College Chemistry Quiz: Net Ionic Equations

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Net Ionic EquationsQuestion 1 of 20

Aluminum metal reacts with copper(II) sulfate solution. If the complete ionic equation is $2Al(s) + 3Cu^{2+}(aq) + 3SO_4^{2-}(aq) \rightarrow 2Al^{3+}(aq) + 3Cu(s) + 3SO_4^{2-}(aq)$ , what is the net ionic equation?

Al(s) + Cu^{2+}(aq) \rightarrow Al^{3+}(aq) + Cu(s)

2Al(s) + 3Cu^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Cu(s)

2Al(s) + 3CuSO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3Cu(s)

3SO_4^{2-}(aq) \rightarrow 3SO_4^{2-}(aq)

Al^{3+}(aq) + SO_4^{2-}(aq) \rightarrow AlSO_4^+(aq)

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College Chemistry Quiz

College Chemistry Quiz: Net Ionic Equations

Practice Net Ionic Equations in College Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Net Ionic Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$Al(s) + Cu^{2+}(aq) \rightarrow Al^{3+}(aq) + Cu(s)$
$2Al(s) + 3Cu^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Cu(s)$ (correct answer)
$2Al(s) + 3CuSO_4(aq) \rightarrow Al_2(SO_4)_3(aq) + 3Cu(s)$
$3SO_4^{2-}(aq) \rightarrow 3SO_4^{2-}(aq)$
$Al^{3+}(aq) + SO_4^{2-}(aq) \rightarrow AlSO_4^+(aq)$

Explanation: When you encounter ionic equations, you're working with reactions involving ions in solution. The key concept here is distinguishing between complete ionic equations (which show all ions present) and net ionic equations (which show only the ions that actually participate in the reaction). To find the net ionic equation, you need to identify and remove spectator ions—ions that appear unchanged on both sides of the equation. Looking at the given complete ionic equation, notice that

SO_4^{2-}

appears identically on both sides:

3SO_4^{2-}(aq)

as reactants and

3SO_4^{2-}(aq)

as products. Since these sulfate ions don't change, they're spectators. Removing the spectator ions leaves you with only the species that actually undergo chemical change:

2Al(s) + 3Cu^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Cu(s)

. This represents the actual redox reaction where aluminum is oxidized and copper(II) is reduced. Choice A is incorrect because it shows unbalanced coefficients—you can't simply reduce all coefficients to 1 when the reaction requires a 2:3 stoichiometric ratio. Choice C is wrong because it shows the molecular equation, not the net ionic equation (it includes the complete compounds rather than individual ions). Choice D is incorrect because it only shows the spectator ions, which by definition don't participate in the net reaction. Remember: net ionic equations only include species that change during the reaction. Always identify spectator ions first, eliminate them, then verify that mass and charge are balanced in your final equation.

Question 2

Nitrous acid ( $HNO_2$ ) is a weak acid that reacts with potassium hydroxide, a strong base. The reaction produces potassium nitrite and water. In writing the net ionic equation, how should the nitrous acid be represented?

As $H^+(aq) + NO_2^-(aq)$ because it's an acid
As $HNO_2(aq)$ because it's a weak acid that remains largely molecular (correct answer)
As $HNO_2(s)$ because weak acids are insoluble
As $N^{3+}(aq) + H^+(aq) + 2O^{2-}(aq)$ showing all component ions
As $NO_2^-(aq)$ only, omitting the hydrogen

Explanation: When writing net ionic equations, you need to understand how different types of compounds behave in aqueous solution. Strong acids and bases dissociate completely into ions, while weak acids and bases remain largely in their molecular form. Nitrous acid (

HNO_2

) is a weak acid, meaning it only partially ionizes in water. Most

HNO_2

molecules stay intact rather than breaking apart into

H^+

and

NO_2^-

ions. Because the vast majority exists as whole molecules, you represent it as

HNO_2(aq)

in net ionic equations. This molecular representation accurately reflects what's actually present in solution. Looking at the wrong answers: Choice A incorrectly treats nitrous acid like a strong acid that completely dissociates. While

HNO_2

does produce some

H^+

and

NO_2^-

ions, the equilibrium heavily favors the molecular form. Choice C suggests weak acids are insoluble, which is false—weak acids dissolve in water but don't fully ionize. The (s) designation would only apply to truly insoluble compounds. Choice D breaks

HNO_2

into individual atoms as if it completely decomposed, which doesn't happen. Additionally, the oxidation states shown (

N^{3+}

and

O^{2-}

) don't match the actual bonding in nitrous acid. Study tip: Remember the key rule for net ionic equations—write strong electrolytes (strong acids, strong bases, soluble salts) as separate ions, but keep weak acids, weak bases, and molecular compounds in their molecular form. This distinction between strong and weak is crucial for correctly representing chemical species in solution.

Question 3

In a solution containing iron(III) nitrate and sodium hydroxide, a reddish-brown precipitate of iron(III) hydroxide forms. If a student writes the net ionic equation as $Fe^{3+}(aq) + OH^-(aq) \rightarrow FeOH^{2+}(aq)$ , what correction is needed?

