Chapter 2: Acids, Bases and Salts

Given data

• Three colourless solutions in separate test-tubes: one is distilled water, one is an acid, one is a base.
• Only red litmus paper is available.

Required  Identify which test-tube contains which substance without mixing the solutions.

Concept used

• Litmus is a natural indicator.
• Action of litmus papers:

Step 1  — Test with red litmus in all three tubes

1. Cut the red litmus paper into three small strips so that the supply is not exhausted.
2. Dip one strip into each test-tube and observe.

Observation:

• In one particular tube the red litmus strip turns blue. Mark this tube as T_B.
• In the remaining two tubes the red litmus strips show no colour change. Mark them temporarily as T₁ and T₂.

Inference:

• The conversion $$\text{red}\;\longrightarrow\;\text{blue}$$ can occur only in a basic medium. Hence T_B contains the basic solution.

Step 2  — Convert one end of the same litmus into blue litmus

The strip removed from T_B is now blue. Rinse it gently with distilled water to remove traces of the base and dry it on filter paper. The strip now has two coloured portions:

• Original red part (untouched end)
• Newly formed blue part (end dipped in base)

This single strip therefore supplies us with both red and blue litmus.

Step 3  — Test the two unknown tubes with the blue end

1. Dip the blue end first into T₁ and observe.
2. Repeat with T₂.

Possibilities:

• If the blue end changes back to red in one of them, that tube is acidic, because $$\text{blue}\;\longrightarrow\;\text{red}$$ occurs only in acids.
• The tube in which even the blue litmus shows no change is neutral, that is, it contains distilled water.

Step 4  — Summarise the identification

• T_B (turned red to blue first)  →  Basic solution.
• Among T₁ and T₂:
• Tube turning blue litmus red  →  Acidic solution.
• Tube producing no change with either colour  →  Distilled water (neutral).

Why the method works

The results rely on the indicator property of litmus, which itself is a weak acid–base dye. In an alkaline medium the excess $$\mathrm{OH^-}$$ ions react with the acidic form of litmus converting it to its blue basic form. Conversely, the acidic medium provides excess $$\mathrm{H^+}$$ ions that protonate the dye, restoring the red colour.

Thus the three solutions are unambiguously identified using only red litmus paper.

Step 1 — Identify the nature of curd / sour food

• Curd contains the weak organic acid lactic acid, $$\mathrm{CH_3CH(OH)COOH}$$.
• Lemon juice, tamarind, pickles, etc. contain other organic acids such as $$\mathrm{C_6H_8O_7}$$ (citric acid) or $$\mathrm{C_4H_6O_6}$$ (tartaric acid).
• Because of their acids, these foods have pH < 7 and behave as acidic substances.

Step 2 — Identify what brass and copper supply

• Pure copper vessels are made of the metal $$\mathrm{Cu}$$.
• Brass is an alloy mainly of copper and zinc; its “active” metals are therefore $$\mathrm{Cu}$$ and $$\mathrm{Zn}$$ at the surface.

Step 3 — Write the acid–metal reactions that can occur

The organic acid present in the food slowly attacks the metal surface. Representative equations are:

• Copper (slow, needs aeration): $$\mathrm{Cu + 2\,H^+ \;\longrightarrow \; Cu^{2+} + H_2\uparrow}$$
• Zinc (readily): $$\mathrm{Zn + 2\,H^+ \;\longrightarrow \; Zn^{2+} + H_2\uparrow}$$
• The metal ions combine with the organic anion of the food acid. For curd (lactic acid):
  $$\mathrm{Cu^{2+} + 2\,CH_3CH(OH)COO^- \;\longrightarrow \; [CH_3CH(OH)COO]_2Cu}$$ (copper lactate)
• For lemon juice / tamarind (citric acid), the corresponding salt is copper citrate; with the zinc of brass, zinc lactate or zinc citrate is formed.

Step 4 — Explain the consequence

• The salts so produced — copper lactate, copper citrate, zinc lactate, zinc citrate, etc. — are soluble in the food.
• Many soluble copper and zinc salts are toxic and may cause food-poisoning, vomiting and other health problems when the food is eaten.
• The food also acquires an unpleasant metallic taste and its quality is spoiled.

Conclusion

Because the organic acids present in curd and other sour substances slowly corrode copper (and the copper/zinc surface of brass), forming poisonous soluble salts such as copper lactate and copper citrate, such foods must never be stored in brass or copper containers.

Step 1 – Identify the gas
When a metal reacts with a dilute acid, the products are a salt and a gas. The gas set free is hydrogen.

Step 2 – Write the general equation
$$\mathrm{Metal + Acid \;\longrightarrow\; Salt + H_2\,(g)}$$

Step 3 – Illustrate with an example
Take a few zinc granules in a test tube and add dilute hydrochloric acid. Bubbles appear, showing a gas is released.
Chemical equation (balanced):
$$\mathrm{Zn + 2HCl \;\longrightarrow\; ZnCl_2 + H_2\,(g)}$$

Step 4 – Test for the evolved gas

• Collect some of the gas in an inverted test tube or bring a burning splint close to the mouth of the reaction tube.
• If the gas is hydrogen, it burns with a small ‘pop’ sound because of its explosive nature in air.

Inference: The ‘pop’ confirms that the liberated gas is $$\mathrm{H_2}$$.

Step 1 | Identify the gas

The question states that a gas is evolved which extinguishes a burning candle. A well-known laboratory test is that carbon dioxide extinguishes flames (while hydrogen burns with a ‘pop’ and oxygen rekindles a glowing splint). Hence

$$\text{Evolved gas} = \mathrm{CO_2}$$

Step 2 | Deduce the type of metal compound A

• Effervescence with a dilute acid and liberation of $$\mathrm{CO_2}$$ normally points to a carbonate or hydrogencarbonate.
• Because a product mentioned in the question is calcium chloride $$\left(\mathrm{CaCl_2}\right)$$, the metal in compound A must be calcium.

