Net Ionic Equations and Spectator Ions

Learn to write complete ionic and net ionic equations, identify spectator ions, and understand precipitation, acid-base, and gas-forming reactions at the ionic level.

Net Ionic Equations and Spectator Ions

Introduction

Why net ionic equations?

Show what's actually happening at the ionic level
Eliminate spectator ions (don't participate)
Focus on the chemical change
More accurate representation of aqueous reactions

Three types of equations:

Molecular equation (complete formula units)
Complete ionic equation (all ions shown)
Net ionic equation (only reacting species)

Types of Equations

Molecular Equation

Shows: Complete formula units of all reactants and products

Example:

$\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}$

Characteristics:

Looks like a typical balanced equation
Doesn't show ionic nature of species
Good for stoichiometry calculations
Doesn't reveal mechanism

Complete Ionic Equation

Shows: All strong electrolytes as separated ions

Rules for complete ionic equations:

Write as ions (aq):

Strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄
Strong bases: Group 1 hydroxides (NaOH, KOH), Ba(OH)₂, Sr(OH)₂, Ca(OH)₂
Soluble ionic compounds (use solubility rules)

Keep as molecules:

Solids (s)
Liquids (l)
Gases (g)
Weak acids and bases
Insoluble compounds
Water

Example (same reaction):

$\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq) + NO3-(aq)}$

All ions separated except AgCl (precipitate)

Net Ionic Equation

Shows: Only species that undergo chemical change

Eliminate: Spectator ions (appear on both sides unchanged)

Example (same reaction):

Complete ionic:

$\ce{Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) -> AgCl(s) + Na+(aq) + NO3-(aq)}$

Identify spectators:

Na⁺ appears on both sides → spectator
NO₃⁻ appears on both sides → spectator

Cancel spectators:

$\ce{Ag+(aq) + \cancel{NO3-(aq)} + \cancel{Na+(aq)} + Cl-(aq) -> AgCl(s) + \cancel{Na+(aq)} + \cancel{NO3-(aq)}}$

Net ionic equation:

$\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$

This shows the essential chemical change!

Spectator Ions

Spectator ions: Ions present in solution but don't participate in reaction

Appear on both sides of complete ionic equation
Remain dissolved and unchanged
Don't affect chemical change
Important for charge balance but not for reaction

How to identify:

Write complete ionic equation
Find ions that appear on both sides
Those are spectators

Example:

$\ce{2KOH(aq) + H2SO4(aq) -> K2SO4(aq) + 2H2O(l)}$

Complete ionic:

$\ce{2K+(aq) + 2OH-(aq) + 2H+(aq) + SO4^{2-}(aq) -> 2K+(aq) + SO4^{2-}(aq) + 2H2O(l)}$

Spectators: K⁺ and SO₄²⁻

Net ionic:

$\ce{2OH-(aq) + 2H+(aq) -> 2H2O(l)}$

Or simplified:

$\ce{OH-(aq) + H+(aq) -> H2O(l)}$

Writing Net Ionic Equations

Step-by-Step Procedure

Step 1: Write balanced molecular equation

Include states (s, l, g, aq)
Balance atoms and charges

Step 2: Write complete ionic equation

Separate strong electrolytes into ions
Keep solids, liquids, gases, and weak electrolytes together

Step 3: Identify spectator ions

Find ions unchanged on both sides

Step 4: Write net ionic equation

Remove spectator ions
Simplify coefficients if possible
Check that atoms and charges balance

Example: Precipitation

Step 1: Molecular

$\ce{Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq)}$

Step 2: Complete ionic

$\ce{Pb^{2+}(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq) -> PbI2(s) + 2K+(aq) + 2NO3-(aq)}$

Step 3: Identify spectators

K⁺: both sides
NO₃⁻: both sides

Step 4: Net ionic

$\ce{Pb^{2+}(aq) + 2I-(aq) -> PbI2(s)}$

Check:

Atoms: 1 Pb, 2 I (both sides) ✓
Charge: +2 + 2(-1) = 0 left, 0 right ✓

Common Types of Net Ionic Equations

1. Precipitation Reactions

Pattern: Cation + Anion → Insoluble salt

Example 1:

$\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$

Example 2:

$\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4(s)}$

Example 3:

$\ce{Ca^{2+}(aq) + CO3^{2-}(aq) -> CaCO3(s)}$

Driving force: Formation of insoluble solid

2. Acid-Base Neutralization

Strong acid + Strong base:

General net ionic equation:

$\ce{H+(aq) + OH-(aq) -> H2O(l)}$

This is the net ionic equation for ALL strong acid + strong base reactions!

