Double Displacement and Precipitation Reactions

What is a Double Displacement Reaction?

In the previous section we saw that in displacement, one element kicks another out. Double displacement is a little different — two compounds exchange their ions (or groups of atoms) with each other.

Definition: A reaction in which the ions of two reactants exchange places to form two new compounds is called a double displacement reaction.

General Form

$$AB + CD \rightarrow AD + CB$$

Explanation:

• Reactant 1 — A and B are paired
• Reactant 2 — C and D are paired
• After reaction — A is now paired with D, and C with B
• The ions have 'changed partners'.

Activity 1.10 — A Famous NCERT Experiment

Procedure:

1. Take 3 mL of sodium sulphate solution ($Na_2SO_4$) in a test tube.
2. Take 3 mL of barium chloride solution ($BaCl_2$) in another test tube.
3. Mix the two solutions.

Observation:

• Instantly, a white precipitate is formed.
• This precipitate is insoluble in water.

Reaction:

$$Na_2SO_4(aq) + BaCl_2(aq) \rightarrow BaSO_4(s)\downarrow + 2NaCl(aq)$$

Analysis:

• $Na^+$ and $Ba^{2+}$ ions swapped partners.
• $Ba^{2+}$ now joins $SO_4^{2-}$ to form $BaSO_4$ (insoluble = white precipitate).
• $Na^+$ now joins $Cl^-$ to form $NaCl$ (stays in solution).

Visual rule: A double displacement reaction always has two reactants → two products.

What is a Precipitate?

When mixing two solutions produces an insoluble solid that settles at the bottom, that solid is called a precipitate. The reaction in which a precipitate forms is called a precipitation reaction.

How Does Precipitation Happen?

When ions meet in solution —

• If the new pair is soluble → no precipitate.
• If the new pair is insoluble → a precipitate forms (settles down).

Example:

$$\underbrace{Ba^{2+}(aq) + SO_4^{2-}(aq)}_{\text{free ions}} \rightarrow \underbrace{BaSO_4(s)\downarrow}_{\text{insoluble precipitate}}$$

Not Every Double Displacement is a Precipitation

This is an important distinction.

• Double displacement + precipitate forms = Precipitation reaction
• Double displacement + both products soluble = simple double displacement (no precipitate)

Colours of Common Precipitates — Essential Board List

Precipitate	Colour	How it forms
$BaSO_4$	White	$BaCl_2 + Na_2SO_4$
$AgCl$	White	$AgNO_3 + NaCl$
$AgBr$	Pale yellow	$AgNO_3 + KBr$
$AgI$	Yellow	$AgNO_3 + KI$
$PbI_2$	Bright yellow	$Pb(NO_3)_2 + KI$
$PbCl_2$	White (soluble in hot water)	$Pb(NO_3)_2 + NaCl$
$Cu(OH)_2$	Blue	$CuSO_4 + NaOH$
$Fe(OH)_3$	Brown	$FeCl_3 + NaOH$
$CaCO_3$	White	$Ca(OH)_2 + CO_2$

[Board Important] Memorise these colours — questions like "What is the colour of the precipitate of …?" are common.

Sub-types of Double Displacement Reactions

Double displacement reactions come in three sub-types —

1. Precipitation Reactions

When one of the products is insoluble (a precipitate).

Examples:

$$AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s)\downarrow + NaNO_3(aq)$$

(white AgCl precipitate)

$$Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s)\downarrow + 2KNO_3(aq)$$

(yellow PbI₂ precipitate — the 'Golden Rain' experiment)

2. Neutralisation Reactions

When acid and base combine to form salt and water.

General form: $$\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}$$

Examples:

$$HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$$

$$2HNO_3(aq) + Ca(OH)_2(aq) \rightarrow Ca(NO_3)_2(aq) + 2H_2O(l)$$

$$H_2SO_4(aq) + 2KOH(aq) \rightarrow K_2SO_4(aq) + 2H_2O(l)$$

All neutralisation reactions are a special case of double displacement.

3. Gas-Evolution Reactions

When one of the products escapes as a gas.

Examples:

$$CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)\uparrow$$

$$Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)\uparrow$$

(Here $H_2CO_3$ is unstable and immediately decomposes to $H_2O + CO_2$ — that's what causes the gas evolution.)

A Summary Table

Sub-type	Feature	Symbol
Precipitation	One product is solid	$\downarrow$
Neutralisation	Acid + Base → Salt + Water	$H^+ + OH^- \rightarrow H_2O$
Gas evolution	One product is a gas	$\uparrow$

Displacement vs. Double Displacement — A Comparison

This distinction is asked very often in board exams. Pay close attention.

