Worksheet titled "Practice Problems on Net Ionic Equations" showing chemical reactions and instructions for writing total and net ionic equations, along with solubility rules.

Let’s solve each problem step by step. We’ll write the total ionic equation (break all aqueous compounds into ions) and then the net ionic equation (remove spectator ions — those that appear unchanged on both sides). If everything cancels out, we say “no reaction” or “all spectator ions.”

We’ll use the solubility rules at the bottom to decide what stays together (solid, liquid, gas) and what breaks apart (aqueous).

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Problem 1:

AgNO₃(aq) + KCl(aq) → AgCl(s) + KNO₃(aq)

Step 1: Break aqueous compounds into ions.
- AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
- KCl(aq) → K⁺(aq) + Cl⁻(aq)
- AgCl(s) → stays as solid (not broken up)
- KNO₃(aq) → K⁺(aq) + NO₃⁻(aq)

Total Ionic Equation:
Ag⁺(aq) + NO₃⁻(aq) + K⁺(aq) + Cl⁻(aq) → AgCl(s) + K⁺(aq) + NO₃⁻(aq)

Step 2: Cancel spectator ions (K⁺ and NO₃⁻ appear on both sides).

Net Ionic Equation:
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

✔ Done.

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Problem 2:

Mg(NO₃)₂(aq) + Na₂CO₃(aq) → MgCO₃(s) + NaNO₃(aq)

First, balance the original equation!

Left: Mg=1, N=2, O=6 from nitrate + 3 from carbonate? Wait — let’s count atoms properly.

Unbalanced:
Mg(NO₃)₂ + Na₂CO₃ → MgCO₃ + NaNO₃

Right side has only 1 Na and 1 NO₃, left has 2 Na and 2 NO₃ → need 2 NaNO₃ on right.

Balanced:
Mg(NO₃)₂(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2NaNO₃(aq)

Now break into ions:

Mg²⁺(aq) + 2NO₃⁻(aq) + 2Na⁺(aq) + CO₃²⁻(aq) → MgCO₃(s) + 2Na⁺(aq) + 2NO₃⁻(aq)

Cancel spectators: Na⁺ and NO₃⁻

Net Ionic:
Mg²⁺(aq) + CO₃²⁻(aq) → MgCO₃(s)

✔ Done.

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Problem 3:

strontium bromide(aq) + potassium sulfate(aq) → strontium sulfate(s) + potassium bromide(aq)

Write formulas first:

Strontium bromide = SrBr₂ (Sr is +2, Br is -1)
Potassium sulfate = K₂SO₄ (K is +1, SO₄ is -2)
Strontium sulfate = SrSO₄ (solid per rule #4)
Potassium bromide = KBr

Unbalanced:
SrBr₂(aq) + K₂SO₄(aq) → SrSO₄(s) + KBr(aq)

Balance: Need 2 KBr on right.

Balanced:
SrBr₂(aq) + K₂SO₄(aq) → SrSO₄(s) + 2KBr(aq)

Break into ions:

Sr²⁺(aq) + 2Br⁻(aq) + 2K⁺(aq) + SO₄²⁻(aq) → SrSO₄(s) + 2K⁺(aq) + 2Br⁻(aq)

Cancel spectators: K⁺ and Br⁻

Net Ionic:
Sr²⁺(aq) + SO₄²⁻(aq) → SrSO₄(s)

✔ Done.

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Problem 4:

manganese(II) chloride(aq) + ammonium carbonate(aq) → manganese(II) carbonate(s) + ammonium chloride(aq)

Formulas:

MnCl₂(aq) + (NH₄)₂CO₃(aq) → MnCO₃(s) + NH₄Cl(aq)

Balance: Right needs 2 NH₄Cl

Balanced:
MnCl₂(aq) + (NH₄)₂CO₃(aq) → MnCO₃(s) + 2NH₄Cl(aq)

Ions:

Mn²⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq) → MnCO₃(s) + 2NH₄⁺(aq) + 2Cl⁻(aq)

Cancel spectators: NH₄⁺ and Cl

Net Ionic:
Mn²⁺(aq) + CO₃²⁻(aq) → MnCO₃(s)

✔ Done.