Example 1: HCl + NaOH

$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$

Net ionic:

$\ce{H+(aq) + OH-(aq) -> H2O(l)}$

Example 2: H₂SO₄ + KOH

$\ce{H2SO4(aq) + 2KOH(aq) -> K2SO4(aq) + 2H2O(l)}$

Net ionic:

$\ce{H+(aq) + OH-(aq) -> H2O(l)}$

Weak acid + Strong base:

Weak acid stays together (not fully ionized)

Example: Acetic acid + NaOH

$\ce{CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)}$

Net ionic:

$\ce{CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)}$

Note: CH₃COOH stays together (weak acid)

3. Gas-Forming Reactions

Carbonate/Bicarbonate + Acid → CO₂ gas

Example 1: Carbonate

$\ce{2HCl(aq) + Na2CO3(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)}$

Net ionic:

$\ce{2H+(aq) + CO3^{2-}(aq) -> H2O(l) + CO2(g)}$

Example 2: Bicarbonate

$\ce{HCl(aq) + NaHCO3(aq) -> NaCl(aq) + H2O(l) + CO2(g)}$

Net ionic:

$\ce{H+(aq) + HCO3-(aq) -> H2O(l) + CO2(g)}$

Sulfite + Acid → SO₂ gas

$\ce{2H+(aq) + SO3^{2-}(aq) -> H2O(l) + SO2(g)}$

Sulfide + Acid → H₂S gas

$\ce{2H+(aq) + S^{2-}(aq) -> H2S(g)}$

Ammonium salt + Strong base → NH₃ gas

$\ce{NH4+(aq) + OH-(aq) -> NH3(g) + H2O(l)}$

4. Oxidation-Reduction (Redox)

Electron transfer reactions

Example: Zn + Cu²⁺

$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$

No spectators (all species participate)

This IS the net ionic equation

Rules for States in Net Ionic Equations

Aqueous (aq):

Strong electrolytes that dissociate
Soluble ionic compounds
Strong acids
Strong bases

Solid (s):

Precipitates (insoluble compounds)
Pure metals

Liquid (l):

Water (when product)
Pure molecular liquids

Gas (g):

Gases that escape (CO₂, H₂S, NH₃, SO₂)

Checking Net Ionic Equations

Two things must balance:

1. Mass Balance (Atoms)

Count each element on both sides

Example:

$\ce{Fe^{3+}(aq) + 3OH-(aq) -> Fe(OH)3(s)}$

Left: 1 Fe, 3 O, 3 H Right: 1 Fe, 3 O, 3 H ✓

2. Charge Balance

Sum of charges must equal on both sides

Example:

$\ce{Fe^{3+}(aq) + 3OH-(aq) -> Fe(OH)3(s)}$

Left: +3 + 3(-1) = 0 Right: 0 (neutral solid) ✓

Common mistake: Forgetting to balance charges!

Special Cases

Case 1: No Reaction

If all ions are spectators → No net ionic equation

Example:

$\ce{NaCl(aq) + KNO3(aq) -> NaNO3(aq) + KCl(aq)}$

All products soluble → all ions are spectators

Net ionic: None (or write "No reaction")

No driving force:

No precipitate
No water formed
No gas formed
No electron transfer

Case 2: Molecular Compounds React

If reactants are molecular (not ionic)

Example:

$\ce{NH3(aq) + HCl(aq) -> NH4Cl(aq)}$

Net ionic:

$\ce{NH3(aq) + H+(aq) -> NH4+(aq)}$

NH₃ stays together (weak base)

Case 3: Polyatomic Ions Stay Together

Common mistake: Breaking up polyatomic ions

Wrong: SO₄²⁻ → S⁶⁺ + 4O²⁻

Right: SO₄²⁻ stays as one unit

Polyatomic ions that stay together:

SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻
ClO₄⁻, ClO₃⁻, MnO₄⁻
NH₄⁺, C₂H₃O₂⁻

