The ${\rm{pH}}$ of ${\rm{0}}{\rm{.1}}\;{\rm{M}}$ solution of the following salts increases in the order

${\rm{HCl}} < N{{\rm{H}}_4}{\rm{Cl}} < NaCl < NaCN$

${\rm{NaCl}} < {\rm{N}}{{\rm{H}}_4}{\rm{Cl}} < NaCN < HCl$

${\rm{NaCN}} < {\rm{N}}{{\rm{H}}_4}{\rm{Cl}} < NaCl < HCl$

${\rm{HCl}} < NaCl < NaCN < {\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$

$10\,{\rm{mL}}$ of a solution contains $0.1\,M\,{\rm{N}}{{\rm{H}}_4}{\rm{Cl}} + 0.01\,M\,{\rm{N}}{{\rm{H}}_4}{\rm{OH}}{\rm{.}}$ Which addition would not change the ${\rm{pH}}$ of the solution?

Adding $5\,{\rm{mL}}$ of$ 0.1\,M\,{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$

Adding $5\,{\rm{mL}}$ of $0.1\,M\,{\rm{N}}{{\rm{H}}_4}{\rm{OH}}$

Adding$10\,{\rm{mL}}$ of $0.1\,M\,{\rm{N}}{{\rm{H}}_4}{\rm{Cl}}$

The equilibrium constant, $K$ for the reaction $2{\rm{HI}}\left( {\rm{g}} \right)\;\leftrightharpoons\;{{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right)$ At room temperature is 2.85 and that at $698\;K$, it is $1.4 \times {10^{ - 2}}$. This implies that

${\rm{HI}}$ is less stable than ${{\rm{H}}_2}$ and ${{\rm{I}}_2}$

${\rm{HI}}$ is resonance stabilised

${\rm{HI}}$ is exothermic compound

${\rm{HI}}$ is stable at room temperature

Equilibrium MCQs for NEET UG CHEMISTRY — Easy, Page 3

61 Easy

61 of 144

Understanding +4 | -1

Boric acid is a

View answer and explanation

Correct option: C

Lewis acid

Option C (Lewis acid) is correct because Boric acid (

B(OH)_3

) accepts a lone pair of electrons from hydroxide ions, thus acting as a Lewis acid.

Solution approach

Correct Answer + Core Reason: Lewis acid. Boric acid accepts electron pairs.

Step-by-Step Solution:

Step 1: Recall the definitions of Arrhenius, Bronsted, and Lewis acids.

Step 2: An Arrhenius acid produces $H^+$ ions in water.

Step 3: A Bronsted acid donates protons.

Step 4: A Lewis acid accepts electron pairs.

Step 5: Boric acid, $B(OH)_3$ , does not donate protons. Instead, it accepts a hydroxide ion ( $OH^-$ ) to form $[B(OH)_4]^-$ .

Step 6: Therefore, boric acid acts as a Lewis acid.

Why other options are wrong:
- Option A: Arrhenius acid is incorrect because boric acid does not directly produce $H^+$ ions in water.
- Option B: Bronsted acid is incorrect because boric acid does not donate protons.
- Option C: Lewis acid is the correct answer because boric acid accepts electron pairs.
- Option D: All of these is incorrect because only Lewis acid is correct.

Key Concepts:
- Arrhenius acids and bases
- Bronsted acids and bases
- Lewis acids and bases
- Electron-pair acceptance

Memory Aids:
- Boron is electron deficient, so it accepts electrons (Lewis acid).

Quick Tips:
- Remember that Lewis acids are electron-pair acceptors, and Bronsted acids are proton donors.

62 Easy

62 of 144

Application +4 | -1

In which of the following reaction

{K_p} > {K_c}

?

View answer and explanation

Correct option: C

2\mathrm{S}{\mathrm{O}_3}\leftrightharpoons{\mathrm{O}_2} + 2\mathrm{S}{\mathrm{O}_2}

Option C is correct because Kp > Kc when Δn_g > 0, and in this reaction, Δn_g = (1+2) - 2 = 1.

Solution approach

Correct Answer + Core Reason: $2\mathrm{SO}_3\leftrightharpoons{\mathrm{O}_2} + 2\mathrm{SO}_2$ . Kp > Kc when $\Delta n_g > 0$ .

Step-by-Step Solution:

Step 1: Recall the relationship between Kp and Kc: $K_p = K_c(RT)^{\Delta n_g}$

Step 2: Determine $\Delta n_g$ for each reaction:
- A: ${\mathrm{N}_2} + 3{\mathrm{H}_2}\leftrightharpoons2\mathrm{NH}_3$ , $\Delta n_g = 2 - (1 + 3) = -2$
- B: ${\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI}$ , $\Delta n_g = 2 - (1 + 1) = 0$
- C: $2\mathrm{SO}_3\leftrightharpoons{\mathrm{O}_2} + 2\mathrm{SO}_2$ , $\Delta n_g = (1 + 2) - 2 = 1$
- D: $\mathrm{PCl}_3 + \mathrm{Cl}_2\leftrightharpoons\mathrm{PCl}_5$ , $\Delta n_g = 1 - (1 + 1) = -1$

Step 3: Kp > Kc when $\Delta n_g > 0$ . Only reaction C has a positive $\Delta n_g$ .

Why other options are wrong:
- Option A: For this reaction, $\Delta n_g < 0$ , so Kp < Kc.
- Option B: For this reaction, $\Delta n_g = 0$ , so Kp = Kc.
- Option D: For this reaction, $\Delta n_g < 0$ , so Kp < Kc.

Key Concepts:
- Relationship between Kp and Kc: $K_p = K_c(RT)^{\Delta n_g}$
- Calculation of $\Delta n_g$

Memory Aids:
- If $\Delta n_g$ is positive, Kp > Kc.
- If $\Delta n_g$ is negative, Kp < Kc.
- If $\Delta n_g$ is zero, Kp = Kc.

Quick Tips:
- Always calculate $\Delta n_g$ carefully, paying attention to the stoichiometric coefficients.

63 Easy

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Understanding +4 | -1

An aqueous solution of ammonium acetate is:

View answer and explanation

Correct option: D

Almost neutral

Option D (Almost neutral) is correct because ammonium acetate is a salt of a weak acid (acetic acid) and a weak base (ammonium hydroxide), and their strengths are comparable, resulting in a nearly neutral solution.