Hence compound A is a calcium carbonate:

$$\mathrm{A = CaCO_3}$$

Step 3 | Write the word equation first

calcium carbonate + hydrochloric acid → calcium chloride + water + carbon dioxide

Step 4 | Convert to a symbolic equation

$$\mathrm{CaCO_3 + HCl \rightarrow CaCl_2 + H_2O + CO_2}$$

Step 5 | Balance the equation

Hydrogen and chlorine are not yet balanced. Put the coefficient 2 before $$\mathrm{HCl}$$:

$$\mathrm{CaCO_3 + 2\,HCl \rightarrow CaCl_2 + H_2O + CO_2}$$

Now every element has the same count on both sides:

• Ca = 1 : 1
• C = 1 : 1
• O = 3 : 3
• H = 2 : 2
• Cl = 2 : 2

The equation is therefore balanced.

Step 6 | State the final balanced chemical equation

$$\mathrm{CaCO_3 + 2\,HCl \rightarrow CaCl_2 + H_2O + CO_2 \uparrow}$$

(The upward arrow indicates the liberation of the gas.)

All common tests for an acid – the blue litmus test, reaction with metals, evolution of $$\mathrm{CO_2}$$ from metal carbonates, conduction of electricity, etc. – depend on the presence of free hydrogen ions (more precisely, hydronium ions $$\mathrm{H_3O^+}$$) in the aqueous solution.

1  Release of hydrogen ions by mineral acids

When hydrogen chloride or nitric acid is dissolved in water, the molecules actually break up (ionise) :

$$\mathrm{HCl(aq)} \;\longrightarrow\; \mathrm{H^+(aq)} + \mathrm{Cl^-(aq)}$$
$$\mathrm{HNO_3(aq)} \;\longrightarrow\; \mathrm{H^+(aq)} + \mathrm{NO_3^-(aq)}$$

The $$\mathrm{H^+}$$ ions immediately combine with water to give hydronium ions: $$\mathrm{H^+ + H_2O \;\to\; H_3O^+}$$. It is these $$\mathrm{H_3O^+}$$ ions that impart all the acidic properties to the solution.

2  No ionisation of alcohol or glucose

Alcohols (e.g. ethanol $$\mathrm{C_2H_5OH}$$) and glucose $$\mathrm{C_6H_{12}O_6}$$ also contain hydrogen atoms, but merely having hydrogen in the formula does not make a substance acidic. In water they stay as intact neutral molecules:

$$\mathrm{C_2H_5OH(aq)} \;\rightleftharpoons\; \mathrm{C_2H_5OH(aq)}$$
$$\mathrm{C_6H_{12}O_6(aq)} \;\rightleftharpoons\; \mathrm{C_6H_{12}O_6(aq)}$$

The O–H bond present in these organic molecules is covalent and non-ionisable under ordinary aqueous conditions; therefore no $$\mathrm{H^+}$$ / $$\mathrm{H_3O^+}$$ ions appear in the solution, so the solution does not turn blue litmus red, does not react with metals, and so on.

3  Conclusion

An aqueous solution shows acidic character only if the solute furnishes free $$\mathrm{H^+}$$ (hydronium) ions. Mineral acids like $$\mathrm{HCl}$$ and $$\mathrm{HNO_3}$$ ionise to give such ions, whereas compounds such as alcohol and glucose do not, hence they are not acidic in water.

In solids, electricity is carried by electrons. In solutions, electricity can pass only if there are free ions that can move between the electrodes.

Pure (anhydrous) acids contain neutral molecules, so they cannot conduct. However, when an acid is mixed with water, each acid molecule ionises (dissociates) according to a reaction such as

$$\mathrm{HCl\; +\; H_2O \; \longrightarrow \; H_3O^+ \; + \; Cl^-}$$

The aqueous solution therefore contains

• positive hydronium ions $$\mathrm{H_3O^+}$$ (often written simply as $$\mathrm{H^+}$$), and
• negative acid-radical ions such as $$\mathrm{Cl^-},\; SO_4^{2-},\; NO_3^-$$, etc.

When the two electrodes of an external circuit are dipped into this solution and a potential difference is applied, these ions migrate:

• $$\mathrm{H_3O^+}$$ (or $$\mathrm{H^+}$$) moves towards the cathode (negative electrode).
• The corresponding anion (e.g. $$\mathrm{Cl^-}$$) moves towards the anode (positive electrode).

The movement of these charged particles constitutes an electric current through the solution, so the aqueous solution of an acid conducts electricity.

Thus, ionisation in water → formation of mobile ions → conduction of electricity.

Concept involved – Arrhenius definition of an acid

An Arrhenius acid is a substance that furnishes hydrogen ions $$\mathrm{(H^+)}$$ (or equivalently hydronium ions $$\mathrm{(H_3O^+)}$$) only when it is in aqueous solution.

Step 1 – Nature of hydrogen chloride gas

Hydrogen chloride in the gaseous state is a covalent molecule. In the absence of water it does not split into ions:

$$\mathrm{HCl_{(g)} \; \xrightarrow[no\;water]{}\; HCl_{(g)}}$$

Step 2 – Why litmus changes colour

Litmus paper changes colour only when it comes in contact with a solution that contains either $$\mathrm{H^+}$$ ions (acid) or $$\mathrm{OH^-}$$ ions (base). These ions interact with the dye molecules present in litmus and bring about the colour change.

