Question

A solution contains one or more of the following ions: Ag+, Ca2+, and Co2+. Lithium bromide is added to the solution and no precipitate forms. An excess of lithium sulfate is then added to the solution and a precipitate forms. The precipitate is filtered off and lithium phosphate is added to the remaining solution, producing a precipitate.A. Which ions are present in the original solution?Ca2+Ag+Co2+B. Write a net ionic equation for the formation of the precipitate observed after the addition of lithium sulfate. Include physical states.C. Write a net ionic equation for the formation of the precipitate observed after the addition of lithium phosphate. Include physical states.

Answer 1

Answer:

#1. Ca²⁺

# 2. Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

#3. 3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

Explanation:

The question above concerns solubility of salts or ions in water.

The solution given contains Ag+, Ca2+, and Co2+ ions.

• In the first case, when Lithium bromide is added to the solution, there is no white precipitate formed.
• In the second case, the addition of Lithium sulfate results in the formation of a precipitate because of the Ca²⁺ in the solution combined with the SO₃²⁻ from lithium sulfate to form an insoluble CaSO₄.

• The net ionic equation for the reaction is;

Ca²⁺(aq) + SO₃²⁻(aq) → CaSO₄(s)

• From the solubility rules, all sulfates are soluble except BaSO₄, CaSO₄, and PbSO₄.

• In the third case, the addition of Lithium phosphate results in the formation of a precipitate because Ag⁺ ions in the solution combine with phosphate ions ( PO₄³⁻) from lithium phosphate to form an insoluble salt, Ag₃PO₄.

• The net ionic equation for the reaction is;

3Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)

• According to solubility rules, all phosphates are insoluble in water except Na₃PO₄, K₃PO₄, and (NH₄)₃PO₄.

Answer 2

A. The ion present in the original solution is $Ca^2^+$

B. $\rm Ca^2^+(aq) + SO_3^2^-(aq) \rightarrow CaSO_4(s)$

C. $\rm 3Ag^+(aq) + PO_4^3^+(aq) \rightarrow Ag_3PO_4(s)$

What is a solution?

A solution is a mixture of a solute to the solvent.

The given solution contains $Ag^+, Ca2^+, and\; Co_2^+ ions.$

Given that, lithium bromide is added to the solution and no precipitate forms.

Then, an excess of lithium sulfate is then added to the solution and a precipitate forms.

The equation forming is

$\rm Ca^2^+(aq) + SO_3^2^-(aq) \rightarrow CaSO_4(s)$  

The precipitate is filtered, and then, lithium phosphate is added to the remaining solution that produces a precipitate

The equation will be

$\rm 3Ag^+(aq) + PO_4^3^+(aq) \rightarrow Ag_3PO_4(s)$

Thus, A. The ion present in the original solution is $Ca^2^+$

B. $\rm Ca^2^+(aq) + SO_3^2^-(aq) \rightarrow CaSO_4(s)$

C. $\rm 3Ag^+(aq) + PO_4^3^+(aq) \rightarrow Ag_3PO_4(s)$

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