My paper has a picture of a beaker and in it it had the names: Na+(aq), C2H302-(aq), and SrSO4(s). The directions say: Your lab partener tells you that she mized two solutions that contain ions. You analyze the solution and find that it contains the ions and precipitate shown in the beaker. Write the molecular, total ionic, and net ionic equations.

Question

My paper has a picture of a beaker and in it it had the names: Na+(aq), C2H302-(aq), and SrSO4(s). The directions say: Your lab partener tells you that she mized two solutions that contain ions. You analyze the solution and find that it contains the ions and precipitate shown in the beaker. Write the molecular, total ionic, and net ionic equations.

Answer 1

To write the molecular, total ionic, and net ionic equations, we need to first identify the reactants and products in the solution.

Given reactants:
1. Na+(aq): Sodium ion in aqueous solution.
2. C2H3O2-(aq): Acetate ion in aqueous solution.
3. SrSO4(s): Strontium sulfate in solid form (precipitate).

Now, let's write the equations step by step:

1. Molecular Equation:
The molecular equation represents the overall reaction, with all reactants and products written as complete compounds. In this case, the reactants are the sodium ion (Na+) and acetate ion (C2H3O2-) in aqueous solution, and the precipitate is strontium sulfate (SrSO4(s)). The equation can be written as:

Na+(aq) + C2H3O2-(aq) + SrSO4(s) → NaC2H3O2(aq) + SrSO4(s)

2. Total Ionic Equation:
The total ionic equation represents the species that are actually present as ions in the solution. In water, many ionic compounds dissociate into their component ions. In this case, the sodium acetate (NaC2H3O2) present as sodium ion (Na+) and acetate ion (C2H3O2-) should be shown as ions. However, since the strontium sulfate (SrSO4) is a solid, it does not dissociate and remains as a solid. Thus, the total ionic equation becomes:

Na+(aq) + C2H3O2-(aq) + Sr2+(aq) + SO42-(aq) → Na+(aq) + C2H3O2-(aq) + SrSO4(s)

3. Net Ionic Equation:
The net ionic equation represents the actual chemical reaction by removing any ions that appear on both sides of the equation. In this case, both sodium and acetate ions appear on both sides of the equation and can be eliminated. Therefore, the net ionic equation becomes:

Sr2+(aq) + SO42-(aq) → SrSO4(s)

In summary:
- Molecular Equation: Na+(aq) + C2H3O2-(aq) + SrSO4(s) → NaC2H3O2(aq) + SrSO4(s)
- Total Ionic Equation: Na+(aq) + C2H3O2-(aq) + Sr2+(aq) + SO42-(aq) → Na+(aq) + C2H3O2-(aq) + SrSO4(s)
- Net Ionic Equation: Sr2+(aq) + SO42-(aq) → SrSO4(s)