Step 3 – Condition with dry $$\mathrm{HCl}$$ and dry litmus

• Because both the gas and the litmus paper are dry, no water is present.
• Without water, $$\mathrm{HCl_{(g)}}$$ cannot ionise to give $$\mathrm{H^+}$$/$$\mathrm{H_3O^+}$$ ions:

$$\mathrm{HCl_{(g)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)}}\quad\text{(does not occur here)}$$

Step 4 – Result

Since no $$\mathrm{H^+}$$ ions are produced, the litmus dye experiences no change in the chemical environment, and its colour remains the same.

Conclusion

Dry $$\mathrm{HCl}$$ gas does not change the colour of dry litmus paper because the absence of water prevents the formation of hydrogen (hydronium) ions that are essential for the acidic behaviour detected by litmus.

Step 1 – Recall the nature of the dilution process
When a concentrated acid such as $$\mathrm{H_2SO_4}$$, $$\mathrm{HCl}$$ or $$\mathrm{HNO_3}$$ is mixed with water, the dissolution is highly exothermic; i.e. it releases a large amount of heat.

Step 2 – Compare the two possible procedures

• Correct method: add concentrated acid slowly into a beaker containing plenty of water, while stirring continuously.
• Wrong method: pour water into a container of concentrated acid.

Step 3 – Explain why the correct method is safe
When a few drops of acid enter a large volume of water, the heat produced at the contact point is absorbed quickly by the surrounding water. The temperature rise in any one spot is therefore small, so the mixture remains below its boiling point, and no liquid splashes out.

Step 4 – Explain the danger in the wrong method
If water is added into concentrated acid, the first drops of water get surrounded by acid. The heat released cannot dissipate fast enough, so the thin layer of water boils instantly, converting to steam. The sudden expansion of steam can violently eject the acid mixture from the container, leading to spattering, burns and damage.

Step 5 – Conclusion
Therefore, to control the heat safely, one must always add acid to water and never the reverse.

Step 1 - Recall how acids produce hydronium ions
An aqueous acid donates protons to water:
$$\mathrm{HA + H_2O \;\longrightarrow\; H_3O^+ + A^-}$$
Every formula unit of the acid that dissociates yields one $$\mathrm{H_3O^+}$$ ion (for a monoprotic acid).

Step 2 - Define concentration
The molar concentration of hydronium ions is the amount (in moles) of $$\mathrm{H_3O^+}$$ present per unit volume of the solution:
$$C = \dfrac{n(\mathrm{H_3O^+})}{V}$$
where

• $$n(\mathrm{H_3O^+})$$ = number of moles of hydronium ions present
• $$V$$ = total volume of the solution

Step 3 - What does dilution mean mathematically?
When water is added, the amount of hydronium ions already present remains essentially the same (no extra acid or base is added), so $$n(\mathrm{H_3O^+})$$ is nearly constant.
Let the initial volume be $$V_{1}$$ and the added water volume be $$\Delta V$$.
The new total volume becomes
$$V_{2}=V_{1}+\Delta V$$.

Step 4 - Compare the two concentrations
Initial concentration:
$$C_{1}=\dfrac{n(\mathrm{H_3O^+})}{V_{1}}$$
Concentration after dilution:
$$C_{2}=\dfrac{n(\mathrm{H_3O^+})}{V_{2}}=\dfrac{n(\mathrm{H_3O^+})}{V_{1}+\Delta V}$$
Adding water increases the denominator (the total volume) while the numerator (moles of $$\mathrm{H_3O^+}$$) stays the same. Hence,
$$V_{2}\;\gt\;V_{1} \;\;\Longrightarrow\;\; \dfrac{n(\mathrm{H_3O^+})}{V_{2}}\;\lt\;\dfrac{n(\mathrm{H_3O^+})}{V_{1}}$$
that is,
$$\boxed{\;C_{2}\;\lt\;C_{1}\;}$$
The hydronium-ion concentration after dilution is strictly less than the concentration before dilution.

Step 5 - Conclusion
Diluting the acidic solution decreases the concentration of hydronium ions ($$C_{2}\;\lt\;C_{1}$$); the greater the amount of water added, the lower the $$[\mathrm{H_3O^+}]$$. (However, even after very heavy dilution the solution remains slightly acidic; $$[\mathrm{H_3O^+}]$$ approaches, but never falls below, $$10^{-7}\,\mathrm{mol\,L^{-1}}$$ at 25 °C because of the auto-ionisation of water.)

Step 1 : Understand what “excess base” means
In NCERT language a “base” like $$\mathrm{NaOH}$$ is a strong base; it dissociates completely in water.

$$\mathrm{NaOH_{(aq)} \;\longrightarrow\; Na^+_{(aq)} + OH^-_{(aq)}}$$

Step 2 : Relate amount dissolved to ion concentration
Because the dissociation is 100 % complete, each formula unit of $$\mathrm{NaOH}$$ added furnishes exactly one hydroxide ion:

For every 1 mol $$\mathrm{NaOH}$$  → 1 mol $$\mathrm{OH^-}$$

Step 3 : What happens when still more base is added?
If an already basic solution gets “excess” (more) $$\mathrm{NaOH}$$, still more hydroxide ions are produced, so the numerical value of $$[\mathrm{OH^-}]$$ rises. (Only when the solution becomes saturated would solid remain undissolved, but in ordinary laboratory concentrations saturation is not reached.)

Step 4 : Effect on hydroxide-ion concentration
Therefore,

$$\text{Excess base added} \;\Longrightarrow\; \uparrow\,[\mathrm{OH^-}]$$

i.e. the concentration of hydroxide ions increases.