Solution approach

Correct Answer + Core Reason: Almost neutral. Ammonium acetate is a salt of a weak acid and a weak base with comparable strengths.

Step-by-Step Solution:

Step 1: Identify the acid and base that form ammonium acetate.

Step 2: Ammonium acetate ( $CH_3COONH_4$ ) is formed from a weak acid (acetic acid, $CH_3COOH$ ) and a weak base (ammonium hydroxide, $NH_4OH$ ).

Step 3: Understand that the pH of a salt of a weak acid and a weak base depends on the relative strengths of the acid and base.

Step 4: Since acetic acid and ammonium hydroxide have comparable strengths, the hydrolysis of the ions will be approximately equal, resulting in a nearly neutral solution.

Why other options are wrong:
- Option A: Faintly acidic: This would be the case if the acid were slightly stronger than the base.
- Option B: Fair acidic: This would be the case if the acid were significantly stronger than the base.
- Option C: Faintly alkaline: This would be the case if the base were slightly stronger than the acid.

Key Concepts:
- Salt hydrolysis
- pH of salt solutions
- Weak acids and bases

Memory Aids:
- Weak Acid + Weak Base: pH depends on relative strengths.

Quick Tips:
- If the weak acid and weak base have similar strengths, the solution will be nearly neutral.

64 Easy

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Analyzing +4 | -1

STATEMENT I: When aqueous, solution of

\mathrm{CH}_3 \mathrm{COONH}_4

is diluted, then its degree of hydrolysis increase.

STATEMENT II : Ammonium acetate is the salt of weak acid and weak base, its degree of hydrolysis depend on the concentration.

View answer and explanation

Correct option: B

Statement I, II are incorrect

Option B (Statement I, II are incorrect) is correct because the degree of hydrolysis for ammonium acetate increases with dilution, and it depends on the ionic product of water, not just concentration.

Solution approach

Correct Answer + Core Reason: Statement I, II are incorrect because the degree of hydrolysis of ammonium acetate increases with dilution and depends on the ionic product of water.

Step-by-Step Solution:

Step 1: Analyze Statement I: When an aqueous solution of $\mathrm{CH}_3 \mathrm{COONH}_4$ is diluted, its degree of hydrolysis increases. This statement is correct.

Step 2: Analyze Statement II: Ammonium acetate is the salt of a weak acid and a weak base; its degree of hydrolysis depends on the concentration. This statement is incorrect. The degree of hydrolysis for salts of weak acids and weak bases is independent of concentration but depends on temperature and the ionic product of water ( $K_w$ ).

Step 3: Conclusion: Statement I is correct, and Statement II is incorrect.

Why other options are wrong:
- Option A: Statement I, II are correct: Incorrect because Statement II is incorrect.
- Option C: Statement I is correct, Statement II is incorrect: Incorrect because Statement I is also incorrect. The degree of hydrolysis for salts of weak acids and weak bases is independent of concentration.
- Option D: Statement I is incorrect, Statement II is correct: Incorrect because Statement II is incorrect.

Key Concepts:
- Hydrolysis of salts
- Salts of weak acids and weak bases
- Degree of hydrolysis

Memory Aids:
- Hydrolysis: Reaction of a salt with water.

Quick Tips:
- The degree of hydrolysis for salts of weak acids and weak bases is independent of concentration.

65 Easy

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Remembering +4 | -1

When the

\mathrm{pH}

of a solution is 2, the hydrogen ion concentration is:

View answer and explanation

Correct option: B

1 \times {10^{ - 2}}\;M

Option B (

1 \times {10^{ - 2}}\;M

) is correct because pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration.

Solution approach

Correct Answer + Core Reason: $1 \times {10^{ - 2}}\;M$ . This is calculated directly from the definition of pH.

Step-by-Step Solution:

Step 1: The definition of pH is $pH = -\log_{10}[H^+]$ , where $[H^+]$ is the hydrogen ion concentration.

Step 2: Given that $pH = 2$ , we can write $2 = -\log_{10}[H^+]$ .

Step 3: To find the hydrogen ion concentration, we can rewrite the equation as $[H^+] = 10^{-pH}$ .

Step 4: Substituting $pH = 2$ , we get $[H^+] = 10^{-2} = 1 \times 10^{-2} \text{ M}$ .

Why other options are wrong:
- **Option A ( $1 \times {10^{ - 14}}\;M$ ): This is the concentration of hydroxide ions in pure water at 25°C, not the hydrogen ion concentration at pH 2.
- Option C ( $1 \times {10^{ - 7}}\;M$ ): This is the hydrogen ion concentration in neutral water at 25°C (pH 7).
- Option D ( $1 \times {10^{ - 12}}\,M$ ): This would correspond to a pH of 12, which is a basic solution.

Key Concepts:
- pH definition
- Hydrogen ion concentration

Memory Aids:**
- $pH = -\log_{10}[H^+]$
- $[H^+] = 10^{-pH}$

Quick Tips:
- Remember the relationship between pH and hydrogen ion concentration. A lower pH indicates a higher hydrogen ion concentration (more acidic).

66 Easy

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Analyzing +4 | -1

Assertion : For a gaseous reaction,

x A+y B \rightleftharpoons l C+m D, K_p=K_C .

Reason : Concentration of gaseous reactant is taken to be unity.

View answer and explanation

Correct option: D

If the assertion and reason both are false.

Option D (If the assertion and reason both are false) is correct. The assertion is false because

K_p = K_c

only when

\Delta n_g = 0

, and the reason is false because the concentration of gaseous reactant is not always unity.

Solution approach

Correct Answer + Core Reason: If the assertion and reason both are false. The condition for $K_p = K_c$ is not met, and concentrations are not always unity.

Step-by-Step Solution:

Step 1: The relationship between $K_p$ and $K_c$ is given by the equation $K_p = K_c (RT)^{\Delta n_g}$ , where $\Delta n_g$ is the change in the number of moles of gas in the reaction (moles of gaseous products - moles of gaseous reactants).

Step 2: For $K_p$ to be equal to $K_c$ , $(RT)^{\Delta n_g}$ must be equal to 1. This happens only when $\Delta n_g = 0$ .

Step 3: The assertion states that $K_p = K_c$ for any gaseous reaction $x A+y B \rightleftharpoons l C+m D$ . This is false because it only holds true when $(l+m) - (x+y) = 0$ .

Step 4: The reason states that the concentration of gaseous reactant is taken to be unity. This is generally not true. Concentrations can vary, and are only unity under specific standard conditions.