Aspect	Displacement	Double Displacement
Participants	One element + one compound	Two compounds
What happens?	An element displaces another element	Two ion-groups swap partners
General form	$A + BC \rightarrow AC + B$	$AB + CD \rightarrow AD + CB$
Driver	Activity series	Solubility / precipitate / gas / water formation
Example	$Fe + CuSO_4 \rightarrow FeSO_4 + Cu$	$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$
Visual cue	Metallic deposit	Precipitate / gas / water
Products	One free element (Cu)	Two compounds

A Simple Identification Rule

If the reaction has any free element (a single element on its own), it is a displacement. If all reactants and all products are compounds, it is a double displacement.

Example:

$Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$ → Zn and Cu appear as free elements → displacement

$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$ → all are compounds → double displacement

A Special Note

The formation of a precipitate in a double displacement is just as 'spectacular' as the colour change in displacement — both give visible proof that a reaction has occurred.

Daily Life Applications of Double Displacement and Precipitation

1. Identification of Ions in the Lab

In chemistry labs, specific precipitation tests are used to identify what ion a solution contains.

Examples:

• Test for $SO_4^{2-}$ — adding $BaCl_2$ produces a white precipitate
• Test for $Cl^-$ — adding $AgNO_3$ produces a white precipitate ($AgCl$)
• Test for $CO_3^{2-}$ — adding $HCl$ produces $CO_2$ effervescence

2. Water Purification

Impurities in drinking water are removed by precipitation —

• Adding alum ($KAl(SO_4)_2 \cdot 12H_2O$) causes fine clay particles to settle out.
• This technique is widely used in chemical purification industries.

3. Digestion

Our stomach contains $HCl$. When we take $NaHCO_3$ (an antacid) with food, neutralisation occurs —

$$NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2\uparrow$$

This relieves acidity — a burp is a sign of the released $CO_2$.

4. Kidney Stones

A precipitate of calcium oxalate, $CaC_2O_4$, can form in the kidneys — that is what a kidney stone is.

5. Cement and Marble

The shiny coating that forms on a whitewashed wall is also a $CaCO_3$ precipitate (we saw this in Section 3).

6. The Lime-Water Test (for $CO_2$)

$$Ca(OH)_2(aq) + CO_2(g) \rightarrow CaCO_3(s)\downarrow + H_2O(l)$$

If passing a gas through lime water turns it milky, the gas is $CO_2$.

[Board Important] Precipitation-based questions appear regularly in experimental, identification, or real-life contexts.

🧠 Memory Capsule

A one-glance recap to revisit just before the board exam — every key idea about double displacement and precipitation in one place.

1. Double Displacement Formula

$$AB + CD \rightarrow AD + CB$$ (two compounds → two compounds; ions swap partners)

2. Precipitate Formula

When two ions combine to form an insoluble solid — $$X^{n+}(aq) + Y^{m-}(aq) \rightarrow XY(s)\downarrow$$

3. Must-Memorise Reactions

Reaction	Precipitate	Colour
$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$	$BaSO_4$	White
$AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$	$AgCl$	White
$Pb(NO_3)_2 + 2KI \rightarrow PbI_2 + 2KNO_3$	$PbI_2$	Yellow
$CuSO_4 + 2NaOH \rightarrow Cu(OH)_2 + Na_2SO_4$	$Cu(OH)_2$	Blue
$FeCl_3 + 3NaOH \rightarrow Fe(OH)_3 + 3NaCl$	$Fe(OH)_3$	Brown

4. Three Sub-types of Double Displacement

Type	Feature	Example
Precipitation	Insoluble product	$BaCl_2 + Na_2SO_4$
Neutralisation	Acid + base → salt + water	$HCl + NaOH$
Gas evolution	Gas product	$CaCO_3 + HCl$

5. Displacement vs Double Displacement — At a Glance

• Displacement: A free element appears ($Fe + CuSO_4$)
• Double Displacement: All are compounds ($BaCl_2 + Na_2SO_4$)

6. Repeatedly Asked Board Questions

1. What is a double displacement reaction? Explain Activity 1.10.
2. What is a precipitation reaction? Give an example.
3. What is the difference between displacement and double displacement?
4. Write the reaction of lead nitrate + potassium iodide. Name and state the colour of the precipitate.
5. Give one example of a neutralisation reaction.

