium hydroxide, Ca(OH)2
B. lead(II) nitrate, Pb(NO3)2 , and potassium iodide, KI
A. H+(aq)+OH−(aq)⟶H2O(l)
and
B. Pb2+(aq)+2I−(aq)⟶PbI2(s)
A. The balanced net ionic equation for the reaction between nitric acid (HNO3) and calcium hydroxide (Ca(OH)2) is:
2H+(aq) + 2OH-(aq) → 2H2O(l)
B. The balanced net ionic equation for the reaction between lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI) is:
Pb2+(aq) + 2I-(aq) → PbI2(s)
A. To write the balanced net ionic equation for the reaction between nitric acid (HNO3) and calcium hydroxide (Ca(OH)2), we first need to write the balanced chemical equation:
HNO3 (aq) + Ca(OH)2 (aq) → Ca(NO3)2 (aq) + 2H2O (l)
The balanced net ionic equation would exclude the spectator ions, which are ions that are present on both the reactant and product sides of the equation without undergoing any chemical change. In this case, the spectator ions are the nitrate ions (NO3-) and the calcium ions (Ca2+).
Net ionic equation:
2H+ (aq) + 2OH- (aq) → 2H2O (l)
B. For the reaction between lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI), we can write the balanced chemical equation:
Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq)
The balanced net ionic equation would exclude the spectator ions, which in this case are the nitrate ions (NO3-) and the potassium ions (K+).
Net ionic equation:
Pb2+ (aq) + 2I- (aq) → PbI2 (s)
To write the balanced net ionic equation for the reactions, we need to first write the balanced molecular equation for each reaction by combining the reactants, followed by identifying the soluble compounds that dissociate into ions, and finally balancing the equation based on the stoichiometry.
A. nitric acid, HNO3, and calcium hydroxide, Ca(OH)2:
Step 1: Write the molecular equation:
HNO3 (aq) + Ca(OH)2 (aq) →
Step 2: Determine the products and if they are soluble or insoluble:
HNO3 is a strong acid, so it ionizes completely in water:
H+ (aq) + NO3- (aq)
Ca(OH)2 is a strong base and also dissociates fully in water:
Ca2+ (aq) + 2OH- (aq)
Step 3: Combine the ions to form the products:
H+ (aq) + NO3- (aq) + Ca2+ (aq) + 2OH- (aq) →
Step 4: Balance the equation to ensure the number of atoms and charges are equal on both sides:
H+ (aq) + OH- (aq) → H2O (l) (This is a special case since H+ and OH- ions combine to form water.)
Therefore, the balanced net ionic equation is:
H+ (aq) + OH- (aq) → H2O (l)
B. lead(II) nitrate, Pb(NO3)2, and potassium iodide, KI:
Step 1: Write the molecular equation:
Pb(NO3)2 (aq) + 2KI (aq) →
Step 2: Determine the products and if they are soluble or insoluble:
Pb(NO3)2 is soluble in water and dissociates into ions:
Pb2+ (aq) + 2NO3- (aq)
KI is also soluble in water and dissociates into ions:
K+ (aq) + I- (aq)
Step 3: Combine the ions to form the products:
Pb2+ (aq) + 2NO3- (aq) + 2K+ (aq) + 2I- (aq) →
Step 4: Balance the equation:
Pb2+ (aq) + 2I- (aq) → PbI2 (s)
Therefore, the balanced net ionic equation is:
Pb2+ (aq) + 2I- (aq) → PbI2 (s)
Please note that in both cases, the spectator ions (H+ and NO3- in the first reaction, K+ in the second reaction) are not included in the net ionic equation because they do not participate in any chemical change.