Given data

• Solution A : pH = 6
• Solution B : pH = 8

Step 1 – Recall the definition of pH

The pH of any aqueous solution is related to its hydrogen-ion concentration by

$$\text{pH}= -\log_{10}[\mathrm H^+]$$

Rearranging, the concentration of hydrogen ions is

$$[\mathrm H^+]=10^{-\text{pH}}\;(\text{mol\ L}^{-1}).$$

Step 2 – Calculate the [H⁺] of each solution

For solution A:

$$[\mathrm H^+]_A = 10^{-6}\;\text{mol\ L}^{-1}$$

For solution B:

$$[\mathrm H^+]_B = 10^{-8}\;\text{mol\ L}^{-1}$$

Step 3 – Compare the hydrogen-ion concentrations

Because $$10^{-6} \gt 10^{-8}$$, solution A has the larger hydrogen-ion concentration. In fact,

$$\frac{[\mathrm H^+]_A}{[\mathrm H^+]_B}=\frac{10^{-6}}{10^{-8}}=10^{2}=100,$$

so A is 100 times more concentrated in $$\mathrm H^+$$ than B.

Step 4 – Classify each solution

• If pH < 7, the solution is acidic.
• If pH > 7, the solution is basic (alkaline).

Therefore:

• Solution A (pH 6 < 7) is acidic.
• Solution B (pH 8 > 7) is basic.

Conclusion

Solution A has the higher hydrogen-ion concentration and is acidic, whereas solution B has the lower hydrogen-ion concentration and is basic.

Step 1 | Recall the pH scale
The pH of a solution is defined as $$\text{pH}= -\log_{10}[\mathrm{H^+(aq)}]$$ where $$[\mathrm{H^+(aq)}]$$ is the molar concentration (in mol L⁻¹) of hydrogen ions in the solution.

Step 2 | Relate [H⁺] to the nature of the solution

• When $$[\mathrm{H^+(aq)}]$$ is high, the numerical value of pH becomes small (pH < 7). Such solutions are called acidic.
• When $$[\mathrm{H^+(aq)}]$$ is exactly $$1.0 \times 10^{-7}\ \text{mol L}^{-1}$$, the pH equals 7 and the solution is neutral.
• When $$[\mathrm{H^+(aq)}]$$ is low (or, equivalently, $$[\mathrm{OH^-(aq)}]$$ is high), the pH becomes > 7 and the solution is basic/alkaline.

Step 3 | Conclude the effect
Therefore, as the concentration of hydrogen ions in an aqueous solution increases, its acidic character increases (pH decreases). Conversely, if the concentration of hydrogen ions decreases, the solution becomes less acidic and may turn neutral or basic depending on how low the concentration becomes.

The dissociation (self-ionisation) of water takes place in every aqueous solution:

$$\mathrm{H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)}$$

In pure water at 25 °C the concentrations are equal,

$$[H^+] = [OH^-] = 1.0 \times 10^{-7}\;\mathrm{mol\,L^{-1}}$$

so the ionic product is

$$K_w = [H^+][OH^-] = 1.0 \times 10^{-14}$$

When a base such as $$\mathrm{NaOH}$$ is dissolved, it furnishes extra hydroxide ions:

$$\mathrm{NaOH(s) \xrightarrow{aq} Na^+(aq) + OH^-(aq)}$$

The added $$OH^-$$ ions increase the value of $$[OH^-]$$. To keep $$K_w$$ constant, $$[H^+]$$ must decrease:

$$[H^+] = \dfrac{K_w}{[OH^-]} \;\;(\text{from } K_w = [H^+][OH^-])$$

Thus a basic solution still contains some $$H^+(aq)$$ ions, but

• $$[OH^-] \gt [H^+]$$, therefore its pH > 7, and the solution shows basic (alkaline) properties.

In short, yes, $$H^+$$ ions are always present, yet the excess of $$OH^-$$ ions makes the solution basic.

Concept involved – Soil pH and neutralisation

• For healthy growth most crop plants need the soil pH to be nearly neutral (about 6.5 – 7.5).
• Sometimes soil becomes acidic (pH < 7) because of factors such as acid rain, over-use of ammonium fertilisers, leaching of basic ions, etc.

Why quick lime, slaked lime or chalk are chosen

• Quick lime : $$\mathrm{CaO}$$ (basic oxide)
• Slaked lime : $$\mathrm{Ca(OH)_2}$$ (strong base)
• Chalk : $$\mathrm{CaCO_3}$$ (weakly basic carbonate)

All three substances are alkaline; they supply $$\mathrm{O^{2-}}$$, $$\mathrm{OH^-}$$ or $$\mathrm{CO_3^{2-}}$$ ions that can neutralise the excess hydrogen ions present in an acidic soil.

Neutralisation reactions (representative)

For slaked lime:
$$\mathrm{Ca(OH)_2 + 2H^+ \;\longrightarrow\; Ca^{2+} + 2H_2O}$$

For chalk:
$$\mathrm{CaCO_3 + 2H^+ \;\longrightarrow\; Ca^{2+} + H_2O + CO_2\uparrow}$$

The hydrogen ions $$\mathrm{(H^+)}$$ responsible for acidity are removed, so the pH of the soil moves towards 7.

Conclusion

A farmer spreads quick lime, slaked lime or chalk on the field when the soil has turned acidic so that these basic materials can neutralise the excess acidity and restore the pH to a range suitable for crop growth.

The given formula is $$\mathrm{Ca(ClO)_2}$$.

1. In the ion $$\mathrm{ClO^-}$$ the oxidation state of chlorine is +1, so the ion is called the hypochlorite ion.

2. Because calcium forms the divalent cation $$\mathrm{Ca^{2+}}$$, two hypochlorite ions are required to balance the charge, giving the salt $$\mathrm{Ca(ClO)_2}$$. Its systematic (IUPAC) name is therefore calcium hypochlorite.