7. A Trick

• When a question asks about precipitate colour, mentally run through the table above — Ba/Ag/Pb-Cl white, Ag/PbI yellow, $Cu(OH)_2$ blue, $Fe(OH)_3$ brown.

The Bottom Line: Two pairs swap partners → double displacement. A solid drops out → precipitation.

Solved Examples

Example 1: NCERT — Activity 1.10

What are the observations when sodium sulphate solution is added to barium chloride solution? Write the balanced reaction.

Solution:

Observations:

• A white precipitate is formed instantly.
• The precipitate is insoluble in water.

Reaction:

$$Na_2SO_4(aq) + BaCl_2(aq) \rightarrow BaSO_4(s)\downarrow + 2NaCl(aq)$$

Explanation:

1. Before mixing — the solutions contain $Na^+, SO_4^{2-}, Ba^{2+}, Cl^-$ ions.
2. $Ba^{2+}$ and $SO_4^{2-}$ combine to form insoluble $BaSO_4$ (white).
3. $Na^+$ and $Cl^-$ remain in solution as soluble $NaCl$.
4. Type of reaction:

• Double displacement — ions swap partners.
• Precipitation — white $BaSO_4$ is formed.

Name of the precipitate: Barium sulphate ($BaSO_4$).

[NCERT textbook experiment — frequently asked]

Example 2: NCERT — Lead Nitrate and Potassium Iodide

In Activity 1.2, you mixed solutions of lead (II) nitrate and potassium iodide. (i) What was the colour of the precipitate? Name the compound formed. (ii) Write the balanced chemical equation for this reaction. (iii) Is this a double displacement reaction?

Solution:

(i) Colour of the precipitate — bright yellow. The compound formed is lead (II) iodide ($PbI_2$).

(ii) Balanced equation:

$$Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s)\downarrow + 2KNO_3(aq)$$

(iii) Is this a double displacement reaction?

Yes — it is clearly a double displacement reaction, because —

• $Pb^{2+}$ and $K^+$ ions swap their 'partners' (counter-ions).
• After the reaction, $Pb^{2+}$ is paired with $I^-$ and $K^+$ with $NO_3^-$.
• All reactants and products are compounds (no free element) — so it is double displacement, not displacement.

Bonus: The bright yellow $PbI_2$ is famous — that is why this experiment is also called the 'Golden Rain'.

[Board Important — 3-mark question]

Example 3: Definition of Precipitation Reaction

What do you understand by a precipitation reaction? Explain with an example.

Solution:

Definition: A reaction in which mixing two solutions forms an insoluble solid (a precipitate) is called a precipitation reaction.

Features:

1. It is always a kind of double displacement reaction.
2. One product must be an insoluble solid that settles at the bottom of the solution.
3. The precipitate is shown with a $\downarrow$ symbol.

Two examples:

(1) Formation of BaSO₄:

$$Na_2SO_4(aq) + BaCl_2(aq) \rightarrow BaSO_4(s)\downarrow + 2NaCl(aq)$$

(2) Formation of AgCl:

$$AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s)\downarrow + NaNO_3(aq)$$

Daily-life uses:

• Identifying ions in the laboratory.
• Water purification.
• Some diseases (e.g., kidney stones).

[Board Important — 3-mark question]

Example 4: NCERT — Sodium Hydroxide and Hydrochloric Acid

A solution of sodium hydroxide reacts with hydrochloric acid solution to form sodium chloride solution and water. Write the balanced equation. What kind of reaction is this?

Solution:

1. Reaction:

$$NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$$

2. Type of reaction:

• Neutralisation reaction — acid and base combine to form salt and water.
• A special case of double displacement — Na and H exchanged 'partners' (Na took Cl, H took OH → H₂O).

2. At the ionic level:

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

This is the essential ionic form of any neutralisation.

4. Visible observations:

• No precipitate (NaCl is soluble).
• Slight rise in temperature (neutralisation is exothermic).

4. Daily life: Antacids like $NaHCO_3$ neutralise stomach $HCl$ in this same way.

[NCERT textbook question]

Example 5: Which Are Double Displacement Reactions?

Which of the following are double displacement reactions? (i) $Zn + CuSO_4 \rightarrow ZnSO_4 + Cu$ (ii) $AgNO_3 + KI \rightarrow AgI + KNO_3$ (iii) $2Mg + O_2 \rightarrow 2MgO$ (iv) $H_2SO_4 + 2KOH \rightarrow K_2SO_4 + 2H_2O$ (v) $Pb(NO_3)_2 + 2NaCl \rightarrow PbCl_2 + 2NaNO_3$

Solution:

Quick rule: If a reaction contains a free element (on either side) → displacement or combination. If everything is a compound → double displacement.