The product should be written as a solid, not aqueous
The stoichiometry is incorrect; three hydroxide ions are needed per iron ion
Both the stoichiometry and phase of the product are incorrect (correct answer)
The iron should be written as Fe²⁺, not Fe³⁺
The hydroxide should be written as H⁺ + O²⁻

Explanation: When writing net ionic equations for precipitation reactions, you need to carefully consider both the stoichiometry (mole ratios) and the physical state of products. The question stem tells you that iron(III) nitrate reacts with sodium hydroxide to form a "reddish-brown precipitate" of iron(III) hydroxide. The student's equation has two major errors. First, the stoichiometry is wrong. Iron(III) has a +3 charge, so it needs three hydroxide ions (each with -1 charge) to form a neutral compound:

Fe^{3+}(aq) + 3OH^-(aq) \rightarrow Fe(OH)_3(s)

. The student only used one hydroxide ion and created an impossible product (

FeOH^{2+}

) that would still carry a charge. Second, the product should be written as a solid (s), not aqueous (aq), since the problem explicitly states a precipitate forms. Precipitates are insoluble solids that fall out of solution. Looking at the answer choices: A) correctly identifies the phase error but misses the stoichiometry problem. B) correctly identifies the stoichiometry error but ignores the phase issue. D) is completely wrong—the problem clearly states iron(III), meaning Fe³⁺ is correct. C) correctly identifies that both the stoichiometry and phase are wrong, making it the complete answer. Study tip: When writing precipitation equations, always check two things: does the charge balance work out (ensuring you have the right mole ratios), and is the insoluble product marked as a solid? Both stoichiometry and phase notation must be correct.

Question 4

Calcium hydroxide solution reacts with phosphoric acid ( $H_3PO_4$ ) to form calcium phosphate precipitate and water. In writing the net ionic equation, which ions should be treated as spectators and eliminated?

$Ca^{2+}$ and $PO_4^{3-}$ ions
$H^+$ and $OH^-$ ions
$Ca^{2+}$ and $OH^-$ ions
No ions are spectators in this reaction (correct answer)
$PO_4^{3-}$ and $H^+$ ions

Explanation: When writing net ionic equations, you need to identify spectator ions—ions that appear unchanged on both sides of the equation and don't participate in the actual chemical change. To do this, first write the complete molecular equation, then the complete ionic equation, and finally eliminate the spectators. The molecular equation is:

3Ca(OH)_2 + 2H_3PO_4 \rightarrow Ca_3(PO_4)_2 + 6H_2O

In the complete ionic equation, you break down all soluble compounds into ions, but keep precipitates and molecular compounds intact. Since calcium phosphate is insoluble (precipitate) and water is molecular, they stay together:

3Ca^{2+} + 6OH^- + 6H^+ + 2PO_4^{3-} \rightarrow Ca_3(PO_4)_2 + 6H_2O

Now examine each ion type. The

Ca^{2+}

ions (choice A) are consumed to form the precipitate—they're not spectators. The

PO_4^{3-}

ions (choice A) are also consumed in precipitate formation. The

H^+

and

OH^-

ions (choice B) react with each other to form water molecules. The

Ca^{2+}

and

OH^-

ions (choice C) are both reactants that get consumed. Since all ions participate in chemical changes—either forming the precipitate or neutralizing to form water—there are no spectator ions in this reaction. Choice D is correct. Remember: precipitation reactions combined with acid-base neutralizations often have no spectator ions because all reactants are consumed in chemical changes. Always check if ions are actually unchanged between reactants and products.

Question 5

A student mixes aqueous solutions of lead(II) nitrate and potassium iodide, resulting in the formation of a yellow precipitate. If the student incorrectly writes the net ionic equation as $Pb^{2+}(aq) + I^-(aq) \rightarrow PbI(s)$ , what error has been made?

The charges on the ions are incorrect
The precipitate should be written as aqueous, not solid
The stoichiometric coefficients are incorrect (correct answer)
Iodide should be written as $I_2$ instead of $I^-$
The reaction arrow direction is incorrect

Explanation: When you encounter precipitation reactions, the key is ensuring that the chemical formula reflects the correct stoichiometric ratios based on the charges of the ions involved. To write the correct net ionic equation, you need to balance the charges. Lead(II) has a +2 charge (

Pb^{2+}

), while iodide has a -1 charge (

I^-

). For the compound to be electrically neutral, you need two iodide ions to balance one lead ion, giving the formula

PbI_2

. The correct net ionic equation should be:

Pb^{2+}(aq) + 2I^-(aq) \rightarrow PbI_2(s)

The student's equation shows a 1:1 ratio, but this violates charge balance and doesn't reflect the actual empirical formula of lead(II) iodide. Looking at the incorrect answers: A) The charges on the ions are actually correct - lead(II) is indeed

Pb^{2+}

and iodide is

I^-

. B) The precipitate is correctly written as solid (s), since

PbI_2

is insoluble in water according to solubility rules. D) Iodide should remain as

I^-

because we're dealing with ionic compounds, not elemental iodine (

I_2

). The error is purely stoichiometric - the coefficients don't balance the charges properly, making C the correct answer. Study tip: Always check that your ionic equations maintain charge neutrality. Count the total positive and negative charges on each side - they must be equal. This charge-balancing requirement often determines the correct stoichiometric coefficients in ionic equations.