Step 5: Therefore, both the assertion and the reason are false.

Why other options are wrong:
- Option A: Both assertion and reason are not true.
- Option B: Both assertion and reason are not true.
- Option C: The assertion is not true.

Key Concepts:
- Relationship between $K_p$ and $K_c$ : $K_p = K_c (RT)^{\Delta n_g}$
- $\Delta n_g$ (change in number of moles of gas)

Memory Aids:
- Remember the formula $K_p = K_c (RT)^{\Delta n_g}$

Quick Tips:
- Always check the value of $\Delta n_g$ before assuming $K_p = K_c$ .

67 Easy

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Understanding +4 | -1

Which of the following statements is not correct

View answer and explanation

Correct option: A

The value of equilibrium constant is changed in the presence of catalyst in the reaction at equilibrium

Option A (The value of equilibrium constant is changed in the presence of catalyst in the reaction at equilibrium) is correct because a catalyst only speeds up the rate at which equilibrium is reached but does not change the value of the equilibrium constant.

Solution approach

Correct Answer + Core Reason: The value of equilibrium constant is changed in the presence of catalyst in the reaction at equilibrium. Catalysts do not change the equilibrium constant.

Step-by-Step Solution:

Step 1: Define a catalyst: A substance that speeds up the rate of a chemical reaction without being consumed in the reaction.

Step 2: Explain the effect of a catalyst on equilibrium: A catalyst lowers the activation energy for both the forward and reverse reactions equally. This means that the reaction reaches equilibrium faster, but the position of equilibrium remains unchanged.

Step 3: Analyze the given statements:
- The value of the equilibrium constant is changed in the presence of a catalyst in the reaction at equilibrium: This statement is incorrect because a catalyst does not change the value of the equilibrium constant.
- Enzymes catalyse mainly bio-chemical reaction: This statement is correct.
- Co-enzymes increases the catalytic activity of enzyme: This statement is correct.
- Catalyst does not initiate any reaction: This statement is correct.

Why other options are wrong:
- Option B: Enzymes catalyse mainly bio-chemical reaction is a correct statement.
- Option C: Co-enzymes increases the catalytic activity of enzyme is a correct statement.
- Option D: Catalyst does not initiate any reaction is a correct statement.

Key Concepts:
- Catalyst
- Equilibrium
- Equilibrium constant
- Enzymes and co-enzymes

Memory Aids:
- Catalysts speed up reactions but do not change the equilibrium constant.

Quick Tips:
- Remember that catalysts only affect the rate of reaction, not the equilibrium position.

68 Easy

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Application +4 | -1

In the reactions,

\mathrm{PC}{\mathrm{l}_5}\leftrightharpoons\mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2},

, the amounts of

\mathrm{PC}{\mathrm{l}_5},\mathrm{PC}{\mathrm{l}_3}

and

\mathrm{C}{\mathrm{l}_2}

at equilibrium are

2

mole each and the total pressure is

3

atm

. The equilibrium constant

{K_p}

is :

View answer and explanation

Correct option: A

1.0\mathrm{ }\;atm

Option A (

1.0\mathrm{ }\;atm

) is correct because it is the value of Kp calculated using partial pressures at equilibrium.

Solution approach

Correct Answer + Core Reason: $1.0\mathrm{ }\;atm$ . The equilibrium constant Kp is calculated using partial pressures.

Step-by-Step Solution:

Step 1: Write the equilibrium reaction: $\mathrm{PC}{\mathrm{l}_5} \leftrightharpoons \mathrm{PC}{\mathrm{l}_3} + \mathrm{C}{\mathrm{l}_2}$

Step 2: Determine the mole fraction of each gas at equilibrium.\nTotal moles = 2 (PCl5) + 2 (PCl3) + 2 (Cl2) = 6 moles\nMole fraction of PCl5 = 2/6 = 1/3\nMole fraction of PCl3 = 2/6 = 1/3\nMole fraction of Cl2 = 2/6 = 1/3

Step 3: Calculate the partial pressure of each gas.\nPartial pressure = (mole fraction) x (total pressure)\nPartial pressure of PCl5 = (1/3) x 3 atm = 1 atm\nPartial pressure of PCl3 = (1/3) x 3 atm = 1 atm\nPartial pressure of Cl2 = (1/3) x 3 atm = 1 atm

Step 4: Calculate Kp.
$K_p = \frac{{P_{PCl3} \times P_{Cl2}}}{{P_{PCl5}}} = \frac{{1 \times 1}}{1} = 1\mathrm{ atm}$

Why other options are wrong:
- Option B (2.0 atm): This is incorrect because it doesn't correctly calculate the partial pressures and apply them to the Kp expression.
- Option C (3.0 atm): This is incorrect because it likely uses the total pressure directly without considering the partial pressures of the individual components.
- Option D (6.0 atm): This is incorrect due to an error in calculating or applying the partial pressures in the Kp expression.

Key Concepts:
- Equilibrium constant (Kp)
- Partial pressure
- Mole fraction

Memory Aids:
- Kp is the equilibrium constant expressed in terms of partial pressures.
- Partial pressure = (mole fraction) x (total pressure)

Quick Tips:
- Always calculate the mole fractions first when dealing with partial pressures.
- Make sure to use the correct partial pressures in the Kp expression.

69 Easy

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Remembering +4 | -1

A reaction attains equilibrium when the Gibbs energy change accompanying the reaction is:

View answer and explanation

Correct option: B

Zero

Option B (Zero) is correct because a reaction attains equilibrium when the Gibbs energy change (ΔG) is zero, indicating that the system is at its lowest energy state.

Solution approach

Correct Answer + Core Reason: Zero. At equilibrium, the Gibbs energy change is zero.

Step-by-Step Solution:

Step 1: Define Gibbs energy (G): It is a thermodynamic potential that measures the amount of energy available in a chemical or physical system to do useful work at a constant temperature and pressure.

Step 2: Define Gibbs energy change (ΔG): It is the change in Gibbs energy during a reaction. It indicates the spontaneity of a reaction.

Step 3: At equilibrium, the system is at its lowest energy state, and there is no driving force for the reaction to proceed in either direction. Therefore, the Gibbs energy change (ΔG) is zero.

