a. To determine the ionic compounds that will precipitate out of solution, we need to consider the solubility rules. According to the solubility rules:
1. All sodium (Na+) salts are soluble, so Na₂SO₄ will remain in solution.
2. Zinc (Zn2+) salts are generally soluble, except for zinc sulfide (ZnS) and zinc hydroxide (Zn(OH)₂). However, in this case, we are adding ZnCl₂ to the solution, which contains chloride (Cl-) ions. Chloride ions form soluble salts with most cations, including Zn2+. Therefore, ZnCl₂ will remain in solution.
3. Barium (Ba2+) salts are generally soluble, except for barium sulfate (BaSO₄) and barium carbonate (BaCO₃). However, in this case, we are adding Ba(CN)₂ to the solution, which contains cyanide (CN-) ions. Cyanide ions form insoluble salts with most cations, including Ba2+. Therefore, Ba(CN)₂ will precipitate out of solution as Ba(CN)₂ is not soluble.
b. Assuming complete precipitation of all insoluble compounds and that volumes are additive, we can calculate the molarity of each ion remaining in the solution.
For Na₂SO₄:
- Sodium (Na+) ion concentration: 2 * 0.100 M = 0.200 M
- Sulfate (SO₄2-) ion concentration: 0.100 M
For ZnCl₂:
- Zinc (Zn2+) ion concentration: 0.300 M
- Chloride (Cl-) ion concentration: 2 * 0.300 M = 0.600 M
For Ba(CN)₂:
- Barium (Ba2+) ion concentration: 0.200 M
- Cyanide (CN-) ion concentration: 2 * 0.200 M = 0.400 M
Therefore, the molarity of each ion remaining in the solution, assuming complete precipitation of all insoluble compounds and that volumes are additive, are as follows:
- Sodium (Na+) ion: 0.200 M
- Sulfate (SO₄2-) ion: 0.100 M
- Zinc (Zn2+) ion: 0.300 M
- Chloride (Cl-) ion: 0.600 M
- Barium (Ba2+) ion: 0.200 M
- Cyanide (CN-) ion: 0.400 M