Exception: Insoluble compounds or molecules

Summary of Net Ionic Equation Types

| Type | Molecular Example | Net Ionic | Driving Force | |------|-------------------|-----------|---------------| | Precipitation | AgNO₃ + NaCl | Ag⁺ + Cl⁻ → AgCl(s) | Solid forms | | Acid-Base | HCl + NaOH | H⁺ + OH⁻ → H₂O | Water forms | | Gas formation | HCl + Na₂CO₃ | 2H⁺ + CO₃²⁻ → H₂O + CO₂(g) | Gas escapes | | Redox | Zn + CuSO₄ | Zn + Cu²⁺ → Zn²⁺ + Cu | Electron transfer |

Practice Strategy

When writing net ionic equations:

✓ Start with balanced molecular equation
✓ Assign correct states
✓ Separate strong electrolytes
✓ Keep weak electrolytes, solids, gases together
✓ Identify and cancel spectators
✓ Check atom balance
✓ Check charge balance
✓ Simplify coefficients if possible

Common strong electrolytes to memorize:

Strong acids: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄
Strong bases: NaOH, KOH, LiOH, Ba(OH)₂, Sr(OH)₂, Ca(OH)₂
Most salts (if soluble)

Common weak electrolytes:

Weak acids: CH₃COOH, HF, H₂CO₃, H₃PO₄
Weak bases: NH₃, amines
Water (H₂O)

📚 Practice Problems

1Problem 1easy

❓ Question:

Write the complete ionic and net ionic equations for: BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)

💡 Show Solution

Solution:

Given molecular equation:

$\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}$

Task: Write complete ionic and net ionic equations

Step 1: Identify which species dissociate

Reactants:

BaCl₂(aq): Soluble ionic compound → dissociates
- Ba²⁺ and Cl⁻ ions
Na₂SO₄(aq): Soluble ionic compound → dissociates
- Na⁺ and SO₄²⁻ ions

Products:

BaSO₄(s): Precipitate (insoluble) → stays together
- Keep as BaSO₄(s)
NaCl(aq): Soluble ionic compound → dissociates
- Na⁺ and Cl⁻ ions

Key point: Solids don't dissociate in ionic equations!

Step 2: Write complete ionic equation

Dissociate all aqueous ionic compounds:

BaCl₂(aq):

Formula: BaCl₂
Ions: 1 Ba²⁺ + 2 Cl⁻

Na₂SO₄(aq):

Formula: Na₂SO₄
Ions: 2 Na⁺ + 1 SO₄²⁻

2NaCl(aq):

Formula: 2 NaCl
Ions: 2 Na⁺ + 2 Cl⁻

Complete ionic equation:

$\ce{Ba^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + SO4^{2-}(aq) -> BaSO4(s) + 2Na^+(aq) + 2Cl^-(aq)}$

Answer - Complete ionic:

$\boxed{\ce{Ba^{2+}(aq) + 2Cl^-(aq) + 2Na^+(aq) + SO4^{2-}(aq) -> BaSO4(s) + 2Na^+(aq) + 2Cl^-(aq)}}$

Step 3: Identify spectator ions

Look for ions appearing on both sides unchanged:

Left side:

Ba²⁺(aq)
2Cl⁻(aq)
2Na⁺(aq)
SO₄²⁻(aq)

Right side:

BaSO₄(s) - not an ion!
2Na⁺(aq)
2Cl⁻(aq)

Spectator ions:

2Na⁺: Appears on both sides as 2Na⁺(aq)
2Cl⁻: Appears on both sides as 2Cl⁻(aq)

Ions that react:

Ba²⁺: Left side as ion, right side in solid
SO₄²⁻: Left side as ion, right side in solid

Step 4: Cancel spectator ions

$\ce{Ba^{2+}(aq) + \cancel{2Cl^-(aq)} + \cancel{2Na^+(aq)} + SO4^{2-}(aq) -> BaSO4(s) + \cancel{2Na^+(aq)} + \cancel{2Cl^-(aq)}}$

Step 5: Write net ionic equation

What remains after canceling:

$\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4(s)}$

Answer - Net ionic:

