Chem 120 Worksheet: Writing Ionic Equations for Aqueous Reactions

Let’s solve each problem step by step. We’ll write:

1. The balanced molecular equation (with states: aq, s, l, g)
2. The total ionic equation (break all soluble strong electrolytes into ions)
3. The net ionic equation (cancel spectator ions — those that appear unchanged on both sides)

We’ll use solubility rules to know what precipitates (forms a solid).

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Problem 2: Ca(NO₃)₂(aq) + K₂SO₄(aq) → ?

Step 1: Predict products

This is a double replacement reaction:

Ca(NO₃)₂ + K₂SO₄ → CaSO₄ + 2KNO₃

Now check solubility:

- CaSO₄: slightly soluble → often treated as solid (s) in these problems (precipitate)
- KNO₃: always soluble → aqueous (aq)

So balanced molecular equation:

Ca(NO₃)₂(aq) + K₂SO₄(aq) → CaSO₄(s) + 2KNO₃(aq)

Step 2: Total Ionic Equation

Break all aqueous compounds into ions:

Ca²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + SO₄²⁻(aq) → CaSO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq)

Step 3: Net Ionic Equation

Cancel spectator ions: K⁺ and NO₃⁻ appear on both sides.

Left with:

Ca²⁺(aq) + SO₄²⁻(aq) → CaSO₄(s)

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Problem 3: Na₃PO₄(aq) + CoCl₂(aq) → ?

Step 1: Predict products

Double replacement:

Na₃PO₄ + CoCl₂ → Co₃(PO₄)₂ + NaCl

Balance it:

Need 3 Co and 2 PO₄ → so 2 Na₃PO₄ and 3 CoCl₂

→ 2Na₃PO₄ + 3CoCl₂ → Co₃(PO)₂ + 6NaCl

Check solubility:

- Co₃(PO₄)₂: phosphates are generally insoluble → solid (s)
- NaCl: soluble → aqueous (aq)

Balanced molecular equation:

2Na₃PO₄(aq) + 3CoCl₂(aq) → Co₃(PO₄)₂(s) + 6NaCl(aq)

Step 2: Total Ionic Equation

Break apart aqueous compounds:

6Na⁺(aq) + 2PO₄³⁻(aq) + 3Co²⁺(aq) + 6Cl⁻(aq) → Co₃(PO)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)

Step 3: Net Ionic Equation

Cancel Na⁺ and Cl⁻ (spectator ions):

Left with:

3Co²⁺(aq) + 2PO₄³⁻(aq) → Co₃(PO₄)₂(s)

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Problem 4: Fe(ClO₄)₃(aq) + NaOH(aq) → ?

Step 1: Predict products

Double replacement:

Fe(ClO₄)₃ + NaOH → Fe(OH)₃ + NaClO₄

Balance:

Fe(ClO₄)₃ + 3NaOH → Fe(OH)₃ + 3NaClO₄

Solubility:

- Fe(OH)₃: hydroxides of transition metals are insoluble → solid (s)
- NaClO₄: perchlorates are always soluble → aqueous (aq)

Balanced molecular equation:

Fe(ClO₄)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaClO₄(aq)

Step 2: Total Ionic Equation

Break aqueous compounds:

Fe³⁺(aq) + 3ClO₄⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) + 3Na⁺(aq) + 3ClO₄⁻(aq)

Step 3: Net Ionic Equation

Cancel Na⁺ and ClO₄⁻:

Left with:

Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)

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Problem 5: AgNO₃(aq) + NH₄I(aq) → ?

Step 1: Predict products

Double replacement:

AgNO₃ + NH₄I → AgI + NH₄NO₃

Already balanced.

Solubility:

- AgI: iodides are usually soluble EXCEPT with Ag⁺, Pb²⁺, Hg₂²⁺ → so AgI is solid (s)
- NH₄NO₃: ammonium and nitrate salts are always soluble → aqueous (aq)

Balanced molecular equation:

AgNO₃(aq) + NH₄I(aq) → AgI(s) + NH₄NO₃(aq)

Step 2: Total Ionic Equation

Break aqueous compounds:

Ag⁺(aq) + NO₃⁻(aq) + NH₄⁺(aq) + I⁻(aq) → AgI(s) + NH₄⁺(aq) + NO₃⁻(aq)

Step 3: Net Ionic Equation

Cancel NH₄⁺ and NO₃⁻:

Left with:

Ag⁺(aq) + I⁻(aq) → AgI(s)

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Final Answer:

2.
Molecular: Ca(NO₃)₂(aq) + K₂SO₄(aq) → CaSO₄(s) + 2KNO₃(aq)
TIE: Ca²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + SO₄²⁻(aq) → CaSO₄(s) + 2K⁺(aq) + 2NO₃⁻(aq)
NIE: Ca²⁺(aq) + SO₄²⁻(aq) → CaSO₄(s)

3.
Molecular: 2Na₃PO₄(aq) + 3CoCl₂(aq) → Co₃(PO₄)₂(s) + 6NaCl(aq)
TIE: 6Na⁺(aq) + 2PO₄³⁻(aq) + 3Co²⁺(aq) + 6Cl⁻(aq) → Co₃(PO₄)₂(s) + 6Na⁺(aq) + 6Cl⁻(aq)
NIE: 3Co²⁺(aq) + 2PO₄³⁻(aq) → Co₃(PO₄)₂(s)

4.
Molecular: Fe(ClO₄)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaClO₄(aq)
TIE: Fe³⁺(aq) + 3ClO₄⁻(aq) + 3Na⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s) + 3Na⁺(aq) + 3ClO₄⁻(aq)
NIE: Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)

5.
Molecular: AgNO₃(aq) + NH₄I(aq) → AgI(s) + NH₄NO₃(aq)
TIE: Ag⁺(aq) + NO₃⁻(aq) + NH₄⁺(aq) + I⁻(aq) → AgI(s) + NH₄⁺(aq) + NO₃⁻(aq)
NIE: Ag⁺(aq) + I⁻(aq) → AgI(s)