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Problem 5:

chromium(III) nitrate(aq) + iron(II) sulfate(aq) → chromium(III) sulfate(aq) + iron(II) nitrate(aq)

Wait — check solubility! Are any products insoluble?

Chromium(III) sulfate: Rule #5 says sulfates are soluble except Ba, Ca, Sr, Pb, Hg, Ag → Cr is not exception → soluble.

Iron(II) nitrate: All nitrates soluble → soluble.

So both products are aqueous → no precipitate → all ions are spectators?

But let’s write it anyway.

Formulas:

Cr(NO₃)₃(aq) + FeSO₄(aq) → Cr₂(SO₄)₃(aq)? Wait — charges don’t match.

Chromium(III) = Cr³⁺, sulfate = SO₄²⁻ → formula is Cr₂(SO₄)₃

Iron(II) = Fe²⁺, nitrate = NO₃⁻ → Fe(NO₃)₂

So unbalanced:

Cr(NO₃)₃ + FeSO₄ → Cr₂(SO₄)₃ + Fe(NO₃)₂

Balance:

Left: Cr=1, right=2 → put 2 Cr(NO₃)₃
Left: Fe=1, right=1 → ok for now
Sulfate: left=1, right=3 → need 3 FeSO₄
Then Fe becomes 3 on left → need 3 Fe(NO₃)₂ on right
Nitrate: left=2×3=6, right=3×2=6 → good

Balanced:
2Cr(NO₃)₃(aq) + 3FeSO₄(aq) → Cr₂(SO₄)₃(aq) + 3Fe(NO₃)₂(aq)

All are aqueous → break all into ions:

Left:
2Cr³⁺(aq) + 6NO₃⁻(aq) + 3Fe²⁺(aq) + 3SO₄²⁻(aq)

Right:
2Cr³⁺(aq) + 3SO₄²⁻(aq) + 3Fe²⁺(aq) + 6NO₃⁻(aq)

Everything appears on both sides → all spectator ions.

→ No reaction occurs. Or: All species are spectator ions.

✔ Answer: All spectator ions — no net reaction.

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Problem 6:

K₃PO₄(aq) + Al(NO₃)₃(aq) → ?

Predict products: swap partners → K with NO₃, Al with PO₄

KNO₃ and AlPO₄

Check solubility:

KNO₃: all group 1 salts soluble → aqueous

AlPO₄: phosphates are insoluble except group 1 and ammonium → Al is not → solid!

So:

K₃PO₄(aq) + Al(NO₃)₃(aq) → AlPO₄(s) + KNO₃(aq)

Balance:

Left: K=3, PO₄=1, Al=1, NO₃=3

Right: AlPO₄ has Al=1, PO₄=1; KNO₃ has K=1, NO₃=1 → need 3 KNO₃

Balanced:
K₃PO₄(aq) + Al(NO₃)₃(aq) → AlPO₄(s) + 3KNO₃(aq)

Total Ionic:

3K⁺(aq) + PO₄³(aq) + Al³(aq) + 3NO₃⁻(aq) → AlPO₄(s) + 3K⁺(aq) + 3NO₃⁻(aq)

Cancel spectators: K⁺ and NO⁻

Net Ionic:
Al³⁺(aq) + PO₄³⁻(aq) → AlPO₄(s)

✔ Done.

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Problem 7:

BeI₂(aq) + Cu₂SO₄(aq) → ?

Wait — copper(I) sulfate? Usually copper is +2, but here it says Cu₂SO₄ → so copper must be +1 (since SO₄ is -2, two Cu⁺ make +2).

Products: swap → Be with SO₄, Cu with I

BeSO₄ and CuI

Solubility:

BeSO₄: sulfates are soluble except Ba, Ca, Sr, Pb, Hg, Ag → Be is not listed → assume soluble? But wait — beryllium is group 2, and rule #5 doesn't list it as exception → so soluble.