$\boxed{\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4(s)}}$

Step 6: Verify the net ionic equation

Check 1: Atom balance

Left side: 1 Ba, 1 S, 4 O Right side: 1 Ba, 1 S, 4 O ✓

Check 2: Charge balance

Left side: (+2) + (-2) = 0 Right side: 0 (neutral solid) ✓

Both balanced! ✓

Summary:

| Type | Equation | |------|----------| | Molecular | BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq) | | Complete Ionic | Ba²⁺(aq) + 2Cl⁻(aq) + 2Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) + 2Na⁺(aq) + 2Cl⁻(aq) | | Net Ionic | Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) | | Spectators | Na⁺, Cl⁻ |

Interpretation:

What the net ionic equation tells us:

Essential change: Ba²⁺ ions + SO₄²⁻ ions combine to form solid BaSO₄
Spectators ignored: Na⁺ and Cl⁻ just "watch" the reaction
Driving force: Formation of insoluble BaSO₄ precipitate

Why this matters:

Net ionic shows ACTUAL chemical change
Same net ionic equation for ANY soluble barium salt + ANY soluble sulfate
Example: Ba(NO₃)₂ + K₂SO₄ gives same net ionic!

General pattern for this type:

$\ce{Ba^{2+} + SO4^{2-} -> BaSO4(s)}$

Any source of Ba²⁺ + any source of SO₄²⁻ → BaSO₄ precipitate

Observable evidence:

Clear solutions mixed
White precipitate forms immediately
Solution becomes cloudy/milky

Applications:

Test for sulfate ions (add Ba²⁺ solution)
Test for barium ions (add SO₄²⁻ solution)
Qualitative analysis

2Problem 2medium

❓ Question:

Write the net ionic equation for the reaction between aqueous solutions of hydrochloric acid and sodium carbonate. The products are sodium chloride, water, and carbon dioxide gas.

💡 Show Solution

Solution:

Given information:

Reactants: HCl(aq) and Na₂CO₃(aq)
Products: NaCl, H₂O, CO₂(g)
This is an acid-carbonate reaction (gas-forming)

Step 1: Write balanced molecular equation

Reactants:

HCl(aq) - hydrochloric acid
Na₂CO₃(aq) - sodium carbonate

Products:

NaCl(aq) - sodium chloride (soluble)
H₂O(l) - water
CO₂(g) - carbon dioxide gas

Unbalanced:

$\ce{HCl(aq) + Na2CO3(aq) -> NaCl(aq) + H2O(l) + CO2(g)}$

Balance:

Check atoms:

H: 1 left, 2 right (from H₂O)
Cl: 1 left, 1 right
Na: 2 left, 1 right
C: 1 left, 1 right
O: 3 left, 3 right (1 in H₂O, 2 in CO₂)

Need 2 HCl and 2 NaCl:

$\ce{2HCl(aq) + Na2CO3(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)}$

Final check:

H: 2, 2 ✓
Cl: 2, 2 ✓
Na: 2, 2 ✓
C: 1, 1 ✓
O: 3, 3 ✓

Balanced molecular equation:

$\boxed{\ce{2HCl(aq) + Na2CO3(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)}}$

Step 2: Write complete ionic equation

Identify what dissociates:

HCl(aq): Strong acid → completely dissociates

H⁺(aq) + Cl⁻(aq)
2HCl → 2H⁺ + 2Cl⁻

Na₂CO₃(aq): Soluble ionic compound → dissociates

2Na⁺(aq) + CO₃²⁻(aq)

NaCl(aq): Soluble ionic compound → dissociates

Na⁺(aq) + Cl⁻(aq)
2NaCl → 2Na⁺ + 2Cl⁻

H₂O(l): Liquid → stays together (molecular)

CO₂(g): Gas → stays together (molecular)

Complete ionic equation:

$\ce{2H+(aq) + 2Cl-(aq) + 2Na+(aq) + CO3^{2-}(aq) -> 2Na+(aq) + 2Cl-(aq) + H2O(l) + CO2(g)}$

Step 3: Identify spectator ions

Compare both sides:

Left side:

2H⁺(aq)
2Cl⁻(aq)
2Na⁺(aq)
CO₃²⁻(aq)

Right side:

2Na⁺(aq)
2Cl⁻(aq)
H₂O(l)
CO₂(g)

Spectator ions (appear unchanged on both sides):

2Na⁺(aq)
2Cl⁻(aq)

Reacting species:

2H⁺ → becomes part of H₂O
CO₃²⁻ → becomes part of H₂O and CO₂

Step 4: Cancel spectator ions

$\ce{2H+(aq) + \cancel{2Cl-(aq)} + \cancel{2Na+(aq)} + CO3^{2-}(aq) -> \cancel{2Na+(aq)} + \cancel{2Cl-(aq)} + H2O(l) + CO2(g)}$