3. Commercially, solid calcium hypochlorite is sold and used as a disinfectant and bleaching agent. In everyday language (and in the NCERT text) it is known as bleaching powder.

Hence, the common name of $$\mathrm{Ca(ClO)_2}$$ is bleaching powder.

Bleaching powder (calcium oxychloride, $$\mathrm{CaOCl_2}$$) is manufactured by passing dry chlorine gas over dry slaked lime.

The reaction is 
$$\mathrm{2Ca(OH)_2 + 2Cl_2 \rightarrow Ca(OCl)_2 + CaCl_2 + 2H_2O}$$

Hence, the substance that reacts with chlorine to give bleaching powder is slaked lime, i.e. calcium hydroxide $$\mathrm{Ca(OH)_2}$$.

Hard water contains dissolved $$\mathrm{Ca^{2+}}$$ and $$\mathrm{Mg^{2+}}$$ ions. To "soften" it we need a compound that will remove these ions by precipitating them as insoluble carbonates:

$$\mathrm{Ca^{2+} +\; CO_3^{2-}\;\longrightarrow\; CaCO_3 \downarrow}$$

The sodium salt that readily supplies the required $$\mathrm{CO_3^{2-}}$$ ions is washing soda, whose chemical formula is

$$\mathrm{Na_2CO_3\,\cdot\,10\,H_2O}$$

Therefore, the sodium compound used for softening hard water is sodium carbonate decahydrate (washing soda).

Step 1 : Recall the substance involved
Sodium hydrocarbonate (also called sodium hydrogen carbonate or baking soda) has the formula $$\mathrm{NaHCO_3}$$.

Step 2 : Observe what heating does
When a salt of a weak acid (here: carbonic acid) is heated, it generally decomposes, giving the corresponding normal salt, water and carbon dioxide. Thus, heating a solution of $$\mathrm{NaHCO_3}$$ causes thermal decomposition.

Step 3 : Write the balanced chemical equation
In aqueous solution the products remain or escape in the indicated physical states:

$$2\mathrm{NaHCO_3(aq)} \xrightarrow{\text{heat}} \mathrm{Na_2CO_3(aq)} + \mathrm{H_2O(l)} + \mathrm{CO_2(g)}$$

Step 4 : Describe the observable change
• The liquid will start effervescing as $$\mathrm{CO_2}$$ gas is released.
• The remaining solution now contains $$\mathrm{Na_2CO_3}$$ (washing soda).
• No solid residue is seen because both $$\mathrm{NaHCO_3}$$ and $$\mathrm{Na_2CO_3}$$ are soluble.

Thus, on heating, sodium hydrocarbonate in solution decomposes into sodium carbonate, water and carbon dioxide.

Step 1 — Recall the formulae
Plaster of Paris is calcium sulphate hemihydrate, whose formula is $$\mathrm{CaSO_4\cdot\tfrac12 H_2O}$$.
On adding water, it is converted back to gypsum, calcium sulphate dihydrate, $$\mathrm{CaSO_4\cdot 2H_2O}$$.

Step 2 — Write the skeletal equation
$$\mathrm{CaSO_4\cdot\tfrac12 H_2O + H_2O \longrightarrow CaSO_4\cdot 2H_2O}$$

Step 3 — Balance the water molecules
Each mole of Plaster of Paris already contains $$\tfrac12$$ mole of water; to reach 2 moles of water in the product, we need an additional $$\tfrac32$$ moles from the reactant side. Hence the balanced molecular equation is

$$\mathrm{CaSO_4\cdot\tfrac12 H_2O + \tfrac32 H_2O \;\longrightarrow\; CaSO_4\cdot 2H_2O}$$

An alternative whole-number form (multiplying throughout by 2):

$$\mathrm{2\,(CaSO_4\cdot\tfrac12 H_2O) + 3H_2O \;\longrightarrow\; 2\,CaSO_4\cdot 2H_2O}$$

Thus, Plaster of Paris combines with water to give gypsum.

Litmus test

• Red litmus paper turns blue only in a basic (alkaline) solution.

Connecting litmus with pH

• The pH scale runs from $$0$$ (strongly acidic) to $$14$$ (strongly basic).
• Neutral solutions have $$\text{pH}=7$$.
• Bases have $$\text{pH} > 7$$.

Examine the four given pH values

Only option (iv) corresponds to a basic solution, which alone can turn red litmus blue.

Therefore, the likely pH is $$10$$.

The observation mentioned in the question can be split into two parts:

• When the unknown solution is added to crushed egg-shells (mainly $$\mathrm{CaCO_3}$$), a gas is produced.
• The gas turns lime-water ($$\mathrm{Ca(OH)_2}$$ solution) milky.

Turning lime-water milky is the standard test for carbon dioxide, because

$$\mathrm{Ca(OH)_2\,(aq) + CO_2\,(g) \rightarrow CaCO_3\,(s) + H_2O\,(l)}$$

Therefore the gas evolved in the reaction must be $$\mathrm{CO_2}$$.

Egg-shells contain calcium carbonate. Calcium carbonate gives off carbon dioxide only when it reacts with an acid:

$$\mathrm{CaCO_3\,(s) + 2H^+\,(aq) \rightarrow Ca^{2+}\,(aq) + CO_2\,(g) + H_2O\,(l)}$$

Let us test the four alternatives to see which one supplies $$\mathrm{H^+}$$ ions:

1. $$\mathrm{NaCl}$$ – a neutral salt; does not furnish $$\mathrm{H^+}$$ ions.
2. $$\mathrm{HCl}$$ – a strong acid; does furnish plenty of $$\mathrm{H^+}$$ ions.
3. $$\mathrm{LiCl}$$ – a neutral salt; does not furnish $$\mathrm{H^+}$$ ions.
4. $$\mathrm{KCl}$$ – a neutral salt; does not furnish $$\mathrm{H^+}$$ ions.

