AP chem equations and reactions summary

Question

Predict whether a precipitation reaction will occur in each of the following cases. If it does, write the complete balanced equation AND the net ionic equation with appropriate state symbols. If there is no reaction, indicate why.

a. CuSO4 (aq) and Na2CO3 (aq)

b. HI (aq) and Zn(NO3)2 (aq)

c. Aqueous silver nitrate and aqueous sodium bromide

Answer 1

In order to determine whether or not a reaction will occur, we need to know the solubilities of the predicted products. If there are no insoluble products produced, the reaction will not occur. Let's start with the first question and walk through it step by step.

A. The predicted products of the reaction between copper sulfate (CuSO₄) and sodium carbonate (Na₂CO₃) are copper carbonate (CuCO₃) and sodium sulfate (Na₂SO₄).

According to the solubility rules for ionic compounds, all sodium salts are soluble, however, most carbonates, including copper carbonate are insoluble. So the reaction will proceed.

https://www.sigmaaldrich.com/US/en/technical-documents/technical-article/materials-science-and-engineering/solid-state-synthesis/solubility-rules-solubility-of-common-ionic-compounds

Next we need to balance the equation. We write the reactants and products as we expect them to appear in the reaction. We also need to keep track of our atom counts, so we will write them below the equation. We can treat polyatomic ions as a unit because their composition is not being altered during the reaction

CuSO₄ (aq) + Na₂CO₃ (aq) → CuCO₃ (s) + Na₂SO₄ (aq)

1 Cu 1

1 SO₄ 1

1 Na 1

1 CO₃ 1

As we can see above, the equation is already balanced, so no coefficients are needed.

We can now write the full ionic equation by separating our aqueous (dissolved) substances into their constituent ions while leaving the solid compound intact.

Cu²⁺ (aq) + SO₄^2- (aq) + 2 Na⁺ (aq) + CO₃^2- (aq) → CuCO₃ (s) + 2 Na⁺ (aq) + SO₄^2- (aq)

Ions that remain dissolved in both the products and reactants (spectator ions) can be eliminated from the equation.

Cu²⁺ (aq) + SO₄^2- (aq) + 2 Na⁺ (aq) + CO₃^2- (aq) → CuCO₃ (s) + 2 Na⁺ (aq) + SO₄^2- (aq)

Removing the spectator ions leaves us with the following net ionic equation:

Cu²⁺ (aq) + CO₃^2- (aq) → CuCO₃ (s)

B. No reaction will occur between hydrogen iodide (HI) and zinc nitrate (Zn(NO₃)₂) because the expected products, zinc iodide (ZnI₂) and nitric acid (HNO₃) are both soluble in water.

C. The reaction between aqueous silver nitrate (AgNO₃) and aqueous sodium bromide (NaBr), produces silver bromide (AgBr) and sodium nitrate (NaNO₃). According to the solubility rules, sodium nitrate is soluble, but silver halides (compounds containing halogens, such as bromine (Br) are not. So the reaction will produce solid silver bromide (AgBr).

Our unbalanced equation is:

AgNO₃ (aq) + NaBr (aq) → AgBr (s) + NaNO₃ (aq)

1 Ag 1

1 NO₃ 1

1 Na 1

1 Br 1

The equation is already balanced.

Writing the full ionic equation gives us:

Ag⁺ (aq) + NO₃^- (aq) + Na⁺ (aq) + Br^- (aq) → AgBr (s) + Na⁺ (aq) + NO₃^- (aq)

Eliminate the spectator ions.

Ag⁺ (aq) + NO₃^- (aq) + Na⁺ (aq) + Br^- (aq) → AgBr (s) + Na⁺ (aq) + NO₃^- (aq)

Our net ionic equation is

Ag⁺ (aq) + Br^- (aq) → AgBr (s)