CuI: iodides are soluble except Ag, Pb, Hg → Cu is not exception? Wait — actually, copper(I) iodide is INSOLUBLE in reality, but according to given rules? Rule #3: halides soluble except Ag, Pb, Hg(I), Cu(I)? The rule says "copper(I)" is an exception? Let me check:

Rule #3: “All salts of halides are soluble except those of silver(I), copper(I), lead(II), and mercury(I).”

YES! Copper(I) is an exception → CuI is insoluble.

So CuI(s) forms.

BeSO₄(aq) remains aqueous.

Unbalanced:

BeI₂(aq) + Cu₂SO₄(aq) → BeSO₄(aq) + CuI(s)

Balance I: left=2, right=1 → need 2 CuI

Then Cu: left=2, right=2 → good

SO₄: left=1, right=1 → good

Be: 1 each → good

Balanced:
BeI₂(aq) + Cu₂SO₄(aq) → BeSO₄(aq) + 2CuI(s)

Total Ionic:

Be²⁺(aq) + 2I⁻(aq) + 2Cu⁺(aq) + SO₄²⁻(aq) → Be²⁺(aq) + SO₄²⁻(aq) + 2CuI(s)

Cancel spectators: Be²⁺ and SO₄²⁻

Net Ionic:
2Cu⁺(aq) + 2I⁻(aq) → 2CuI(s)

Can simplify by dividing by 2:

Cu⁺(aq) + I⁻(aq) → CuI(s)

✔ Done.

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Problem 8:

Ni(NO₃)₂(aq) + KBr(aq) → ?

Swap: Ni with Br, K with NO₃ → NiBr₂ and KNO₃

Both are soluble?

NiBr₂: bromides soluble except Ag, Pb, Hg, Cu(I) → Ni not exception → soluble

KNO₃: always soluble

So both products aqueous → no precipitate → all spectator ions.

But let’s write balanced equation:

Ni(NO₃)₂(aq) + 2KBr(aq) → NiBr₂(aq) + 2KNO₃(aq)

Total Ionic:

Ni²⁺(aq) + 2NO₃⁻(aq) + 2K⁺(aq) + 2Br⁻(aq) → Ni²⁺(aq) + 2Br⁻(aq) + 2K⁺(aq) + 2NO₃⁻(aq)

Everything cancels → all spectator ions.

→ No reaction.

✔ Answer: All spectator ions — no net reaction.

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Problem 9:

cobalt(III) bromide + potassium sulfide →

Formulas:

CoBr₃ (Co³⁺, Br⁻) + K₂S (K⁺, S²⁻) → Co₂S₃? and KBr

Cobalt(III) sulfide: Co³⁺ and S²⁻ → Co₂S₃

Potassium bromide: KBr

Unbalanced:

CoBr₃ + K₂S → Co₂S₃ + KBr

Balance:

Co: left=1, right=2 → 2 CoBr₃

S: left=1, right=3 → 3 K₂S

Then K: left=6, right=1 → 6 KBr

Br: left=6, right=6 → good

Balanced:
2CoBr₃(aq) + 3K₂S(aq) → Co₂S₃(s) + 6KBr(aq)

Check solubility: Co₂S₃ — sulfides are insoluble except group 1, 2, ammonium → Co is not → solid.

KBr: soluble.

Total Ionic:

2Co³⁺(aq) + 6Br⁻(aq) + 6K⁺(aq) + 3S²⁻(aq) → Co₂S₃(s) + 6K⁺(aq) + 6Br⁻(aq)

Cancel spectators: K⁺ and Br⁻

Net Ionic:
2Co³⁺(aq) + 3S²⁻(aq) → Co₂S₃(s)

✔ Done.