Question 6

Silver nitrate solution is added to a solution containing both chloride and bromide ions. Which statement best describes what happens in terms of net ionic equations?

Only one net ionic equation applies: $Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)$
Only one net ionic equation applies: $Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)$
Two separate net ionic equations apply simultaneously for the formation of both precipitates (correct answer)
The net ionic equation is: $Ag^+(aq) + Cl^-(aq) + Br^-(aq) \rightarrow AgClBr(s)$
No net ionic equation can be written because multiple anions are present

Explanation: When you encounter precipitation reactions involving multiple anions and a common cation, think about solubility rules and whether the anions compete or coexist in their reactions. Silver halides follow clear solubility patterns: AgCl and AgBr are both insoluble in water, with AgBr being slightly less soluble than AgCl. When

Ag^+

ions are added to a solution containing both

Cl^-

and

Br^-

ions, both precipitation reactions occur simultaneously because both products are insoluble. This gives us two separate net ionic equations:

Ag^+(aq) + Cl^-(aq) \rightarrow AgCl(s)

Ag^+(aq) + Br^-(aq) \rightarrow AgBr(s)

Option A is incorrect because it ignores the bromide ions present in solution. Even though AgCl formation occurs, the bromide ions don't just disappear—they also react with silver ions. Option B makes the same error in reverse, ignoring the chloride ions that are definitely present and will form AgCl precipitate. Option D suggests formation of a mixed compound AgClBr, but this doesn't occur. Silver forms separate, distinct compounds with each halide rather than a combined salt containing both anions. The key insight is that precipitation reactions are driven by thermodynamics—if a product is insoluble, the reaction will proceed. Since both AgCl and AgBr are insoluble, both reactions happen independently and simultaneously until one of the reactants is consumed. Study tip: When multiple anions can form insoluble compounds with the same cation, assume both reactions occur unless told otherwise or unless one product is significantly more insoluble.

Question 7

Zinc metal is placed in a solution of copper(II) chloride, causing the zinc to dissolve and copper metal to precipitate. Which net ionic equation correctly represents this single displacement reaction?

$Zn(s) + CuCl_2(aq) \rightarrow ZnCl_2(aq) + Cu(s)$
$Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$ (correct answer)
$Zn^{2+}(aq) + Cu(s) \rightarrow Zn(s) + Cu^{2+}(aq)$
$Zn(s) + 2Cl^-(aq) \rightarrow ZnCl_2(aq)$
$Zn(s) + Cu^{2+}(aq) + 2Cl^-(aq) \rightarrow Zn^{2+}(aq) + Cu(s) + 2Cl^-(aq)$

Explanation: When you encounter single displacement reactions, you need to identify what's actually happening at the ionic level and then write the simplest equation that shows only the species that change. In this reaction, zinc metal displaces copper from copper(II) chloride solution. The zinc atoms lose electrons to become

Zn^{2+}

ions (oxidation), while

Cu^{2+}

ions gain electrons to become copper metal (reduction). The chloride ions remain unchanged throughout the reaction - they're spectator ions. Option B correctly shows this electron transfer:

Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)

. This net ionic equation includes only the species that actually participate in the chemical change, with proper charges and states of matter. Option A shows the complete molecular equation, not the net ionic equation. While chemically accurate, it includes the spectator chloride ions and doesn't represent the simplest ionic form requested. Option C reverses the reaction direction entirely. This would represent copper displacing zinc, which contradicts the problem statement that zinc dissolves while copper precipitates. Option D only shows zinc reacting with chloride ions, completely omitting the copper species. This misses the essential displacement aspect of the reaction and doesn't show the reduction of copper ions. Remember: net ionic equations strip away spectator ions to show only the chemical species that undergo change. Always identify which ions are actively participating in electron transfer versus those just "watching" from the sidelines.

Question 8

When aqueous ammonia ( $NH_3$ ) reacts with hydrochloric acid, ammonium chloride forms. The complete ionic equation is: $NH_3(aq) + H^+(aq) + Cl^-(aq) \rightarrow NH_4^+(aq) + Cl^-(aq)$ . What is the net ionic equation?

$NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$
$NH_3(aq) + H^+(aq) \rightarrow NH_4^+(aq)$ (correct answer)
$NH_4^+(aq) + Cl^-(aq) \rightarrow NH_4Cl(s)$
$NH_3(aq) + H^+(aq) + Cl^-(aq) \rightarrow NH_4Cl(s) + Cl^-(aq)$
$H^+(aq) + Cl^-(aq) \rightarrow HCl(aq)$

Explanation: When you encounter net ionic equations, you're looking for the essential chemical change that occurs when you strip away the "spectator ions" - ions that appear unchanged on both sides of the equation. To find the net ionic equation, start with the complete ionic equation and cancel out any ions that appear identically on both reactant and product sides. In this reaction,

Cl^-

appears on both sides unchanged, making it a spectator ion. When you remove

Cl^-

from both sides, you're left with:

NH_3(aq) + H^+(aq) \rightarrow NH_4^+(aq)

. This shows the fundamental process: ammonia accepting a proton from the acid to form the ammonium ion. Looking at the wrong answers: Choice A shows the molecular equation, not the net ionic equation - it doesn't break compounds into their constituent ions. Choice C represents a precipitation reaction that isn't happening here; ammonium chloride remains dissolved in solution as separate ions, not as a solid precipitate. Choice D is essentially the complete ionic equation with an incorrect product - it shows solid ammonium chloride forming while still having a chloride ion as a separate product, which is chemically impossible. The correct answer is B because it captures only the ions that actually undergo change during the reaction. Study tip: Always identify spectator ions first by finding what appears unchanged on both sides, then cancel them out. The remaining species show you the actual chemical transformation occurring - that's your net ionic equation.

Question 9

When sodium sulfide solution is mixed with cadmium nitrate solution, a yellow precipitate of cadmium sulfide forms. The complete ionic equation shows: $2Na^+(aq) + S^{2-}(aq) + Cd^{2+}(aq) + 2NO_3^-(aq) \rightarrow CdS(s) + 2Na^+(aq) + 2NO_3^-(aq)$ . Which species participate in the net ionic equation?

Only $Na^+$ and $NO_3^-$ ions
Only $S^{2-}$ and $Cd^{2+}$ ions (correct answer)
$Cd^{2+}$ , $S^{2-}$ , $Na^+$ , and $NO_3^-$ ions
Only $Na^+$ , $S^{2-}$ , and $Cd^{2+}$ ions
Only $Cd^{2+}$ , $S^{2-}$ , and $NO_3^-$ ions

Explanation: When you encounter precipitation reactions, the key concept being tested is your ability to identify spectator ions and write net ionic equations. A net ionic equation shows only the species that actually participate in the chemical change, excluding spectator ions that remain unchanged. To find the net ionic equation, you need to identify which ions are spectators. Looking at the complete ionic equation, notice that

Na^+

and

NO_3^-

appear on both sides of the equation in identical forms and quantities. These are spectator ions - they're present in solution but don't participate in forming the precipitate. When you remove the spectators, you're left with the net ionic equation:

S^{2-}(aq) + Cd^{2+}(aq) \rightarrow CdS(s)

This shows that only

S^{2-}

and

Cd^{2+}

ions actually participate in the reaction, making B correct. Option A is backwards -

Na^+

and

NO_3^-

are the spectator ions that specifically don't participate in the net ionic equation. Option C includes all ions, which describes the complete ionic equation, not the net ionic equation. Option D incorrectly includes

Na^+

, which is a spectator ion that cancels out when writing the net equation. Remember this pattern: spectator ions appear unchanged on both sides of the complete ionic equation and get canceled out. The net ionic equation includes only the ions that combine to form products (like precipitates) or undergo actual chemical change. Always look for what cancels versus what reacts.

Question 10

When barium hydroxide solution is mixed with sulfuric acid, both neutralization and precipitation occur simultaneously. The complete ionic equation is: $Ba^{2+}(aq) + 2OH^-(aq) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s) + 2H_2O(l)$ . What can be concluded about the net ionic equation?

It is the same as the complete ionic equation because no spectator ions are present (correct answer)
It should only show the precipitation: $Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)$
It should only show the neutralization: $2H^+(aq) + 2OH^-(aq) \rightarrow 2H_2O(l)$
It cannot be written because two different reactions occur simultaneously
It should show $Ba^{2+}(aq) + 2OH^-(aq) \rightarrow Ba(OH)_2(aq)$

Explanation: When you encounter reactions involving both acid-base neutralization and precipitation, remember that the net ionic equation shows all ions and molecules that actually participate in chemical change—there's no rule requiring you to separate simultaneous reactions. Let's examine what's happening here. In the complete ionic equation, every species shown undergoes a chemical transformation:

Ba^{2+}

and

SO_4^{2-}

combine to form solid

BaSO_4

, while

H^+

and

OH^-

combine to form water. Since no ions remain unchanged (no spectator ions), the net ionic equation is identical to the complete ionic equation. This makes choice A correct. Choice B is wrong because it ignores the neutralization reaction entirely. While

Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)

does represent the precipitation, the

H^+

and

OH^-

ions are also reacting to form water—they're not spectators. Choice C makes the opposite error, showing only the acid-base neutralization while ignoring the precipitation. The

Ba^{2+}

and

SO_4^{2-}

ions are actively forming a precipitate, not acting as spectators. Choice D reflects a common misconception that simultaneous reactions somehow prevent writing a net ionic equation. In reality, net ionic equations can represent multiple reaction types occurring together, as long as you include all the reacting species. Study tip: When multiple reaction types occur simultaneously, don't try to separate them—include all species that undergo chemical change in your net ionic equation. Only exclude true spectator ions that remain unchanged throughout the entire process.