#	Reaction	Type	Reason
(i)	$Zn + CuSO_4$	Displacement	Zn is a free element
(ii)	$AgNO_3 + KI$	Double displacement ✓	All compounds; ion exchange
(iii)	$Mg + O_2$	Combination	Free element, single product
(iv)	$H_2SO_4 + KOH$	Double displacement (neutralisation) ✓	Acid + base
(v)	$Pb(NO_3)_2 + NaCl$	Double displacement (precipitation) ✓	$PbCl_2$ precipitate

Answer: (ii), (iv), and (v) are double displacement reactions.

Example 6: Identification of Silver Chloride

A student added $NaCl$ solution to $AgNO_3$ solution. What are the observations? Write the reaction.

Solution:

1. Reaction:

$$AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s)\downarrow + NaNO_3(aq)$$

2. Observations:

• A white milky precipitate appears at once — silver chloride ($AgCl$).
• It settles at the bottom of the solution.

2. Special test:

• If this white precipitate is exposed to sunlight, it turns grey/black (photochemical decomposition — see Section 4):

$$2AgCl(s) \xrightarrow{\text{sunlight}} 2Ag(s) + Cl_2(g)$$

4. Type of reaction:

• Double displacement (Ag and Na swapped partners)
• Precipitation (white $AgCl$ precipitate)

4. Use: This is the easiest test for the $Cl^-$ ion.

Lesson: A famous identification test — add $AgNO_3$ to an unknown solution and see whether a white precipitate forms.

Example 7: Numerical — Mass of Precipitate

From the complete reaction of $BaCl_2$ and $Na_2SO_4$, 23.3 g of $BaSO_4$ precipitate is formed. Find the mass of $BaCl_2$ used. ($BaCl_2 = 208$, $BaSO_4 = 233$)

Solution:

1. Reaction:

$$BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s)\downarrow + 2NaCl(aq)$$

2. Mole ratio: 1 mol $BaCl_2$ → 1 mol $BaSO_4$
3. Mass ratio: 208 g $BaCl_2$ → 233 g $BaSO_4$
4. Given: $BaSO_4 = 23.3$ g
5. Calculation:

$$\text{Mass of BaCl}_2 = \frac{208}{233} \times 23.3 = 20.8 \text{ g}$$

6. Answer: Mass of $BaCl_2$ used = 20.8 g.

Lesson: In stoichiometry, the mass of a precipitate ties directly to the mole ratio.

Example 8: NCERT — One Question, Three Reactions

Write balanced equations for the following and identify the type of each reaction. (a) Potassium bromide (aq) + Barium iodide (aq) → Potassium iodide (aq) + Barium bromide (s) (b) Zinc carbonate (s) → Zinc oxide (s) + Carbon dioxide (g) (c) Hydrogen (g) + Chlorine (g) → Hydrogen chloride (g) (d) Magnesium (s) + Hydrochloric acid (aq) → Magnesium chloride (aq) + Hydrogen (g)

Solution:

(a): $\quad 2KBr(aq) + BaI_2(aq) \rightarrow 2KI(aq) + BaBr_2(s)$

• Type: Double displacement (precipitation) reaction.

(b): $\quad ZnCO_3(s) \xrightarrow{\Delta} ZnO(s) + CO_2(g)$

• Type: Decomposition (thermal) reaction.

(c): $\quad H_2(g) + Cl_2(g) \rightarrow 2HCl(g)$

• Type: Combination reaction.

(d): $\quad Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)\uparrow$

• Type: Displacement reaction (Mg displaced H).

[NCERT textbook question — frequently asked in board]

Example 9: Copper Sulphate and Sodium Hydroxide

What happens when sodium hydroxide solution is added to copper sulphate solution? Write the reaction.

Solution:

1. Reaction:

$$CuSO_4(aq) + 2NaOH(aq) \rightarrow Cu(OH)_2(s)\downarrow + Na_2SO_4(aq)$$

2. Observation:

• A blue precipitate is formed — copper hydroxide.

2. Type of reaction:

• Double displacement — Cu and Na swapped partners.
• Precipitation — blue $Cu(OH)_2$.

