Pb(NO3)2(aq) + NaSO4(aq) → PbSO4(s)+ 2NaNO3(aq)
Identify the ions that are present in each solution, and calculate their concentrations.
How in the world do you calculate concns with no numbers to start? My crystal ball is hazy today.
buhu
To identify the ions present in each solution, we need to consider the dissociation of the compounds in water.
Pb(NO3)2 (aq):
- Pb(NO3)2 dissociates into Pb2+ ions and 2 NO3- ions.
- So, the ions present are Pb2+ and NO3-.
NaSO4 (aq):
- NaSO4 dissociates into Na+ ions and SO4 2- ions.
- So, the ions present are Na+ and SO4 2-.
Now, let's calculate their concentrations. To do this, we need to know the initial concentration or amount of Pb(NO3)2 and NaSO4 that are mixed.
Please provide the initial concentration or amount of Pb(NO3)2 and NaSO4.
To identify the ions present in each solution, we need to first understand the compounds involved in the chemical equation:
Pb(NO3)2 (aqueous) - This compound is lead(II) nitrate. It dissociates in water to form cations (Pb2+) and anions (NO3-).
NaSO4 (aqueous) - This compound is sodium sulfate. It also dissociates in water to form cations (Na+) and anions (SO4^2-).
Now, let's break down the reaction and identify the ions involved:
Pb(NO3)2(aq) + NaSO4(aq) → PbSO4(s) + 2NaNO3(aq)
In solution, Pb(NO3)2 will dissociate into Pb2+ cations and 2 NO3- anions:
Pb(NO3)2(aq) → Pb2+(aq) + 2NO3-(aq)
Similarly, NaSO4 will dissociate into Na+ cations and SO4^2- anions:
NaSO4(aq) → Na+(aq) + SO4^2-(aq)
After the reaction, a solid precipitate of PbSO4 is formed, which is insoluble in water. This means that it will not dissociate into ions.
Now, let's calculate the concentrations of the ions in each solution. We need the initial concentrations of the compounds and the reaction stoichiometry.
Assuming equal initial concentrations (C) for both Pb(NO3)2 and NaSO4, we can write:
Initial concentrations:
[Pb2+] = C
[NO3-] = 2C
[Na+] = C
[SO4^2-] = C
After the reaction, all the Pb(NO3)2 will be consumed, resulting in the disappearance of Pb2+ and NO3-. The concentration of Na+ and SO4^2- remains unchanged since NaSO4 is a spectator ion.
Final concentrations:
[Pb2+] = 0
[NO3-] = 0
[Na+] = C
[SO4^2-] = C
Therefore, the concentrations of the ions in each solution after the reaction are as follows:
In the solution containing Pb(NO3)2:
[Pb2+] = 0
[NO3-] = 0
In the solution containing NaSO4:
[Na+] = C
[SO4^2-] = C
It is important to note that the concentration values depend on the initial concentration (C) assumed for the reactants. You can substitute any value for C to obtain specific concentration values.