Why other options are wrong:
- Option A: Positive and large is incorrect because a positive ΔG indicates a non-spontaneous reaction.
- Option C: Negative and large is incorrect because a negative ΔG indicates a spontaneous reaction, not equilibrium.
- Option D: Negative and small is incorrect because a negative ΔG indicates a spontaneous reaction, not equilibrium.

Key Concepts:
- Gibbs energy (G)
- Gibbs energy change (ΔG)
- Chemical equilibrium
- Spontaneity of reactions

Memory Aids:
- At equilibrium, ΔG = 0.

Quick Tips:
- Understand the relationship between ΔG and the spontaneity of a reaction.

70 Easy

70 of 144

Understanding +4 | -1

The compound whose 0.1 M solution is basic is

View answer and explanation

Correct option: D

Sodium acetate

Option D (Sodium acetate) is correct because sodium acetate is a salt of a strong base (NaOH) and a weak acid (acetic acid), resulting in a basic solution due to hydrolysis of the acetate ion.

Solution approach

Correct Answer + Core Reason: Sodium acetate. Sodium acetate is the salt of a strong base and a weak acid, leading to a basic solution.

Step-by-Step Solution:

Step 1: Identify the nature of each salt.

Step 2: Ammonium acetate is a salt of a weak acid and a weak base. Its pH depends on the relative strengths of the acid and base.

Step 3: Ammonium chloride is a salt of a weak base and a strong acid, resulting in an acidic solution.

Step 4: Ammonium sulphate is a salt of a weak base and a strong acid, resulting in an acidic solution.

Step 5: Sodium acetate is a salt of a strong base and a weak acid, resulting in a basic solution due to the hydrolysis of acetate ions.

Step 6: The acetate ion (CH3COO-) reacts with water to form acetic acid (CH3COOH) and hydroxide ions (OH-), increasing the pH.

Why other options are wrong:
- Option A: Ammonium acetate: While it can be basic, it depends on the relative strengths of the weak acid and weak base. It's not always basic.
- Option B: Ammonium chloride: This is a salt of a weak base and a strong acid, so it will be acidic.
- Option C: Ammonium sulphate: This is a salt of a weak base and a strong acid, so it will be acidic.

Key Concepts:
- Salt hydrolysis
- Salts of strong acids and strong bases
- Salts of weak acids and strong bases
- Salts of strong acids and weak bases
- Salts of weak acids and weak bases

Memory Aids:
- Strong base + Weak acid = Basic solution
- Strong acid + Weak base = Acidic solution

Quick Tips:
- Remember the common strong acids (HCl, H2SO4, HNO3) and strong bases (NaOH, KOH).

71 Easy

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Understanding +4 | -1

For homogeneous gas reaction

4\,\,N{H_3} + 5{O_2}4\,\,NO + 6{H_2}O

. The equilibrium constant
has the unit of

View answer and explanation

Correct option: A

{\left( {Conc} \right)^1}

Option A (

{\left( {Conc} \right)^1}

) is correct because the unit of equilibrium constant K is determined by the change in the number of moles in the reaction.

Solution approach

Correct Answer + Core Reason: ${\left( {Conc} \right)^1}$ . The unit of K is determined by the change in the number of moles.

Step-by-Step Solution:

Step 1: Write the expression for the equilibrium constant, $K_c$ , for the given reaction:
$4NH_3 + 5O_2 \rightleftharpoons 4NO + 6H_2O$
$K_c = \frac{{[NO]^4[H_2O]^6}}{{[NH_3]^4[O_2]^5}}$

Step 2: Determine the change in the number of moles, $\Delta n = \sum n_{products} - \sum n_{reactants}$ :
$\Delta n = (4 + 6) - (4 + 5) = 10 - 9 = 1$

Step 3: The unit of $K_c$ is $(Conc)^{\Delta n}$ . Therefore, the unit of $K_c$ is $(Conc)^1$ .

Why other options are wrong:
- Option B: ${\left( {Conc} \right)^{ - 1}}$ : This would be the unit if the change in the number of moles was -1, which is not the case here.
- Option C: ${\left( {Conc} \right)^9}$ : This is incorrect because it represents the total number of moles of reactants, not the change in moles.
- Option D: ${\left( {Conc} \right)^{10}}$ : This is incorrect because it represents the total number of moles of products, not the change in moles.

Key Concepts:
- Equilibrium constant expression
- Change in the number of moles in a reaction
- Units of equilibrium constant

Memory Aids:
- $\Delta n = n_{products} - n_{reactants}$

Quick Tips:
- Always calculate the change in the number of moles correctly to determine the correct unit of the equilibrium constant.

72 Easy

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Remembering +4 | -1

Which is Lewis base?

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{H}_3}

Option D (

\mathrm{N}{\mathrm{H}_3}

) is correct because ammonia has a lone pair of electrons that it can donate, making it a Lewis base.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_3}$ . Ammonia is a Lewis base because it can donate a lone pair of electrons.

Step-by-Step Solution:

Step 1: Define a Lewis base: A Lewis base is a substance that can donate an electron pair.

Step 2: Analyze each option:\nA. $\mathrm{HCl}$ : Acts as a Lewis acid (proton donor).\nB. $\mathrm{HN}{\mathrm{O}_3}$ : Acts as a Lewis acid (proton donor).\nC. $\mathrm{HF}$ : Acts as a Lewis acid (proton donor).\nD. $\mathrm{N}{\mathrm{H}_3}$ : Has a lone pair of electrons on the nitrogen atom, making it a Lewis base.

Step 3: Conclusion: $\mathrm{N}{\mathrm{H}_3}$ is the Lewis base.

Why other options are wrong:
- Option A: $\mathrm{HCl}$ is a strong acid and acts as a Lewis acid by donating a proton.
- Option B: $\mathrm{HN}{\mathrm{O}_3}$ is a strong acid and acts as a Lewis acid by donating a proton.
- Option C: $\mathrm{HF}$ is a weak acid and acts as a Lewis acid by donating a proton.

Key Concepts:
- Lewis acids and bases
- Electron pair donation/acceptance

Memory Aids:
- Lewis Acid: Accepts electrons (LA-AE)
- Lewis Base: Donates electrons (LB-DE)

Quick Tips:
- Look for lone pairs of electrons to identify Lewis bases.

73 Easy

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Understanding +4 | -1

Salting out action of soap is based on:

View answer and explanation

Correct option: C

Solubility product

Option C (Solubility product) is correct because salting out involves decreasing the solubility of soap by adding a common ion, which is related to the solubility product principle.