2. Analysis:

• Initial solution had $Cu^{2+}, SO_4^{2-}, Na^+, OH^-$ ions.
• $Cu^{2+}$ and $OH^-$ combine to form insoluble $Cu(OH)_2$.
• $Na^+$ and $SO_4^{2-}$ remain in solution as $Na_2SO_4$.

2. Special note: This is one method of testing for a base — $Cu^{2+}$ ion is identified by adding $NaOH$ (blue precipitate).

Example 10: The Lime-Water Test

Lime water turns milky on passing a particular gas through it. Which gas is it? Write the reaction.

Solution:

1. Identification of the gas: Carbon dioxide ($CO_2$).
2. Reaction:

$$Ca(OH)_2(aq) + CO_2(g) \rightarrow CaCO_3(s)\downarrow + H_2O(l)$$

3. Type of reaction:

• Double displacement (precipitation) — white $CaCO_3$ precipitate.

3. Why milky? $CaCO_3$ is insoluble — it disperses as fine particles in the solution, making it appear 'milky' (turbid).
4. Curious fact — passing more $CO_2$:

• With excess $CO_2$, the precipitate dissolves to form $Ca(HCO_3)_2$ (calcium hydrogen carbonate, which is soluble):

$$CaCO_3(s) + CO_2(g) + H_2O(l) \rightarrow Ca(HCO_3)_2(aq)$$

(This reaction explains the formation of stalactites and stalagmites in caves — Class 11 level reference.)

6. Use: This is the most famous test for $CO_2$ — for example, to verify the gas exhaled during respiration.

[Board Important — 3-mark question]

Example 11: A Neutralisation Example

Doctors prescribe magnesium hydroxide (milk of magnesia) to neutralise stomach acid. Write its reaction with $HCl$.

Solution:

1. Reaction:

$$Mg(OH)_2(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + 2H_2O(l)$$

2. Type of reaction:

• Neutralisation reaction — base ($Mg(OH)_2$) + acid ($HCl$) → salt ($MgCl_2$) + water.
• A special case of double displacement.

2. Daily-life use:

• Treating acidity: Milk of magnesia is a popular antacid.
• When the stomach produces excess $HCl$ (heartburn/discomfort), $Mg(OH)_2$ neutralises it.

2. Other famous antacids:

• Baking soda ($NaHCO_3$) — $NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$
• Calcium carbonate ($CaCO_3$) — found in tablets.

2. Note: This is a preview of Class 10 Chapter 2 (Acids, Bases and Salts).

[Board + Daily-Life]

Example 12: A Challenging Question — Identification by Precipitate Colour

A lab has four solutions — $NaCl, Na_2SO_4, KI, KNO_3$. How would you identify each solution based only on the colour of the precipitate formed?

Solution:

Reactions and precipitates:

1. To identify $NaCl$ — add $AgNO_3$:

$$AgNO_3 + NaCl \rightarrow AgCl(s)\downarrow + NaNO_3$$

White milky precipitate.

2. To identify $Na_2SO_4$ — add $BaCl_2$:

$$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4(s)\downarrow + 2NaCl$$

White precipitate (insoluble in acid — different from $AgCl$).

3. To identify $KI$ — add $Pb(NO_3)_2$:

$$Pb(NO_3)_2 + 2KI \rightarrow PbI_2(s)\downarrow + 2KNO_3$$

Bright yellow precipitate — most distinctive.

4. To identify $KNO_3$ — no common reagent gives a precipitate. So the absence of a precipitate is itself the identifier for $KNO_3$.

Summary table:

Solution	Reagent	Precipitate	Colour
NaCl	$AgNO_3$	$AgCl$	White
$Na_2SO_4$	$BaCl_2$	$BaSO_4$	White
KI	$Pb(NO_3)_2$	$PbI_2$	Yellow
$KNO_3$	(none)	—	—

[Lab-based question — Board Important]

Example 13: NCERT — A Concise Question

Give an example of a double displacement reaction other than the one in Activity 1.10.

Solution:

Activity 1.10 was: $Na_2SO_4 + BaCl_2 \rightarrow BaSO_4\downarrow + 2NaCl$

Other examples:

(1) Silver nitrate + sodium chloride:

$$AgNO_3(aq) + NaCl(aq) \rightarrow AgCl(s)\downarrow + NaNO_3(aq)$$

White $AgCl$ precipitate — test for $Cl^-$ ions.

(2) Lead nitrate + potassium iodide:

$$Pb(NO_3)_2(aq) + 2KI(aq) \rightarrow PbI_2(s)\downarrow + 2KNO_3(aq)$$

Bright yellow $PbI_2$ precipitate.