Step 5: Write net ionic equation

$\ce{2H+(aq) + CO3^{2-}(aq) -> H2O(l) + CO2(g)}$

Answer:

$\boxed{\ce{2H+(aq) + CO3^{2-}(aq) -> H2O(l) + CO2(g)}}$

Step 6: Verify balance

Atom balance:

H: 2 left, 2 right (in H₂O) ✓
C: 1 left, 1 right (in CO₂) ✓
O: 3 left, 3 right (1 in H₂O, 2 in CO₂) ✓

Charge balance:

Left: 2(+1) + (-2) = +2 - 2 = 0
Right: 0 (all neutral molecules) ✓

Both balanced! ✓

Interpretation and Analysis:

What does this equation show?

Essential change: H⁺ ions react with CO₃²⁻ ions
Products: Water and CO₂ gas
Driving forces:
1. Formation of water (stable molecule)
2. Formation of gas (escapes solution)

Why CO₂ forms:

Step-by-step mechanism:

First H⁺ reacts with CO₃²⁻:

$\ce{H+ + CO3^{2-} -> HCO3-}$ (bicarbonate)

Second H⁺ reacts with HCO₃⁻:

$\ce{H+ + HCO3- -> H2CO3}$ (carbonic acid)

Carbonic acid unstable, decomposes:

$\ce{H2CO3 -> H2O + CO2(g)}$

Net result: Same as our net ionic equation!

General applicability:

This net ionic equation applies to:

ANY acid + carbonate
Doesn't matter what the spectator ions are

Examples with same net ionic:

Sulfuric acid + potassium carbonate:

$\ce{H2SO4 + K2CO3 -> K2SO4 + H2O + CO2}$

Net ionic: Same! $\ce{2H+ + CO3^{2-} -> H2O + CO2}$

Nitric acid + calcium carbonate:

$\ce{2HNO3 + CaCO3 -> Ca(NO3)2 + H2O + CO2}$

Net ionic: Same! $\ce{2H+ + CO3^{2-} -> H2O + CO2}$

Observable evidence:

In the lab, you would see:

Vigorous bubbling (CO₂ gas escaping)
"Fizzing" or "effervescence"
Gas can be tested:
- Turns limewater cloudy
- Extinguishes burning splint

This reaction is used for:

Antacids (neutralize stomach acid)
- CaCO₃ + HCl → CO₂ (burping)
Baking: baking soda (NaHCO₃) + acid
Cleaning: vinegar + baking soda
Testing for carbonates in qualitative analysis

Comparison to bicarbonate:

If using NaHCO₃ (bicarbonate) instead:

$\ce{HCl + NaHCO3 -> NaCl + H2O + CO2}$

Net ionic:

$\ce{H+(aq) + HCO3-(aq) -> H2O(l) + CO2(g)}$

Notice: Only need 1 H⁺ (not 2) for bicarbonate

Summary:

| Equation Type | Equation | |---------------|----------| | Molecular | 2HCl + Na₂CO₃ → 2NaCl + H₂O + CO₂ | | Complete Ionic | 2H⁺ + 2Cl⁻ + 2Na⁺ + CO₃²⁻ → 2Na⁺ + 2Cl⁻ + H₂O + CO₂ | | Net Ionic | 2H⁺ + CO₃²⁻ → H₂O + CO₂ | | Spectators | Na⁺, Cl⁻ | | Driving Forces | Water formation, gas evolution |

Key takeaway: Any strong acid + carbonate → water + CO₂ gas

3Problem 3hard

❓ Question:

Consider the following three reactions. For each: (i) write the complete ionic equation, (ii) identify spectator ions, (iii) write the net ionic equation. (a) Fe(NO₃)₃(aq) + 3NaOH(aq) → Fe(OH)₃(s) + 3NaNO₃(aq), (b) HNO₃(aq) + KOH(aq) → KNO₃(aq) + H₂O(l), (c) NH₄Cl(aq) + NaOH(aq) → NaCl(aq) + NH₃(g) + H₂O(l)

💡 Show Solution

Solution:

Given: Three balanced molecular equations

Task: For each reaction, write complete ionic and net ionic equations

Reaction (a): Fe(NO₃)₃ + 3NaOH → Fe(OH)₃ + 3NaNO₃

Type: Precipitation reaction

(i) Complete ionic equation

Identify what dissociates:

Fe(NO₃)₃(aq): Soluble ionic compound

Fe³⁺(aq) + 3NO₃⁻(aq)

3NaOH(aq): Strong base (soluble)

3Na⁺(aq) + 3OH⁻(aq)

Fe(OH)₃(s): Precipitate (insoluble) - stays together!

