Step 1: Write the skeleton equation for the reaction:
NaOH(aq) + Pb(NO3)2(aq) → ?
Step 2: Determine the products formed:
When aqueous sodium hydroxide (NaOH) reacts with lead(II) nitrate (Pb(NO3)2), the products formed are solid lead(II) hydroxide (Pb(OH)2(s)) and aqueous sodium nitrate (NaNO3(aq)).
NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s) + NaNO3(aq)
Step 3: Balance the equation:
The balanced chemical equation for the reaction is:
2 NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s) + 2 NaNO3(aq)
Step 4: Additional reaction:
When additional aqueous hydroxide (OH-) is added to the precipitate, the solid lead(II) hydroxide (Pb(OH)2) redissolves to form a soluble [Pb(OH)4]2–(aq) complex ion.
Pb(OH)2(s) + ? → [Pb(OH)4]2–(aq)
Step 5: Determine the products formed:
When solid lead(II) hydroxide (Pb(OH)2) reacts with additional aqueous hydroxide (OH-), the product formed is the soluble complex ion [Pb(OH)4]2–(aq).
Pb(OH)2(s) + 4 OH-(aq) → [Pb(OH)4]2–(aq)
Step 6: Balance the equation:
The balanced chemical equation for the additional reaction is:
Pb(OH)2(s) + 4 OH-(aq) → [Pb(OH)4]2–(aq)
So, the overall balanced chemical equation for the reactions involving aqueous sodium hydroxide and lead(II) nitrate is:
2 NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s) + 2 NaNO3(aq)
And the balanced chemical equation for the additional reaction where the precipitate redissolves is:
Pb(OH)2(s) + 4 OH-(aq) → [Pb(OH)4]2–(aq)