Question 11

Solid calcium carbonate reacts with nitric acid to produce carbon dioxide, water, and calcium nitrate. Which statement about writing the net ionic equation for this reaction is correct?

All carbonates should be written in ionic form as $Ca^{2+} + CO_3^{2-}$
The solid calcium carbonate should remain in molecular form as $CaCO_3(s)$ (correct answer)
The nitric acid should remain in molecular form as $HNO_3$
The carbon dioxide should be written in ionic form as $C^{4+} + 2O^{2-}$
All compounds should be written in molecular form

Explanation: When writing net ionic equations, you need to determine which compounds should be written in ionic form (separated into ions) versus molecular form (as complete compounds). The key rule is that only strong electrolytes that are dissolved in solution should be written as separated ions. Let's examine this reaction:

CaCO_3(s) + 2HNO_3(aq) \rightarrow CO_2(g) + H_2O(l) + Ca(NO_3)_2(aq)

The correct answer is B because solid calcium carbonate should remain as

CaCO_3(s)

. Even though calcium carbonate is an ionic compound, it exists as a solid with low solubility in water, so it doesn't dissociate into ions in solution. Solids, gases, and weak electrolytes always stay in molecular form in net ionic equations. Choice A is incorrect because carbonates should only be written in ionic form when they're dissolved. Since

CaCO_3

is a solid reactant with limited solubility, it remains molecular. Choice C is wrong because nitric acid (

HNO_3

) is a strong acid that completely ionizes in aqueous solution, so it should be written as

H^+ + NO_3^-

, not in molecular form. Choice D is incorrect because

CO_2

is a gas that exists as molecular units, not as separated ions. The notation

C^{4+} + 2O^{2-}

would represent impossible free ions in solution. Study tip: Remember the acronym "SWIG" - Solids, Weak electrolytes, Insoluble compounds, and Gases all stay in molecular form in net ionic equations. Only strong electrolytes dissolved in solution get separated into ions.

Question 12

A student observes that when sodium phosphate solution is added to iron(III) chloride solution, a yellowish precipitate forms. The student writes: $Na_3PO_4(aq) + FeCl_3(aq) \rightarrow FePO_4(s) + 3NaCl(aq)$ . To convert this to a net ionic equation, what must be done?

Remove the $Na^+$ and $Cl^-$ ions that act as spectators
Convert all compounds to their ionic forms first, then remove spectators (correct answer)
Remove only the $Na^+$ ions because sodium compounds are always soluble
Remove only the $Cl^-$ ions because chlorides are generally soluble
The equation is already in net ionic form

Explanation: When you encounter precipitation reactions, writing a net ionic equation requires a systematic approach that follows specific rules about which compounds dissociate in solution. To write a net ionic equation correctly, you must first convert all soluble ionic compounds to their dissociated ionic forms, while keeping insoluble compounds (precipitates) in their molecular form. For this reaction,

Na_3PO_4

FeCl_3

, and

NaCl

are all soluble and should be written as separate ions, while

FePO_4

remains molecular as the precipitate. The complete ionic equation becomes:

3Na^+(aq) + PO_4^{3-}(aq) + Fe^{3+}(aq) + 3Cl^-(aq) \rightarrow FePO_4(s) + 3Na^+(aq) + 3Cl^-(aq)

. Then you remove the spectator ions (

Na^+

and

Cl^-

) that appear unchanged on both sides, leaving:

Fe^{3+}(aq) + PO_4^{3-}(aq) \rightarrow FePO_4(s)

. Answer choice A incorrectly suggests you can identify spectators directly from the molecular equation without first writing the complete ionic form. Choice C is wrong because while sodium compounds are typically soluble, you must still show the dissociation step and remove all spectators, not just

Na^+

. Choice D fails because it ignores the

Na^+

spectators and doesn't address the required dissociation step. Remember this sequence: molecular equation → complete ionic equation (dissociate all soluble compounds) → net ionic equation (remove spectators). This systematic approach prevents errors and ensures you capture the actual chemical change occurring.

Question 13

Hydrobromic acid ( $HBr$ ) reacts with lithium hydroxide ( $LiOH$ ) in aqueous solution. Both compounds are strong electrolytes that completely dissociate. What is the net ionic equation for this acid-base neutralization?

$HBr(aq) + LiOH(aq) \rightarrow LiBr(aq) + H_2O(l)$
$H^+(aq) + Br^-(aq) + Li^+(aq) + OH^-(aq) \rightarrow Li^+(aq) + Br^-(aq) + H_2O(l)$
$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$ (correct answer)
$Li^+(aq) + Br^-(aq) \rightarrow LiBr(s)$
$HBr(aq) + OH^-(aq) \rightarrow Br^-(aq) + H_2O(l)$

Explanation: When you encounter acid-base neutralization problems involving strong electrolytes, you need to write net ionic equations by focusing on the actual chemical change occurring and eliminating spectator ions. Start with the molecular equation:

HBr(aq) + LiOH(aq) \rightarrow LiBr(aq) + H_2O(l)

. Since both reactants are strong electrolytes, they completely dissociate in solution. The complete ionic equation shows all ions:

H^+(aq) + Br^-(aq) + Li^+(aq) + OH^-(aq) \rightarrow Li^+(aq) + Br^-(aq) + H_2O(l)

. To get the net ionic equation, remove spectator ions (ions that appear unchanged on both sides). Here,

Li^+

and

Br^-

are spectators, leaving:

H^+(aq) + OH^-(aq) \rightarrow H_2O(l)

. Answer C correctly shows this net ionic equation, representing the essential reaction between hydrogen and hydroxide ions to form water. Answer A shows the molecular equation, not the net ionic equation. Answer B is the complete ionic equation with all ions present, including spectators that should be removed. Answer D incorrectly suggests lithium bromide precipitates as a solid, but

LiBr

is highly soluble in water and remains dissolved as ions. Remember that net ionic equations for strong acid-strong base neutralizations always simplify to

H^+(aq) + OH^-(aq) \rightarrow H_2O(l)

, regardless of which specific acid and base are involved. Focus on identifying and removing spectator ions to reveal the actual chemical change.

Question 14

Ammonium sulfate reacts with barium chloride in aqueous solution to form a white precipitate of barium sulfate and ammonium chloride. When writing the net ionic equation, which approach is correct for handling the ammonium ion ( $NH_4^+$ )?

Write it as $NH_3 + H^+$ because ammonia is a weak base
Write it as $NH_4^+(aq)$ because it acts as a spectator ion in this reaction (correct answer)
Omit it entirely because it doesn't participate in precipitation
Write it as $NH_4^+(s)$ because it's part of a salt
Write it in molecular form as $NH_4OH$ because it's basic

Explanation: When writing net ionic equations for precipitation reactions, you need to identify which ions actually participate in forming the precipitate and which remain dissolved as spectator ions. Let's examine this reaction systematically. The molecular equation is:

(NH_4)_2SO_4(aq) + BaCl_2(aq) \rightarrow BaSO_4(s) + 2NH_4Cl(aq)

To write the net ionic equation, you break down all soluble compounds into their constituent ions, then eliminate spectators. The ammonium ion (

NH_4^+

) starts in solution as part of ammonium sulfate and ends in solution as part of ammonium chloride. Since it remains aqueous throughout and doesn't participate in the precipitation of barium sulfate, it's a spectator ion and should be written as

NH_4^+(aq)

before being canceled out. Option A incorrectly breaks

NH_4^+

into

NH_3 + H^+

. While ammonium can act as a weak acid, in net ionic equations you write ions as they actually exist in solution, not as their dissociation products. Option C is wrong because spectator ions do appear in the complete ionic equation—they're just omitted from the final net ionic equation after canceling. Option D incorrectly shows

NH_4^+

as solid. The ammonium remains dissolved in solution throughout this reaction. The correct answer is B: write ammonium as

NH_4^+(aq)

since it's a spectator ion. Remember: in precipitation reactions, focus on which ions combine to form the insoluble product. All other ions that remain dissolved are spectators, regardless of their acid-base properties.

Question 15

Perchloric acid ( $HClO_4$ ) is a strong acid that reacts with aqueous ammonia ( $NH_3$ ), a weak base. The products are ammonium perchlorate and water. Which represents the correct net ionic equation for this acid-base reaction?

$HClO_4(aq) + NH_3(aq) \rightarrow NH_4ClO_4(aq)$
$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$
$H^+(aq) + NH_3(aq) \rightarrow NH_4^+(aq)$ (correct answer)
$HClO_4(aq) + NH_3(aq) \rightarrow NH_4^+(aq) + ClO_4^-(aq)$
$NH_3(aq) + ClO_4^-(aq) \rightarrow NH_3ClO_4(aq)$

Explanation: When you encounter acid-base reactions, especially those asking for net ionic equations, focus on identifying what species actually participate in the chemical change and how they appear in solution. Let's work through this systematically. Perchloric acid (

HClO_4

) is a strong acid, meaning it completely ionizes in water to

H^+

and

ClO_4^-

. Ammonia (

NH_3

) is a weak base that accepts protons. The key insight is that in acid-base reactions, the essential process is proton transfer from acid to base. The correct answer is C:

H^+(aq) + NH_3(aq) \rightarrow NH_4^+(aq)

. This shows the fundamental chemical change occurring—a proton from the strong acid is transferred to the ammonia molecule, forming the ammonium ion. This captures the essence of the acid-base reaction. Option A shows molecular compounds rather than the ionic forms these species actually take in solution, missing the point of a net ionic equation. Option B represents the neutralization of a strong acid with a strong base (producing

OH^-

), but ammonia doesn't generate

OH^-

ions—it directly accepts protons. Option D includes the spectator ion

ClO_4^-

, which doesn't participate in the actual chemical change and should be omitted from net ionic equations. Remember: net ionic equations show only the species that undergo chemical change. Strong acids and bases should be written as ions, weak bases like

NH_3

stay molecular, and spectator ions are excluded. Focus on the proton transfer mechanism.