3NaNO₃(aq): Soluble ionic compound

3Na⁺(aq) + 3NO₃⁻(aq)

Complete ionic:

$\boxed{\ce{Fe^{3+}(aq) + 3NO3-(aq) + 3Na+(aq) + 3OH-(aq) -> Fe(OH)3(s) + 3Na+(aq) + 3NO3-(aq)}}$

(ii) Identify spectator ions

Compare both sides:

Spectators (appear unchanged):

3Na⁺(aq): Both sides
3NO₃⁻(aq): Both sides

Reacting species:

Fe³⁺ and OH⁻ → form Fe(OH)₃(s)

(iii) Net ionic equation

Cancel spectators:

$\ce{Fe^{3+}(aq) + \cancel{3NO3-(aq)} + \cancel{3Na+(aq)} + 3OH-(aq) -> Fe(OH)3(s) + \cancel{3Na+(aq)} + \cancel{3NO3-(aq)}}$

Net ionic:

$\boxed{\ce{Fe^{3+}(aq) + 3OH-(aq) -> Fe(OH)3(s)}}$

Verification:

Atoms: 1 Fe, 3 O, 3 H (both sides) ✓
Charge: (+3) + 3(-1) = 0 left, 0 right ✓

Driving force: Formation of insoluble Fe(OH)₃ precipitate (rust-colored)

Reaction (b): HNO₃ + KOH → KNO₃ + H₂O

Type: Acid-base neutralization (strong acid + strong base)

(i) Complete ionic equation

Identify what dissociates:

HNO₃(aq): Strong acid

H⁺(aq) + NO₃⁻(aq)

KOH(aq): Strong base

K⁺(aq) + OH⁻(aq)

KNO₃(aq): Soluble ionic compound

K⁺(aq) + NO₃⁻(aq)

H₂O(l): Liquid - stays together!

Complete ionic:

$\boxed{\ce{H+(aq) + NO3-(aq) + K+(aq) + OH-(aq) -> K+(aq) + NO3-(aq) + H2O(l)}}$

(ii) Identify spectator ions

Spectators (appear unchanged):

K⁺(aq): Both sides
NO₃⁻(aq): Both sides

Reacting species:

H⁺ and OH⁻ → form H₂O

(iii) Net ionic equation

Cancel spectators:

$\ce{H+(aq) + \cancel{NO3-(aq)} + \cancel{K+(aq)} + OH-(aq) -> \cancel{K+(aq)} + \cancel{NO3-(aq)} + H2O(l)}$

Net ionic:

$\boxed{\ce{H+(aq) + OH-(aq) -> H2O(l)}}$

Verification:

Atoms: 2 H, 1 O (both sides) ✓
Charge: (+1) + (-1) = 0 left, 0 right ✓

Key insight: This is the SAME net ionic equation for ALL strong acid + strong base reactions!

HCl + NaOH → same net ionic
H₂SO₄ + KOH → same net ionic
Spectators change, but essence is always: H⁺ + OH⁻ → H₂O

Driving force: Formation of water (very stable molecule)

Reaction (c): NH₄Cl + NaOH → NaCl + NH₃ + H₂O

Type: Gas-forming reaction (weak base formation)

(i) Complete ionic equation

Identify what dissociates:

NH₄Cl(aq): Soluble ionic compound

NH₄⁺(aq) + Cl⁻(aq)

NaOH(aq): Strong base

Na⁺(aq) + OH⁻(aq)

NaCl(aq): Soluble ionic compound

Na⁺(aq) + Cl⁻(aq)

NH₃(g): Gas - stays together! (molecular)

H₂O(l): Liquid - stays together!