Solution approach

Correct Answer + Core Reason: Solubility product. Salting out reduces soap solubility via the common ion effect, which is related to the solubility product.

Step-by-Step Solution:

Step 1: Define salting out: It is the process of decreasing the solubility of a substance (like soap) by adding a salt containing a common ion.

Step 2: Soaps are salts of fatty acids. When a salt like NaCl is added, it increases the concentration of the common ion (e.g., Na+). This shifts the equilibrium of the soap's dissolution according to Le Chatelier's principle.

Step 3: The solubility product (Ksp) is the equilibrium constant for the dissolution of a sparingly soluble salt. The common ion effect reduces the solubility of the salt to maintain the Ksp value.

Why other options are wrong:
- Option A: Complex ion formation: This involves the formation of complex ions, which is not the primary mechanism in salting out.
- Option B: Common ion effect: While the common ion effect is involved, the underlying principle connecting it to solubility is the solubility product.
- Option D: Acid-base neutralization: This is not directly related to the salting-out effect of soaps.

Key Concepts:
- Solubility product (Ksp)
- Common ion effect
- Salting out
- Le Chatelier's principle

Memory Aids:
- Salting out: Adding salt reduces solubility.

Quick Tips:
- Remember that salting out is a consequence of the common ion effect and the solubility product principle.

74 Easy

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Understanding +4 | -1

The formation of

\mathrm{S}{\mathrm{O}_3}

takes place according to the following reaction,

2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2}\leftrightharpoons2\mathrm{S}{\mathrm{O}_3},\;\Delta H = - 45.2\;\mathrm{kcal}

. The formation of

\mathrm{S}{\mathrm{O}_3}

is favoured by

View answer and explanation

Correct option: B

Increase in pressure

Option B (Increase in pressure) is correct because, according to Le Chatelier's principle, increasing the pressure favors the side with fewer moles of gas, which is the product side in this reaction.

Solution approach

Correct Answer + Core Reason: Increase in pressure. Increasing pressure favors the side with fewer moles of gas.

Step-by-Step Solution:

Step 1: Analyze the given reaction: $2\mathrm{S}{\mathrm{O}_2} + {\mathrm{O}_2} \rightleftharpoons 2\mathrm{S}{\mathrm{O}_3}, \Delta H = - 45.2\mathrm{kcal}$

Step 2: Determine the number of moles of gas on each side of the equation:
- Reactant side: 2 moles of $SO_2$ + 1 mole of $O_2$ = 3 moles of gas
- Product side: 2 moles of $SO_3$ = 2 moles of gas

Step 3: Apply Le Chatelier's principle: Increasing the pressure will shift the equilibrium towards the side with fewer moles of gas, which is the product side (2 moles of $SO_3$ ).

Step 4: Since the reaction is exothermic (ΔH < 0), decreasing the temperature will also favor the formation of $SO_3$ . However, the options only address pressure and volume changes directly.

Why other options are wrong:
- Option A: Increase of volume is incorrect because it favors the side with more moles of gas (reactants).
- Option C: Increase in temperature is incorrect because the reaction is exothermic, so increasing temperature favors the reactants.
- Option D: Removal of oxygen is incorrect because it shifts the equilibrium to the left, decreasing the formation of $SO_3$ .

Key Concepts:
- Le Chatelier's principle
- Effect of pressure on equilibrium
- Effect of temperature on equilibrium

Memory Aids:
- Increasing pressure favors the side with fewer moles of gas.
- For exothermic reactions, decreasing temperature favors the products.

Quick Tips:
- Apply Le Chatelier's principle to predict the effect of changes in pressure, temperature, and concentration on equilibrium.

75 Easy

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Analyzing +4 | -1

The

\mathrm{pH}

of

\mathrm{0}\mathrm{.1}\;\mathrm{M}

solution of the following salts increases in the order

View answer and explanation

Correct option: B

\mathrm{HCl} < N{\mathrm{H}_4}\mathrm{Cl} < NaCl < NaCN

Option B (

\mathrm{HCl} < N{\mathrm{H}_4}\mathrm{Cl} < NaCl < NaCN

) is correct because it represents the increasing order of pH from acidic to basic salts.

Solution approach

Correct Answer + Core Reason: $\mathrm{HCl} < N{\mathrm{H}_4}\mathrm{Cl} < NaCl < NaCN$ is the correct order because it reflects the increasing basicity of the solutions.

Step-by-Step Solution:

Step 1: Analyze each salt solution:
- $\mathrm{HCl}$ : Strong acid, so it has the lowest pH.
- $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ : Salt of a weak base ( $\mathrm{N}{\mathrm{H}_3}$ ) and a strong acid ( $\mathrm{HCl}$ ), so it is acidic.
- $\mathrm{NaCl}$ : Salt of a strong acid ( $\mathrm{HCl}$ ) and a strong base ( $\mathrm{NaOH}$ ), so it is neutral (pH = 7).
- $\mathrm{NaCN}$ : Salt of a strong base ( $\mathrm{NaOH}$ ) and a weak acid ( $\mathrm{HCN}$ ), so it is basic.

Step 2: Arrange the pH in increasing order: Acidic < Neutral < Basic. Therefore, $\mathrm{HCl} < N{\mathrm{H}_4}\mathrm{Cl} < NaCl < NaCN$ .

Why other options are wrong:
- Option A: $\mathrm{NaCl} < \mathrm{N}{\mathrm{H}_4}\mathrm{Cl} < NaCN < HCl$ : Incorrect because HCl is the most acidic and should have the lowest pH.
- Option C: $\mathrm{NaCN} < \mathrm{N}{\mathrm{H}_4}\mathrm{Cl} < NaCl < HCl$ : Incorrect because HCl is the most acidic and NaCN is the most basic.
- Option D: $\mathrm{HCl} < NaCl < NaCN < \mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ : Incorrect because $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ is acidic and NaCN is basic.

Key Concepts:
- pH scale
- Acidic, basic, and neutral salts
- Hydrolysis of salts

Memory Aids:
- Salts of strong acids and weak bases are acidic.
- Salts of strong bases and weak acids are basic.
- Salts of strong acids and strong bases are neutral.

Quick Tips:
- Identify the acid and base that form the salt to determine its acidity or basicity.