Question 16

Consider the reaction between lead(II) acetate and potassium chromate, which produces a yellow precipitate of lead(II) chromate. If the molecular equation is $Pb(CH_3COO)_2(aq) + K_2CrO_4(aq) \rightarrow PbCrO_4(s) + 2KCH_3COO(aq)$ , which step in obtaining the net ionic equation requires the most careful attention to solubility rules?

Determining whether $Pb(CH_3COO)_2$ should be written in ionic form
Determining whether $PbCrO_4$ should be written as a solid or aqueous
Determining whether $K_2CrO_4$ dissociates completely in solution
Determining whether $KCH_3COO$ remains in solution or precipitates (correct answer)
Determining the correct stoichiometric coefficients for the balanced equation

Explanation: When writing net ionic equations, you must carefully apply solubility rules to determine which compounds dissociate into ions and which remain as molecular units. This process requires identifying what actually dissolves versus what precipitates or remains intact in solution. The correct answer is D because determining whether

KCH_3COO

(potassium acetate) remains in solution requires the most careful attention to solubility rules. You need to know that acetates are generally soluble, and since potassium compounds are also highly soluble,

KCH_3COO

will indeed remain dissolved as

K^+

and

CH_3COO^-

ions. This step is crucial because incorrectly assuming it precipitates would fundamentally change your net ionic equation. Option A is straightforward because lead(II) acetate is clearly marked as aqueous in the equation, and acetates are generally soluble. Option B is already given in the molecular equation—

PbCrO_4

is shown as a solid precipitate, and this aligns with the solubility rule that most chromates are insoluble except for alkali metals and ammonium. Option C is also relatively simple since

K_2CrO_4

contains potassium, and all Group 1 compounds are soluble and dissociate completely. The key study tip: When writing net ionic equations, pay special attention to the products, especially when both products could theoretically be ionic compounds. Always double-check solubility rules for the products—this is where most errors occur because students focus on the obviously soluble reactants but forget to verify that products like acetates, nitrates, or other "generally soluble" ions actually remain in solution.

Question 17

Iron metal reacts with hydrochloric acid to produce hydrogen gas and iron(II) chloride. A student writes: $Fe(s) + 2H^+(aq) + 2Cl^-(aq) \rightarrow Fe^{2+}(aq) + 2Cl^-(aq) + H_2(g)$ . What type of equation has the student written?

The correct net ionic equation
The complete ionic equation
The molecular equation
An incorrect equation with wrong stoichiometry
An equation that needs spectator ions removed (correct answer)

Explanation: When analyzing chemical equations, you need to distinguish between three types: molecular equations (showing complete formulas), complete ionic equations (showing all ions), and net ionic equations (showing only species that actually change). The equation given shows all species present in solution, including the chloride ions that appear on both sides. This is the hallmark of a complete ionic equation - it displays every ion and molecule present, even those that don't participate in the actual reaction (spectator ions). To get the net ionic equation, you'd cancel the

2Cl^-(aq)

from both sides, leaving:

Fe(s) + 2H^+(aq) \rightarrow Fe^{2+}(aq) + H_2(g)

. This shows only the species that actually undergo change. Looking at the wrong answers: Choice A is incorrect because this isn't the net ionic equation - it still includes the spectator chloride ions that should be removed. Choice C is wrong because the molecular equation would show

Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)

with complete compound formulas, not separated ions. Choice D is incorrect because the stoichiometry is actually correct - iron does form

Fe^{2+}

, requiring 2 moles of

H^+

and producing 1 mole of

H_2

. The student has written the complete ionic equation, showing all dissolved species as separate ions while maintaining correct stoichiometric relationships. Strategy tip: Always identify spectator ions first - they appear unchanged on both sides and help you distinguish between complete ionic and net ionic equations.

Question 18

In the reaction between copper(II) sulfate and sodium carbonate solutions, a blue-green precipitate forms. If the complete ionic equation shows $Cu^{2+}(aq) + SO_4^{2-}(aq) + 2Na^+(aq) + CO_3^{2-}(aq) \rightarrow CuCO_3(s) + 2Na^+(aq) + SO_4^{2-}(aq)$ , what is the correct net ionic equation?

$Cu^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CuCO_3(s)$ (correct answer)
$CuSO_4(aq) + Na_2CO_3(aq) \rightarrow CuCO_3(s) + Na_2SO_4(aq)$
$Cu^{2+}(aq) + SO_4^{2-}(aq) + CO_3^{2-}(aq) \rightarrow CuCO_3(s) + SO_4^{2-}(aq)$
$2Na^+(aq) + SO_4^{2-}(aq) \rightarrow Na_2SO_4(aq)$
$Cu^{2+}(aq) + 2Na^+(aq) + CO_3^{2-}(aq) \rightarrow CuCO_3(s) + 2Na^+(aq)$

Explanation: When you encounter precipitation reactions, you need to identify which ions actually participate in forming the precipitate versus which ones remain unchanged in solution. The key is distinguishing between complete ionic equations (which show all dissolved ions) and net ionic equations (which show only the ions that react). To find the net ionic equation, you remove spectator ions—those that appear unchanged on both sides of the equation. In the given complete ionic equation,

Na^+

and

SO_4^{2-}

ions appear identically on both reactant and product sides, making them spectators. Only