Complete ionic:

$\boxed{\ce{NH4+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) -> Na+(aq) + Cl-(aq) + NH3(g) + H2O(l)}}$

(ii) Identify spectator ions

Spectators (appear unchanged):

Na⁺(aq): Both sides
Cl⁻(aq): Both sides

Reacting species:

NH₄⁺ and OH⁻ → form NH₃ and H₂O

(iii) Net ionic equation

Cancel spectators:

$\ce{NH4+(aq) + \cancel{Cl-(aq)} + \cancel{Na+(aq)} + OH-(aq) -> \cancel{Na+(aq)} + \cancel{Cl-(aq)} + NH3(g) + H2O(l)}$

Net ionic:

$\boxed{\ce{NH4+(aq) + OH-(aq) -> NH3(g) + H2O(l)}}$

Verification:

Atoms: 5 H, 1 N, 1 O (both sides) ✓
Charge: (+1) + (-1) = 0 left, 0 right ✓

Driving forces:

Formation of gas (NH₃ escapes)
Formation of water

Observable evidence:

Smell of ammonia (pungent, characteristic)
If heated, reaction faster and more gas evolves
Can test with damp red litmus paper → turns blue (NH₃ is basic)

Summary Table

| Reaction | Complete Ionic | Spectators | Net Ionic | Type | |----------|----------------|------------|-----------|------| | (a) Fe(NO₃)₃ + 3NaOH | Fe³⁺ + 3NO₃⁻ + 3Na⁺ + 3OH⁻ → Fe(OH)₃(s) + 3Na⁺ + 3NO₃⁻ | Na⁺, NO₃⁻ | Fe³⁺ + 3OH⁻ → Fe(OH)₃(s) | Precipitation | | (b) HNO₃ + KOH | H⁺ + NO₃⁻ + K⁺ + OH⁻ → K⁺ + NO₃⁻ + H₂O | K⁺, NO₃⁻ | H⁺ + OH⁻ → H₂O | Neutralization | | (c) NH₄Cl + NaOH | NH₄⁺ + Cl⁻ + Na⁺ + OH⁻ → Na⁺ + Cl⁻ + NH₃(g) + H₂O | Na⁺, Cl⁻ | NH₄⁺ + OH⁻ → NH₃(g) + H₂O | Gas formation |

Key Observations and Patterns

Reaction (a) - Precipitation Pattern

General pattern: Metal ion + OH⁻ → Metal hydroxide precipitate

Works for:

Fe³⁺ + 3OH⁻ → Fe(OH)₃(s) [rust-colored]
Al³⁺ + 3OH⁻ → Al(OH)₃(s) [white]
Cu²⁺ + 2OH⁻ → Cu(OH)₂(s) [blue]
Mg²⁺ + 2OH⁻ → Mg(OH)₂(s) [white]

Exceptions (soluble hydroxides):

Group 1 metals (NaOH, KOH)
Ba(OH)₂, Sr(OH)₂, Ca(OH)₂

Reaction (b) - Universal Neutralization

This net ionic equation is universal for:

ALL strong acid + strong base reactions
H⁺ + OH⁻ → H₂O

Examples:

HCl + NaOH
H₂SO₄ + KOH
HBr + LiOH
HI + Ca(OH)₂

All have same essence: Neutralization of H⁺ and OH⁻

Reaction (c) - Ammonium + Base Pattern

General pattern: NH₄⁺ + OH⁻ → NH₃(g) + H₂O

This is how you prepare ammonia gas in lab!

Applications:

Qualitative test for ammonium ions
- Add strong base + heat
- Smell ammonia or test with litmus
Industrial ammonia production (different method)

Additional Practice

Can you predict net ionic equations for these?

Al₂(SO₄)₃ + 6KOH → 2Al(OH)₃ + 3K₂SO₄
- Net ionic: Al³⁺ + 3OH⁻ → Al(OH)₃(s)
HCl + NaOH → NaCl + H₂O
- Net ionic: H⁺ + OH⁻ → H₂O
CaCl₂ + 2AgNO₃ → Ca(NO₃)₂ + 2AgCl
- Net ionic: Ag⁺ + Cl⁻ → AgCl(s)

Notice: Different molecular equations, but similar patterns in net ionic!

Key Takeaway:

Net ionic equations reveal:

Actual chemical change (not spectators)
Driving forces (precipitate, water, gas)
Universal patterns across different reactions
What's really happening at ionic level

Spectator ions are important for:

Electrical neutrality
Complete molecular equation
But don't participate in reaction itself

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