Only option 2, $$\mathrm{HCl}$$, is acidic and therefore capable of reacting with $$\mathrm{CaCO_3}$$ in the egg-shells to release carbon dioxide, which in turn turns lime-water milky.

Hence, the solution must contain $$\mathrm{HCl}$$.

Given data

• 10 mL of the given $$\mathrm{NaOH}$$ solution is exactly neutralised by 8 mL of the given $$\mathrm{HCl}$$ solution.

Step 1 – Write the neutralisation equation

$$\mathrm{NaOH + HCl \;\rightarrow\; NaCl + H_2O}$$

The reaction shows a 1 : 1 molar (or normal) ratio between $$\mathrm{NaOH}$$ and $$\mathrm{HCl}$$.

Step 2 – Relate concentration and volume for the first titration

If $$C_{\mathrm{NaOH}}$$ and $$C_{\mathrm{HCl}}$$ are the concentrations (in the same units) of the two solutions, then

$$C_{\mathrm{NaOH}} \times 10\;\text{mL} = C_{\mathrm{HCl}} \times 8\;\text{mL}$$

Solve for one concentration in terms of the other:

$$C_{\mathrm{HCl}} = \frac{10}{8}\,C_{\mathrm{NaOH}} = 1.25\,C_{\mathrm{NaOH}}$$

Step 3 – Set up the equation for 20 mL of the same $$\mathrm{NaOH}$$ solution

Let $$V$$ be the volume (in mL) of the same $$\mathrm{HCl}$$ solution now required.

$$C_{\mathrm{NaOH}} \times 20\;\text{mL} = C_{\mathrm{HCl}} \times V$$

Substitute $$C_{\mathrm{HCl}} = 1.25\,C_{\mathrm{NaOH}}$$.

$$C_{\mathrm{NaOH}} \times 20 = 1.25\,C_{\mathrm{NaOH}} \times V$$

Cancel $$C_{\mathrm{NaOH}}$$ (it is common on both sides):

$$20 = 1.25\,V$$

$$V = \frac{20}{1.25} = 16\;\text{mL}$$

Step 4 – Choose the correct option

The required volume of the $$\mathrm{HCl}$$ solution is 16 mL → option (iv).

Indigestion generally occurs when excess hydrochloric acid (HCl) is secreted in the stomach. The burning sensation is relieved by neutralising this excess acid.

To decide which type of medicine is appropriate, recall the function of each:

• Antibiotic – kills or inhibits the growth of disease-causing bacteria.
• Analgesic – reduces or eliminates pain.
• Antacid – neutralises excess stomach acid.
• Antiseptic – prevents infection by stopping the growth of micro-organisms on living tissue.

Because indigestion is caused by excess acid, the correct medicine must bring the pH closer to neutral by reaction with the acid. Antacids do exactly this. Common examples are:

• Milk of magnesia: $$\mathrm{Mg(OH)_2}$$
• Baking soda: $$\mathrm{NaHCO_3}$$

They react with the stomach’s $$\mathrm{HCl}$$, forming salt, water and sometimes carbon dioxide, thereby giving quick relief.

Hence, the medicine used for treating indigestion is an antacid.

Activity – Do all hydrogen-containing compounds behave as acids?

Aim : To show that although ethanol $$\mathrm{(C_2H_5OH)}$$ and glucose $$\mathrm{(C_6H_{12}O_6)}$$ contain hydrogen, they do not furnish $$\mathrm{H^+}$$ ions in water and therefore are not acidic.

Apparatus & chemicals

• Three 100 mL beakers
• 6 V battery, connecting wires, key (switch)
• Two sharpened graphite rods (taken from pencil leads) as electrodes
• Small torch bulb / LED fixed in the circuit
• Dropper
• Distilled water, dilute $$\mathrm{HCl}$$ (approx. 1 M)
• 10 % (w / v) aqueous glucose solution
• 10 % (v / v) aqueous ethanol solution

Circuit diagram to draw: a simple series circuit containing the battery, key, bulb and two graphite electrodes whose free ends dip into the test solution kept in the beaker.

Procedure

1. Set up the circuit; keep the key open.
2. Pour about 50 mL dilute $$\mathrm{HCl}$$ into the first beaker, dip the electrodes, close the key and note whether the bulb glows.
3. Wash electrodes with distilled water, place them in 50 mL of glucose solution in the second beaker, close the key and note the bulb.
4. Again wash the electrodes, place them in 50 mL of ethanol solution in the third beaker, close the key and observe.

Observations

Explanation

• For a liquid to conduct electricity, it must supply free ions. When $$\mathrm{HCl}$$ dissolves in water it ionises completely:
  $$\mathrm{HCl(aq)} \;\longrightarrow\; H^+ + Cl^-$$
  The mobile $$\mathrm{H^+}$$ and $$\mathrm{Cl^-}$$ ions complete the circuit, so the bulb glows.
• Glucose and ethanol dissolve without ionisation; their molecules remain electrically neutral:
  $$\mathrm{C_6H_{12}O_6(aq)} \;\text{no ionisation}$$
  $$\mathrm{C_2H_5OH(aq)} \;\text{no ionisation}$$
  Hence almost no ions are present, the solution is practically non-conducting, and the bulb stays off.

Inference

The presence of hydrogen atoms in a compound is not enough for acidic behaviour. A substance shows acidic character only if it can release $$\mathrm{H^+}$$ ions in aqueous solution. Because glucose and ethanol cannot do so, they are not classified as acids.

To allow the passage of an electric current through a liquid, the liquid must contain mobile ions that can carry charge from one electrode to the other.