76 Easy

76 of 144

Remembering +4 | -1

The Henderson Hasselbalch equation explains the relationship between

View answer and explanation

Correct option: C

\mathrm{pH}\text{ and }\mathrm{pKa}

Option C (

\mathrm{pH}\text{ and }\mathrm{pKa}

) is correct because the Henderson-Hasselbalch equation directly relates pH to pKa and the ratio of conjugate base and acid concentrations.

Solution approach

Correct Answer + Core Reason: $\mathrm{pH}\text{ and }\mathrm{pKa}$ . The Henderson-Hasselbalch equation directly relates pH and pKa.

Step-by-Step Solution:

Step 1: Recall the Henderson-Hasselbalch equation.

Step 2: The Henderson-Hasselbalch equation is: $\mathrm{pH} = \mathrm{pKa} + \log \frac{{\left[ {\mathrm{A^-}} \right]}}{{\left[ {\mathrm{HA}} \right]}}$

Step 3: From the equation, it's clear that the equation explains the relationship between pH and pKa.

Why other options are wrong:
- Option A: $\mathrm{pH} \text{ and }\mathrm{pOH}$ : While related ( $\mathrm{pH} + \mathrm{pOH} = 14$ ), the Henderson-Hasselbalch equation doesn't directly explain their relationship.
- Option B: $\mathrm{pH}\text{ and }\log \mathrm{Ka}$ : The equation uses pKa, not log Ka directly.
- Option D: $\mathrm{pOH}\text{ and }\mathrm{pKa}$ : The Henderson-Hasselbalch equation in its standard form relates pH to pKa, not pOH to pKa. A modified version can relate pOH to pKb.

Key Concepts:
- Henderson-Hasselbalch equation
- pH
- pKa

Memory Aids:
- Henderson-Hasselbalch: pH = pKa + log([A-]/[HA])

Quick Tips:
- Know the Henderson-Hasselbalch equation and its components.

77 Easy

77 of 144

Understanding +4 | -1

What is the correct representation for the solubility product of

\mathrm{Sn}{\mathrm{S}_2}

?

View answer and explanation

Correct option: B

\left[ {\mathrm{S}{\mathrm{n}^{4 + }}} \right]{\left[ {{\mathrm{S}^{2 - }}} \right]^2}

Option B (

{\mathrm{S}{\mathrm{n}^{4 + }}} {{\left[ {{\mathrm{S}^{2 - }}} \right]}^2}

) is correct because it accurately represents the solubility product expression for

\mathrm{Sn}{\mathrm{S}_2}

considering the correct stoichiometry and ion charges.

Solution approach

Correct Answer + Core Reason: $\left[ {\mathrm{S}{\mathrm{n}^{4 + }}} \right]{\left[ {{\mathrm{S}^{2 - }}} \right]^2}$ . The solubility product expression is determined by the stoichiometry of the dissolution reaction.

Step-by-Step Solution:

Step 1: Write the dissolution reaction for $\mathrm{Sn}{\mathrm{S}_2}$ :
$\mathrm{Sn}{\mathrm{S}_2}(s) \rightleftharpoons \mathrm{Sn}^{4+}(aq) + 2{\mathrm{S}^{2-}}(aq)$

Step 2: Define the solubility product, $K_{sp}$ , as the product of the ion concentrations raised to the power of their stoichiometric coefficients:
$K_{sp} = \left[ {\mathrm{S}{\mathrm{n}^{4 + }}} \right]{\left[ {{\mathrm{S}^{2 - }}} \right]^2}$

Step 3: The correct representation for the solubility product of $\mathrm{Sn}{\mathrm{S}_2}$ is $\left[ {\mathrm{S}{\mathrm{n}^{4 + }}} \right]{\left[ {{\mathrm{S}^{2 - }}} \right]^2}$ .

Why other options are wrong:
- Option A: $\left[ {\mathrm{S}{\mathrm{n}^{2 + }}} \right]{\left[ {{\mathrm{S}^{2 - }}} \right]^2}$ : This is incorrect because it uses the wrong charge for tin (Sn). Tin exists as Sn $^{4+}$ in this compound, not Sn $^{2+}$ .
- Option C: $\left[ {\mathrm{S}{\mathrm{n}^{2 + }}} \right]\left[ {2{\mathrm{S}^{2 - }}} \right]$ : This is incorrect because it uses the wrong charge for tin and incorrectly represents the concentration of sulfide ions. The concentration of sulfide should be squared, not multiplied by 2 inside the brackets.
- Option D: $\left[ {\mathrm{S}{\mathrm{n}^{4 + }}} \right]{\left[ {2{\mathrm{S}^{2 - }}} \right]^2}$ : This is incorrect because it incorrectly represents the concentration of sulfide ions. The stoichiometric coefficient should be applied as an exponent to the concentration, not by multiplying the concentration inside the brackets.

Key Concepts:
- Solubility product ( $K_{sp}$ )
- Dissolution equilibrium
- Stoichiometry

Memory Aids:
- Remember to raise the ion concentrations to the power of their stoichiometric coefficients in the balanced dissolution equation.

Quick Tips:
- Always write the balanced dissolution equation first to determine the correct stoichiometry.
- Ensure the charges of the ions are correct when writing the solubility product expression.

78 Easy

78 of 144

Understanding +4 | -1

{P^H}

of water is 7.0 at

{25^0}C

. If water is heated to

{70^0}C

, then

View answer and explanation

Correct option: D

{P^H}

will decrease but solution will be neutral

Option D (

{P^H}

will decrease but solution will be neutral) is correct because as temperature increases, the autoionization of water increases, leading to a higher concentration of both H+ and OH- ions, thus decreasing pH, but the solution remains neutral as [H+] = [OH-].

Solution approach

Correct Answer + Core Reason: ${P^H}$ will decrease but solution will be neutral. As temperature increases, the autoionization of water increases, leading to a decrease in pH, but the solution remains neutral.

Step-by-Step Solution:

Step 1: Explain the autoionization of water.\nWater undergoes autoionization, where it acts as both an acid and a base:
$H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$

Step 2: Explain the effect of temperature on autoionization.\nThe autoionization of water is an endothermic process. Therefore, as temperature increases, the equilibrium shifts to the right, increasing the concentrations of both $H^+$ and $OH^-$ ions.

Step 3: Explain the effect on pH.\nSince $pH = -log[H^+]$ , an increase in $[H^+]$ leads to a decrease in pH.

Step 4: Explain the effect on neutrality.\nEven though the pH decreases, the solution remains neutral because $[H^+] = [OH^-]$ . The neutrality condition is that the concentrations of hydrogen and hydroxide ions are equal.