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Problem 10:

barium nitrate + ammonium phosphate →

Formulas:

Ba(NO₃)₂ + (NH₄)₃PO₄ → Ba(PO₄)₂ and NH₄NO₃

Barium phosphate: Ba³⁺? No — Ba is +2, PO₄ is -3 → Ba₃(PO₄)₂

Ammonium nitrate: NH₄NO₃

Unbalanced:

Ba(NO₃)₂ + (NH₄)₃PO₄ → Ba₃(PO₄)₂ + NH₄NO₃

Balance:

Ba: left=1, right=3 → 3 Ba(NO₃)₂

PO₄: left=1, right=2 → 2 (NH₄)₃PO₄

Then NH₄: left=6, right=1 → 6 NH₄NO₃

NO₃: left=6, right=6 → good

Balanced:
3Ba(NO₃)₂(aq) + 2(NH₄)₃PO₄(aq) → Ba₃(PO₄)₂(s) + 6NH₄NO₃(aq)

Solubility: Ba₃(PO₄)₂ — phosphates insoluble except group 1, ammonium → Ba not → solid.

NH₄NO₃: soluble.

Total Ionic:

3Ba²⁺(aq) + 6NO₃⁻(aq) + 6NH₄⁺(aq) + 2PO₄³⁻(aq) → Ba₃(PO₄)₂(s) + 6NH₄⁺(aq) + 6NO₃⁻(aq)

Cancel spectators: NH₄⁺ and NO₃⁻

Net Ionic:
3Ba²⁺(aq) + 2PO₄³⁻(aq) → Ba₃(PO)₂(s)

✔ Done.

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Problem 11:

calcium hydroxide + iron(III) chloride →

Formulas:

Ca(OH)₂ + FeCl₃ → CaCl₂ + Fe(OH)₃

Hydroxides: rule #6 — insoluble except group 1, Ca, Sr, Ba → Ca(OH)₂ is slightly soluble, but often treated as aqueous? Wait — the rule says “except those of group 1A, calcium, strontium and barium” — meaning they ARE soluble? Actually, no — re-read:

Rule #6: “All oxides and hydroxides are insoluble except for those of group 1A, calcium, strontium and barium.”

That means Ca(OH)₂ is considered soluble? But in reality, it's sparingly soluble. However, per this rule, since Ca is listed as exception, we treat Ca(OH)₂ as soluble? But wait — the rule says “insoluble EXCEPT” — so if it’s excepted, it’s soluble.

Similarly, Fe(OH)₃ — iron is not group 1 or Ca/Sr/Ba → insoluble.

So:

Ca(OH)₂(aq) + FeCl₃(aq) → CaCl₂(aq) + Fe(OH)₃(s)

Balance:

Ca: 1 each

OH: left=2, right=3 → LCM=6 → 3 Ca(OH)₂ and 2 Fe(OH)₃

Then Fe: left=2? Wait — start over.

Set FeCl₃ coefficient to 2 → then Fe(OH)₃ = 2

Then OH needed: 6 → so 3 Ca(OH)₂

Then Ca: 3 → so CaCl₂ = 3

Cl: left=6 (from 2 FeCl₃), right=6 (from 3 CaCl₂) → good

Balanced:
3Ca(OH)₂(aq) + 2FeCl₃(aq) → 3CaCl₂(aq) + 2Fe(OH)₃(s)

Total Ionic:

3Ca²⁺(aq) + 6OH⁻(aq) + 2Fe³⁺(aq) + 6Cl⁻(aq) → 3Ca²⁺(aq) + 6Cl⁻(aq) + 2Fe(OH)₃(s)

Cancel spectators: Ca²⁺ and Cl⁻

Net Ionic:
2Fe³⁺(aq) + 6OH⁻(aq) → 2Fe(OH)₃(s)

Simplify by dividing by 2:

Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)

✔ Done.

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Problem 12:

rubidium fluoride + copper(II) sulfate →

Formulas:

RbF + CuSO₄ → Rb₂SO₄ + CuF₂

Rubidium is group 1 → all salts soluble.

Copper(II) fluoride: fluorides — rule #3 doesn’t mention fluoride exceptions? Rule #3 is for halides — includes F? Typically, fluorides have different rules, but here rule #3 says “halides” — which usually includes F, Cl, Br, I.