1. Nature of distilled water
   Distilled water is obtained by boiling water and condensing the vapour, so all dissolved solids and gases are removed. It therefore contains almost no ions. Only the extremely small self-ionisation
   $$\mathrm{2\,H_2O \;\rightleftharpoons\; H_3O^+ + OH^-}$$
   takes place, giving $$[\mathrm{H^+}] = [\mathrm{OH^-}] \approx 1\times10^{-7}\,\text{mol L}^{-1}$$, which is far too low to produce an observable current on ordinary laboratory apparatus. Hence distilled water behaves practically as an insulator.
2. Nature of rain water
   (a) As rain falls through the atmosphere it dissolves atmospheric $$\mathrm{CO_2}$$:
   $$\mathrm{CO_2 + H_2O \;\rightarrow\; H_2CO_3}$$
   Carbonic acid formed above partially ionises:
   $$\mathrm{H_2CO_3 \;\rightleftharpoons\; H^+ + HCO_3^-}$$
   (b) Rain water also picks up traces of mineral salts (for example $$\mathrm{Na^+}$$, $$\mathrm{Cl^-}$$, $$\mathrm{SO_4^{2-}}$$) blown up as aerosols or leached from rocks and soil dust carried by wind.
   Because of these processes, rain water contains a measurable concentration of ions $$\left(>10^{-4}\,\text{mol L}^{-1}\text{ in total}\right)$$, so it can carry charge and therefore conducts electricity.
3. Conclusion
   Distilled water lacks free ions, whereas rain water contains dissolved acidic and mineral ions; hence only rain water conducts electricity.

Key idea – A substance shows acidic properties only when it furnishes free $$\mathrm{H^+}$$ (or $$\mathrm{H_3O^+}$$) ions in aqueous solution.

Ionisation in water

$$\mathrm{HCl(g) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)}$$

The above reaction occurs only when water molecules are present; the ions produced are responsible for all observable acidic behaviours such as turning blue litmus red, reacting with metals, conducting electricity, etc.

What happens with dry gas?

• Dry $$\mathrm{HCl}$$ contains no water.
• Without water the covalent $$\mathrm{H\!\!Cl}$$ bond does not break, so no $$\mathrm{H^+}$$ (or $$\mathrm{H_3O^+}$$) ions are formed.
• Because these ions are absent, the gas cannot exhibit the usual acidic characteristics.

Therefore, dry $$\mathrm{HCl}$$ gas does not show acidic behaviour; it becomes acidic only after dissolving in water, when it can ionise to release hydrogen ions.

Given data

Step 1 : Recall the meaning of pH values

• $$ ext{pH} = 7$$  →  neutral (pure water).
• $$ ext{pH} < 7$$  →  acidic; smaller pH means stronger acid because $$[ ext{H}^+]$$ is larger.
• $$ ext{pH} > 7$$  →  basic / alkaline; larger pH means stronger base because $$[ ext{H}^+]$$ is smaller.

Step 2 : Classify each solution

• Solution D has pH 7 → neutral.
• Solution C has pH 11 → much >7, so it is strongly alkaline.
• Solution B has pH 1 → far <7, so it is strongly acidic.
• Solution A has pH 4 → just below 7, so it is weakly acidic.
• Solution E has pH 9 → just above 7, so it is weakly alkaline.

Step 3 : Order of hydrogen-ion concentration

The relation $$ ext{pH} = -\log_{10}[\text{H}^+]$$ shows that when pH decreases, $$[\text{H}^+]$$ increases. Therefore, to list the pH values in increasing order of $$[\text{H}^+]$$, we must write the pH values from the highest number (lowest $$[\text{H}^+]$$) to the lowest number (highest $$[\text{H}^+]$$).

Thus :

$$11 \;{(C)} < 9 \;{(E)} < 7 \;{(D)} < 4 \;{(A)} < 1 \;{(B)}$$

(arrows show that hydrogen-ion concentration increases from left to right)

Step 1 – Write the chemical reactions

When magnesium metal reacts with an acid, hydrogen gas is liberated. For the two acids given:

In test tube A (hydrochloric acid):
$$\mathrm{Mg\;(s) + 2\,HCl\;(aq) \;\longrightarrow\; MgCl_2\;(aq) + H_2\;(g)}$$

In test tube B (acetic acid):
$$\mathrm{Mg\;(s) + 2\,CH_3COOH\;(aq) \;\longrightarrow\; (CH_3COO)_2Mg\;(aq) + H_2\;(g)}$$

In both cases the visible evidence of the reaction is fizzing caused by bubbles of $$\mathrm{H_2}$$ gas.

Step 2 – Compare the nature of the two acids

• Hydrochloric acid, $$\mathrm{HCl}$$, is a strong acid. It is almost completely ionised in aqueous solution:
  $$\mathrm{HCl\;(aq) \;\longrightarrow\; H^+\;(aq) + Cl^-\;(aq)}$$
• Acetic acid, $$\mathrm{CH_3COOH}$$, is a weak acid. It is only partially ionised in aqueous solution:
  $$\mathrm{CH_3COOH\;(aq) \rightleftharpoons\; H^+\;(aq) + CH_3COO^-\;(aq)}$$
  At the same concentration, acetic acid therefore provides far fewer $$\mathrm{H^+}$$ ions than hydrochloric acid.

Step 3 – Relate rate of reaction to concentration of $$\mathrm{H^+}$$ ions

The rate at which magnesium loses electrons and hydrogen gas is produced depends on the concentration of hydrogen ions present:

• Higher $$[\mathrm{H^+}]$$ → faster attack on the magnesium surface → faster production of $$\mathrm{H_2}$$ → more vigorous fizzing.
• Lower $$[\mathrm{H^+}]$$ → slower reaction → milder fizzing.