Why other options are wrong:
- Option A: ${P^H}$ will decrease and solution become acidic is incorrect because the solution remains neutral even though the pH decreases.
- Option B: ${P^H}$ will increase is incorrect because the pH decreases with increasing temperature.
- Option C: ${P^H}$ will remain constant as 7 is incorrect because the pH changes with temperature.

Key Concepts:
- Autoionization of water
- pH
- Temperature dependence of equilibrium
- Neutrality

Memory Aids:
- Higher temperature, higher autoionization, lower pH, but still neutral.

Quick Tips:
- Remember that neutrality is defined by $[H^+] = [OH^-]$ , not by pH = 7.

79 Easy

79 of 144

Analyzing +4 | -1

Assertion : The equilibrium constant for the reaction

\begin{aligned} & {CaSO}_4 \cdot 5 {H}_2 {O}({s})={CaSO}_4 \cdot 3 {H}_2 {O}({s})+2 {H}_2 {O}({g}) \\ & \text { is } {K}_{{C}}=\frac{\left[{CaSO}_4 \cdot 3 {H}_2 {O}\right]\left[{H}_2 {O}\right]^2}{\left[{CaSO}_4 \cdot 5 {H}_2 {O}\right]} \end{aligned}

Reason : Equilibrium constant is the ratio of the product of molar concentration of the substances produced to the product of the molar concentrations of reactants with each concentrations term raised to the power equal to the respective stoichiometric constant.

View answer and explanation

Correct option: D

If assertion is false but reason is true

Option D (If assertion is false but reason is true) is correct because the assertion incorrectly formulates the equilibrium constant for a heterogeneous reaction, while the reason correctly states the general definition of the equilibrium constant.

Solution approach

Correct Answer + Core Reason: If assertion is false but reason is true. The assertion incorrectly includes solid concentrations in the equilibrium constant expression, while the reason correctly defines the equilibrium constant.

Step-by-Step Solution:

Step 1: Analyze the given reaction: ${CaSO}_4 \cdot 5 {H}_2 {O}({s}) \rightleftharpoons {CaSO}_4 \cdot 3 {H}_2 {O}({s})+2 {H}_2 {O}({g})$ .

Step 2: Recall that for heterogeneous equilibria, the concentrations of pure solids and pure liquids are considered constant and are not included in the equilibrium constant expression.

Step 3: Write the correct equilibrium constant expression: ${K}_{{C}} = [{H}_2 {O}]^2$ . Only the gaseous water is included.

Step 4: Compare the given assertion with the correct expression. The assertion is incorrect because it includes the solid concentrations.

Step 5: Evaluate the reason. The reason correctly states the general definition of the equilibrium constant.

Step 6: Conclude that the assertion is false, but the reason is true.

Why other options are wrong:
- Option A: The assertion is false, so this option is incorrect.
- Option B: The assertion is false, so this option is incorrect.
- Option C: The assertion is false, so this option is incorrect.
- Option D: This is the correct answer because the assertion is false, and the reason is true.

Key Concepts:
- Equilibrium constant
- Heterogeneous equilibria
- Activity of pure solids and liquids

Memory Aids:
- Solids and liquids don't appear in Keq expressions.

Quick Tips:
- Always exclude pure solids and liquids from the equilibrium constant expression.

80 Easy

80 of 144

Analyzing +4 | -1

Which of the following is not a buffer solution?

View answer and explanation

Correct option: C

50\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ C}{\mathrm{H}_3}\mathrm{COOH}\; + \;100\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ NaOH}

Option C (

50\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ C}{\mathrm{H}_3}\mathrm{COOH}\; + \;100\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ NaOH}

) is not a buffer solution because the strong base (NaOH) is in excess, completely neutralizing the weak acid (CH3COOH) and resulting in a basic solution, not a buffer.

Solution approach

Correct Answer + Core Reason: $50\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ C}{\mathrm{H}_3}\mathrm{COOH}\; + \;100\;\mathrm{mL}\;0.1\;\mathrm{M}\;\mathrm{ NaOH}$ . This is not a buffer solution because the strong base is in excess.

Step-by-Step Solution:

Step 1: Define a buffer solution: A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid.

Step 2: Analyze each option:
- Option A: 100 mL 0.1 M CH3COOH + 50 mL 0.1 M CH3COONa: This is a mixture of a weak acid (CH3COOH) and its conjugate base (CH3COONa). It is a buffer solution.
- Option B: 100 mL 0.1 M CH3COOH + 50 mL 0.1 M NaOH: The moles of CH3COOH are greater than the moles of NaOH. The NaOH will react with the CH3COOH to form CH3COONa, leaving some CH3COOH unreacted. This will result in a mixture of CH3COOH and CH3COONa, which is a buffer solution.
- Option C: 50 mL 0.1 M CH3COOH + 100 mL 0.1 M NaOH: The moles of NaOH are greater than the moles of CH3COOH. The CH3COOH will be completely neutralized by the NaOH, and there will be excess NaOH in the solution. This is not a buffer solution.
- Option D: 100 mL 0.1 M NH4OH + 50 mL 0.1 M HCl: The moles of NH4OH are greater than the moles of HCl. The HCl will react with the NH4OH to form NH4Cl, leaving some NH4OH unreacted. This will result in a mixture of NH4OH and NH4Cl, which is a buffer solution.

Step 3: Conclusion: Option C is not a buffer solution because the strong base is in excess.

Why other options are wrong:
- Option A: This is a buffer solution.
- Option B: This is a buffer solution.
- Option D: This is a buffer solution.

Key Concepts:
- Buffer solutions
- Weak acids and bases
- Conjugate acid-base pairs
- Neutralization reactions

Memory Aids:
- A buffer solution must contain a weak acid/base and its conjugate.

Quick Tips:
- Check the moles of acid and base to determine if a buffer solution is formed.

81 Easy

81 of 144

Application +4 | -1

The

pH

of a solution is

5.0

. To this solution sufficient acid is added to decrease the

pH\;to\;2.0

. The increase in hydrogen ion concentration is:

View answer and explanation

Correct option: A

1000\;times

Option A (1000 times) is correct because a decrease in pH from 5.0 to 2.0 represents a 1000-fold increase in hydrogen ion concentration.

Solution approach

Correct Answer + Core Reason: 1000 times. The pH scale is logarithmic, so each unit decrease represents a tenfold increase in hydrogen ion concentration.