But rule #3 exceptions: Ag, Cu(I), Pb, Hg(I) — copper(II) is not listed → so CuF₂ should be soluble? But in reality, CuF₂ is soluble.

However, let’s check the given rules again.

Rule #3: “All salts of halides are soluble except those of silver(I), copper(I), lead(II), and mercury(I).”

Copper(II) is not mentioned → so CuF₂ is soluble.

Rb₂SO₄: group 1 → soluble.

So both products soluble → no precipitate → all spectator ions.

Balanced equation:

2RbF(aq) + CuSO₄(aq) → Rb₂SO₄(aq) + CuF₂(aq)

Total Ionic:

2Rb⁺(aq) + 2F⁻(aq) + Cu²⁺(aq) + SO₄²⁻(aq) → 2Rb⁺(aq) + SO₄²⁻(aq) + Cu²⁺(aq) + 2F⁻(aq)

Everything cancels → all spectator ions.

→ No reaction.

✔ Answer: All spectator ions — no net reaction.

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## Final Answers:

1. Total Ionic: Ag⁺(aq) + NO₃⁻(aq) + K⁺(aq) + Cl⁻(aq) → AgCl(s) + K⁺(aq) + NO₃⁻(aq)
Net Ionic: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)

2. Total Ionic: Mg²⁺(aq) + 2NO₃⁻(aq) + 2Na⁺(aq) + CO₃²⁻(aq) → MgCO₃(s) + 2Na⁺(aq) + 2NO₃⁻(aq)
Net Ionic: Mg²⁺(aq) + CO₃²⁻(aq) → MgCO₃(s)

3. Total Ionic: Sr²⁺(aq) + 2Br⁻(aq) + 2K⁺(aq) + SO₄²⁻(aq) → SrSO₄(s) + 2K⁺(aq) + 2Br⁻(aq)
Net Ionic: Sr²⁺(aq) + SO₄²⁻(aq) → SrSO₄(s)

4. Total Ionic: Mn²⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq) → MnCO₃(s) + 2NH₄⁺(aq) + 2Cl⁻(aq)
Net Ionic: Mn²⁺(aq) + CO₃²⁻(aq) → MnCO₃(s)

5. All species are spectator ions — no net reaction.

6. Total Ionic: 3K⁺(aq) + PO₄³⁻(aq) + Al³⁺(aq) + 3NO₃⁻(aq) → AlPO₄(s) + 3K⁺(aq) + 3NO₃⁻(aq)
Net Ionic: Al³⁺(aq) + PO₄³⁻(aq) → AlPO₄(s)

7. Total Ionic: Be²⁺(aq) + 2I⁻(aq) + 2Cu⁺(aq) + SO₄²⁻(aq) → Be²⁺(aq) + SO₄²⁻(aq) + 2CuI(s)
Net Ionic: Cu⁺(aq) + I⁻(aq) → CuI(s)

8. All species are spectator ions — no net reaction.

9. Total Ionic: 2Co³⁺(aq) + 6Br⁻(aq) + 6K⁺(aq) + 3S²⁻(aq) → Co₂S₃(s) + 6K⁺(aq) + 6Br⁻(aq)
Net Ionic: 2Co³⁺(aq) + 3S²⁻(aq) → Co₂S₃(s)

10. Total Ionic: 3Ba²⁺(aq) + 6NO₃⁻(aq) + 6NH₄⁺(aq) + 2PO₄³⁻(aq) → Ba₃(PO₄)₂(s) + 6NH₄⁺(aq) + 6NO₃⁻(aq)
Net Ionic: 3Ba²⁺(aq) + 2PO₄³⁻(aq) → Ba₃(PO₄)₂(s)

11. Total Ionic: 3Ca²⁺(aq) + 6OH⁻(aq) + 2Fe³⁺(aq) + 6Cl⁻(aq) → 3Ca²⁺(aq) + 6Cl⁻(aq) + 2Fe(OH)₃(s)
Net Ionic: Fe³⁺(aq) + 3OH⁻(aq) → Fe(OH)₃(s)

12. All species are spectator ions — no net reaction.