Write the balanced chemical equation for each of these reactions. Include phases. When aqueous sodium hydroxide is added to a solution containing lead(II) nitrate, a solid precipitate forms. However, when additional aqueous hydroxide is added the precipitate redissolves forming a soluble [Pb(OH)4]2–(aq) complex ion.
Just follow the written statements.
2NaOH(aq) + Pb(NO3)2(aq) --> Pb(OH)2(s) + 2NaNO3(aq)
Pb(OH)2(s) + 2NaOH(aq) --> [Pb(OH)4]^2-(aq) + 2Na^+(aq). However, this leaves it in ionic form and I think the problem wants it in molecular form so I would make the product Na2Pb(OH)4(aq)
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)
To write balanced chemical equations, we need to follow a set of steps:
Step 1: Write the molecular formula for each of the reactants and products involved in the reaction.
In this case, we have:
Reactants:
- Aqueous sodium hydroxide: NaOH(aq)
- Lead(II) nitrate: Pb(NO3)2(aq)
Products:
- Solid precipitate: Pb(OH)2(s)
Step 2: Write the unbalanced equation by combining the reactants and products.
NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s)
Step 3: Balance the equation by adding coefficients in front of the compounds as needed.
NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s)
Since there is only one Na and one OH on the left side, we have:
2NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s)
Step 4: Include phases (states) for each compound.
2NaOH(aq) + Pb(NO3)2(aq) → Pb(OH)2(s)
Now, let's consider the second part of the reaction where additional aqueous hydroxide is added and the precipitate redissolves to form a soluble [Pb(OH)4]2–(aq) complex ion.
Step 1: Write the molecular formula for each component involved in the reaction.
Reactant:
- Solid precipitate: Pb(OH)2(s)
Product:
- [Pb(OH)4]2–(aq)
Step 2: Write the unbalanced equation.
Pb(OH)2(s) + OH–(aq) → [Pb(OH)4]2–(aq)
Step 3: Balance the equation by adding coefficients.
Pb(OH)2(s) + 4OH–(aq) → [Pb(OH)4]2–(aq)
Step 4: Include phases (states) for each compound.
Pb(OH)2(s) + 4OH–(aq) → [Pb(OH)4]2–(aq)
These are the balanced chemical equations for the reactions you described.