Cu^{2+}

and

CO_3^{2-}

actually combine to form the blue-green precipitate

CuCO_3(s)

. Choice A correctly shows

Cu^{2+}(aq) + CO_3^{2-}(aq) \rightarrow CuCO_3(s)

, representing only the ions that participate in the precipitation reaction. Choice B is the molecular equation, not the net ionic equation—it shows complete compounds rather than individual ions. Choice C incorrectly includes

SO_4^{2-}

ions, which are spectators that should be eliminated from the net ionic equation. Choice D shows only the spectator ions forming sodium sulfate, which completely misses the actual precipitation reaction between copper and carbonate ions. Remember: net ionic equations focus on the "action"—strip away everything that doesn't change, and you'll see the essential reaction. Always identify spectator ions first, then eliminate them to reveal what's actually happening chemically.

Question 19

When hydrochloric acid reacts with sodium hydroxide, the complete ionic equation is: $H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightarrow Na^+(aq) + Cl^-(aq) + H_2O(l)$ . What is the net ionic equation for this reaction?

$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$
$H^+(aq) + OH^-(aq) \rightarrow H_2O(l)$ (correct answer)
$Na^+(aq) + Cl^-(aq) \rightarrow NaCl(s)$
$H^+(aq) + Cl^-(aq) + OH^-(aq) \rightarrow H_2O(l) + Cl^-(aq)$
$H^+(aq) + Na^+(aq) + OH^-(aq) \rightarrow NaOH(aq) + H^+(aq)$

Explanation: When you encounter questions about net ionic equations, you're being tested on your ability to identify which ions actually participate in a chemical reaction versus those that remain unchanged as spectator ions. To find the net ionic equation, you need to eliminate spectator ions from the complete ionic equation. Looking at the given equation:

H^+(aq) + Cl^-(aq) + Na^+(aq) + OH^-(aq) \rightarrow Na^+(aq) + Cl^-(aq) + H_2O(l)

Notice that

Na^+

and

Cl^-

appear on both sides of the equation in identical forms - these are spectator ions that don't actually participate in the reaction. When you remove them, you're left with:

H^+(aq) + OH^-(aq) \rightarrow H_2O(l)

. This shows the essence of the acid-base neutralization reaction. Answer A represents the molecular equation, not the net ionic equation - it shows complete compounds rather than the specific ions involved. Answer C is incorrect because it suggests sodium chloride precipitates as a solid, but NaCl remains dissolved in solution (hence the spectator ions). Answer D fails to eliminate all spectator ions, still including

Cl^-

on both sides unnecessarily. Answer B correctly shows only the ions that actually react: the hydrogen ion from the acid combines with the hydroxide ion from the base to form water. Study tip: For net ionic equations, always cross out any ion that appears unchanged on both sides of the arrow. What remains shows you the actual chemical change occurring. Most acid-base neutralizations in aqueous solution reduce to this same net ionic equation.

Question 20

Magnesium carbonate reacts with hydrochloric acid to produce carbon dioxide gas, water, and magnesium chloride. Which represents the correct net ionic equation for this acid-carbonate reaction?

$MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l) + CO_2(g)$
$CO_3^{2-}(aq) + 2H^+(aq) \rightarrow H_2O(l) + CO_2(g)$
$MgCO_3(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H_2O(l) + CO_2(g)$ (correct answer)
$Mg^{2+}(aq) + 2Cl^-(aq) \rightarrow MgCl_2(aq)$
$MgCO_3(s) + 2H^+(aq) + 2Cl^-(aq) \rightarrow Mg^{2+}(aq) + 2Cl^-(aq) + H_2O(l) + CO_2(g)$

Explanation: When you encounter acid-carbonate reactions, you need to write net ionic equations that show only the species that actually participate in the chemical change, excluding spectator ions that remain unchanged. To find the correct net ionic equation, start with the complete molecular equation and identify what's actually reacting. Magnesium carbonate is an insoluble solid, so it appears as

MgCO_3(s)

rather than dissociated ions. When it reacts with

H^+

ions from the acid, the carbonate portion breaks down to form water and carbon dioxide, while the magnesium dissolves as

Mg^{2+}

ions. The correct answer is C:

MgCO_3(s) + 2H^+(aq) \rightarrow Mg^{2+}(aq) + H_2O(l) + CO_2(g)

. This equation shows the actual chemical transformation - solid magnesium carbonate reacting with hydrogen ions to produce dissolved magnesium ions, water, and carbon dioxide gas. Option A is the complete molecular equation, not the net ionic equation, since it includes

HCl

and

MgCl_2

as complete compounds rather than showing the active ions. Option B incorrectly shows carbonate as

CO_3^{2-}(aq)

, but magnesium carbonate is insoluble, so the carbonate doesn't exist as free ions in solution. Option D represents the formation of an ionic compound from its ions, which isn't the primary reaction occurring here - it's missing the acid-carbonate interaction entirely. Remember: for net ionic equations involving insoluble compounds, write the compound in its molecular form, not as separate ions. Focus on what's actually changing during the reaction.