Compare the solubility of calcium sulfite in each of the following aqueous solutions: a. 0.10 M Ca(CH3COO)2
b. 0.10 M K2SO3
c. 0.10 M NaNO3
d. 0.10 M KCH3COO
1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.

0.10 M Ca(CH3COO)2- Less soluble than in pure water.

0.10 M K2SO3- Less soluble than in pure water.

0.10 M NaNO3 - More soluble than in pure water.

0.10 M KCH3COO- Similar solubility as in pure water.

Explanation:

We have to cast our minds back to the idea of common ions effect. If any ion is already present in solution, the presence of that ion in solution prevents any solute containing a common ion with the solution from dissolving in that solution. In order words, the presence of a common ion makes a solute less soluble in a solvent than it is in pure water.

For instance, 0.10 M Ca(CH3COO)2 and K2SO3 both contain Ca^2+ and SO3^2- ions respectively which are also contained in the solute calcium sulfite.

The presence of these common ions in solution makes calcium sulfite less soluble in these solutions than it is in pure water because the equilibrium position for the dissolution of the solute lies towards the left hand side.

However, calcium sulfite is more soluble in 0.10 M NaNO3 than in pure water due to displacement reaction between the ions in solution.

The solubility of calcium sulfite and 0.10 M KCH3COO in pure water is quite comparable.

answer; /// i do believe that the correct answer is (c)nuclear bond

answer; natural resources;