Step 4 – Draw the conclusion

Because hydrochloric acid is fully ionised, test tube A contains a much higher concentration of $$\mathrm{H^+}$$ ions than test tube B, even though the nominal (listed) concentrations of the two acids are the same. Hence, the reaction in test tube A proceeds faster and produces hydrogen gas more quickly.

Therefore, the fizzing will be more vigorous in test tube A than in test tube B.

Given data
Fresh milk has $$\text{pH}=6$$.

Step 1 : Recall the pH scale
On the pH scale (0 to 14) a solution is

• neutral at $$\text{pH}=7$$,
• acidic if $$\text{pH}<7$$,
• basic if $$\text{pH}>7$$.

Because $$6<7$$, fresh milk is already slightly acidic.

Step 2 : What happens during curdling?
During curd formation the bacterium Lactobacillus converts the milk sugar (lactose) $$\mathrm{C_{12}H_{22}O_{11}}$$ into lactic acid $$\mathrm{CH_3CH(OH)COOH}$$.

$$\mathrm{C_{12}H_{22}O_{11}\;(lactose)} \;\longrightarrow\; 4\,\mathrm{CH_3CH(OH)COOH}\;(\text{lactic acid})$$

Step 3 : Effect on $$[H^+]$$ and pH
Lactic acid is a weak acid, yet its accumulation increases the hydrogen-ion concentration, $$[H^+]$$, in the milk.

The definition $$\text{pH}=-\log_{10}[H^+]$$ shows that when $$[H^+]$$ increases, the pH value decreases.

Step 4 : Resulting pH of curd
In practice the pH of curd lies around $$4.5$$. Since $$4.5<6$$, curd is considerably more acidic than fresh milk.

Conclusion
As milk turns into curd its pH falls from about $$6$$ to a value below $$6$$ (≈ $$4.5$$). The decrease occurs because lactic acid, produced by bacterial action, increases the acidity of the medium.

Step 1  Identify the substance
Plaster of Paris is chemically calcium sulphate hemihydrate: $$\mathrm{CaSO_4\,\cdot\,\tfrac12\,H_2O}$$.

Step 2  Describe what happens in the presence of moisture
When it comes in contact with water vapour (moisture) it combines with additional water molecules and is converted into gypsum, calcium sulphate di-hydrate:

$$\mathrm{CaSO_4\,\cdot\,\tfrac12\,H_2O\; +\; \tfrac32\,H_2O\;\longrightarrow\; CaSO_4\,\cdot\,2H_2O}$$

Step 3  Consequences of this reaction

• The product gypsum sets into a hard, solid mass.
• Once this setting occurs, the material loses its ability to again form a smooth paste on mixing with water, so it becomes useless for making casts, models, etc.

Step 4  Reason for moisture-proof storage
To prevent the above hydration (setting) before actual use, Plaster of Paris must be kept away from atmospheric moisture; hence it is packed and stored in airtight, moisture-proof containers.

Definition

When an acid reacts stoichiometrically with a base, the characteristic acidic ions $$\mathrm{H^+}$$ combine with the basic ions $$\mathrm{OH^-}$$ to form the neutral molecule water; the other ions present produce an ionic salt. Such a reaction is called a neutralisation reaction.

Symbolically,

$$\mathrm{Acid\,(H^+) + Base\,(OH^-) \;\longrightarrow\; H_2O +\; Salt}$$

Illustrative examples

1. Between the strong acid hydrochloric acid and the strong base sodium hydroxide:
   $$\mathrm{HCl + NaOH \;\rightarrow\; NaCl + H_2O}$$
2. Between the strong acid sulphuric acid and the weak base calcium hydroxide:
   $$\mathrm{H_2SO_4 + Ca(OH)_2 \;\rightarrow\; CaSO_4 + 2H_2O}$$

In both cases the acidic $$\mathrm{H^+}$$ and basic $$\mathrm{OH^-}$$ ions neutralise each other to give water, while the remaining ions form the corresponding salt.

Step 1 : Recall the common names and formulae

• Washing soda = sodium carbonate decahydrate, $$\mathrm{Na_2CO_3\,\cdot\,10H_2O}$$
• Baking soda = sodium hydrogen-carbonate, $$\mathrm{NaHCO_3}$$

Step 2 : State two important uses of each compound

1. Washing soda ( $$\mathrm{Na_2CO_3\,\cdot\,10H_2O}$$ )
   • It is a key cleaning agent in laundries; the alkaline $$\mathrm{CO_3^{2-}}$$ ions emulsify grease and dirt, helping to wash clothes.
   • It is used to remove the permanent hardness of water by precipitating the calcium and magnesium ions as their carbonates: $$\mathrm{Ca^{2+}+Na_2CO_3 \rightarrow CaCO_3 \downarrow +2Na^+}$$ $$\mathrm{Mg^{2+}+Na_2CO_3 \rightarrow MgCO_3 \downarrow +2Na^+}$$
2. Baking soda ( $$\mathrm{NaHCO_3}$$ )
   • Baking powder is a mixture of baking soda and a mild edible acid (e.g. $$\mathrm{tartaric\,acid}$$); on heating it makes cakes and bread fluffy by releasing carbon dioxide: $$\mathrm{2NaHCO_3 \xrightarrow{\Delta} Na_2CO_3 + CO_2 \uparrow + H_2O}$$
   • It serves as a mild antacid; the hydrogen-carbonate ion neutralises excess stomach acid: $$\mathrm{NaHCO_3 + HCl \rightarrow NaCl + H_2O + CO_2}$$

Thus, each salt has at least two everyday applications that arise directly from its chemical properties.