(3) Copper sulphate + sodium hydroxide:

$$CuSO_4(aq) + 2NaOH(aq) \rightarrow Cu(OH)_2(s)\downarrow + Na_2SO_4(aq)$$

Blue $Cu(OH)_2$ precipitate.

(4) Sodium hydroxide + hydrochloric acid:

$$NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)$$

Neutralisation — no precipitate, but it is double displacement.

[NCERT textbook question]

Example 14: Numerical — Silver Chloride Precipitate

How much $AgCl$ precipitate will be formed by reacting 170 g of $AgNO_3$ with excess $NaCl$ solution? ($AgNO_3 = 170$, $AgCl = 143.5$)

Solution:

1. Reaction:

$$AgNO_3 + NaCl \rightarrow AgCl + NaNO_3$$

2. Mole ratio: 1 mol $AgNO_3$ → 1 mol $AgCl$

3. Mass ratio: 170 g $AgNO_3$ → 143.5 g $AgCl$

4. Given: 170 g $AgNO_3$

5. Calculation: Direct from the ratio — $AgCl = 143.5$ g

6. Verification: Since 170 g = 1 mol of AgNO₃, and per mole 143.5 g of AgCl is produced.

Answer: 143.5 g of silver chloride precipitate will be formed.

Lesson: In a 1:1 stoichiometry, the ratios are immediate from the molar masses — no calculator needed.

Example 15: Sodium Carbonate and Hydrochloric Acid

What happens when $HCl$ is added to a sodium carbonate solution? Write the reaction. What kind of reaction is it?

Solution:

1. Reaction:

$$Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)\uparrow$$

2. Observations:

• Brisk effervescence ($CO_2$ gas).
• If passed through lime water, it turns milky.

2. Type of reaction:

• Double displacement — Na and H swapped partners.
• Gas evolution — one product ($CO_2$) leaves as gas.

2. Analysis:

• First — $Na_2CO_3 + HCl \rightarrow NaCl + NaHCO_3$ (with limited acid).
• Then — $NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2$.
• Combined as the equation above.

2. Daily life:

• Washing soda: $Na_2CO_3$ is used in laundry.
• Food industry: Baking soda/$Na_2CO_3$ has many uses.

2. Identification: This reaction is the famous test for the carbonate ion ($CO_3^{2-}$).

[Board Important]

Example 16: A Tabular Question — Identifying All Reaction Types

Classify the following equations into combination, decomposition, displacement, or double displacement. Give a reason for each —

(a) $4Na + O_2 \rightarrow 2Na_2O$ (b) $2KClO_3 \xrightarrow{\Delta} 2KCl + 3O_2$ (c) $Mg + 2HCl \rightarrow MgCl_2 + H_2$ (d) $AgNO_3 + KI \rightarrow AgI + KNO_3$ (e) $CaO + H_2O \rightarrow Ca(OH)_2$ (f) $2HgO \xrightarrow{\Delta} 2Hg + O_2$

Solution:

#	Reaction	Type	Reason
(a)	$4Na + O_2 \rightarrow 2Na_2O$	Combination	Two reactants → one product
(b)	$2KClO_3 \rightarrow 2KCl + 3O_2$	Decomposition	One → two products
(c)	$Mg + HCl \rightarrow MgCl_2 + H_2$	Displacement	Mg displaced H
(d)	$AgNO_3 + KI \rightarrow AgI + KNO_3$	Double displacement (precipitation)	Ion exchange; yellow AgI precipitate
(e)	$CaO + H_2O \rightarrow Ca(OH)_2$	Combination	Two → one
(f)	$2HgO \rightarrow 2Hg + O_2$	Decomposition (thermal)	One → two

Summary table — five types of reactions covered so far:

Type	Form	Famous Example
Combination	$A + B \rightarrow AB$	$CaO + H_2O \rightarrow Ca(OH)_2$
Decomposition	$AB \rightarrow A + B$	$CaCO_3 \rightarrow CaO + CO_2$
Displacement	$A + BC \rightarrow AC + B$	$Fe + CuSO_4 \rightarrow FeSO_4 + Cu$
Double displacement	$AB + CD \rightarrow AD + CB$	$BaCl_2 + Na_2SO_4 \rightarrow BaSO_4 + 2NaCl$
Neutralisation	Acid + Base → Salt + Water	$HCl + NaOH \rightarrow NaCl + H_2O$

[Board exam 5-mark question]