Step-by-Step Solution:

Step 1: Define pH: $pH = -log[H^+]$

Step 2: Calculate $[H^+]$ at pH 5.0: $[H^+]_1 = 10^{-5} M$

Step 3: Calculate $[H^+]$ at pH 2.0: $[H^+]_2 = 10^{-2} M$

Step 4: Calculate the increase in hydrogen ion concentration: $\frac{{[H^+]_2}}{{[H^+]_1}} = \frac{{10^{-2}}}{{10^{-5}}} = 10^3 = 1000$

Step 5: Final Answer: The increase in hydrogen ion concentration is 1000 times.

Why other options are wrong:
- Option B (5/2 times): This is incorrect because it is a simple division of the pH values and does not account for the logarithmic nature of the pH scale.
- Option C (100 times): This is incorrect because it represents a change of 2 pH units, not 3.
- Option D (5 times): This is incorrect because it is a simple subtraction of the pH values and does not account for the logarithmic nature of the pH scale.

Key Concepts:
- pH scale
- Logarithmic relationship between pH and hydrogen ion concentration

Memory Aids:
- pH = -log[H+]
- Each pH unit change represents a factor of 10 change in [H+]

Quick Tips:
- Remember that pH is a logarithmic scale.
- A decrease in pH means an increase in hydrogen ion concentration.

82 Easy

82 of 144

Analyzing +4 | -1

10\,\mathrm{mL}

of a solution contains

0.1\,M\,\mathrm{N}{\mathrm{H}_4}\mathrm{Cl} + 0.01\,M\,\mathrm{N}{\mathrm{H}_4}\mathrm{OH}\mathrm{.}

Which addition would not change the

\mathrm{pH}

of the solution?

View answer and explanation

Correct option: A

Adding

1\,\mathrm{mL}

water

Option A (Adding

1\,{\mathrm{mL}

water) is correct because adding water to a buffer solution does not change the ratio of the weak acid/base and its conjugate, thus maintaining the pH.

Solution approach

Correct Answer + Core Reason: Adding $1\,{\mathrm{mL}$ water. Adding water to a buffer solution does not change the pH because it does not change the ratio of the weak acid/base and its conjugate.

Step-by-Step Solution:

Step 1: Understand the properties of a buffer solution. A buffer solution resists changes in pH upon the addition of small amounts of acid or base, or upon dilution.

Step 2: Analyze each option:
- Adding 1 mL water: This dilutes the solution, but the ratio of [NH4Cl] to [NH4OH] remains the same. Therefore, the pH will not change significantly.
- Adding 5 mL of 0.1 M NH4Cl: This increases the concentration of the conjugate acid, which will decrease the pH.
- Adding 5 mL of 0.1 M NH4OH: This increases the concentration of the weak base, which will increase the pH.
- Adding 10 mL of 0.1 M NH4Cl: This increases the concentration of the conjugate acid, which will decrease the pH.

Step 3: Conclusion: Adding water will not significantly change the pH of the buffer solution.

Why other options are wrong:
- Option B: Adding NH4Cl will change the pH of the solution.
- Option C: Adding NH4OH will change the pH of the solution.
- Option D: Adding NH4Cl will change the pH of the solution.

Key Concepts:
- Buffer solutions
- Dilution
- pH changes

Memory Aids:
- Dilution does not change the pH of a buffer.

Quick Tips:
- Adding water to a buffer does not change the pH, but adding acid or base does.

83 Easy

83 of 144

Application +4 | -1

The percentage of pyridine

({C_5}{H_5}N)

that forms pyridinium ion

({C_5}{H_5}{N^ + }H)

in a 0.10 M aqueous pyridine solution

({K_b}\,\mathrm{for}\,{C_5}{H_5}N = 1.7 \times {10^{ - 9}})

is

View answer and explanation

Correct option: C

0.00013

Option C (0.00013) is correct because it represents the calculated percentage of pyridine that forms pyridinium ion based on the given Kb value and concentration.

Solution approach

Correct Answer + Core Reason: 0.00013. The percentage of pyridinium ion formed is calculated using the Kb expression and the given concentration.

Step-by-Step Solution:

Step 1: Write the equilibrium reaction: ${C_5}{H_5}N + H_2O \rightleftharpoons {C_5}{H_5}{N^ + }H + O{H^ - }$

Step 2: Write the Kb expression: ${K_b} = \frac{{[{C_5}{H_5}{N^ + }H][O{H^ - }]}}{{[C{}_5{H_5}N]}}$

Step 3: Let x be the concentration of ${C_5}{H_5}{N^ + }H$ and $O{H^ - }$ at equilibrium. Then, the concentration of ${C_5}{H_5}N$ at equilibrium is 0.10 - x.

Step 4: Substitute into the Kb expression: $1.7 \times {10^{ - 9}} = \frac{{{x^2}}}{{0.10 - x}}$

Step 5: Since Kb is very small, we can assume that x << 0.10, so 0.10 - x ≈ 0.10. Therefore, $1.7 \times {10^{ - 9}} = \frac{{{x^2}}}{{0.10}}$

Step 6: Solve for x: ${x^2} = 1.7 \times {10^{ - 10}}$ , so $x = \sqrt {1.7 \times {{10}^{ - 10}}} = 1.304 \times {10^{ - 5}}$

Step 7: Calculate the percentage of pyridine that forms pyridinium ion: $\frac{x}{{0.10}} \times 100 = \frac{{1.304 \times {{10}^{ - 5}}}}{{0.10}} \times 100 = 0.0001304\% \approx 0.00013\%$

Why other options are wrong:
- Option A (0.016): This is incorrect because it likely involves a miscalculation or misunderstanding of the Kb expression and the approximation made.
- Option B (0.00006): This is incorrect and likely arises from an error in the square root calculation or a different incorrect assumption.
- Option D (0.0077): This is incorrect and may result from using an incorrect formula or not taking the square root when solving for x.

Key Concepts:
- Weak base equilibrium
- Kb expression
- Approximation for small Kb values
- Percentage calculation

Memory Aids:
- Kb is the base dissociation constant.
- Small Kb values imply weak bases.

Quick Tips:
- Always check if the approximation (x << initial concentration) is valid.
- Remember to convert the concentration to percentage at the end.

84 Easy

84 of 144

Analyzing +4 | -1

The equilibrium constant,

K

for the reaction

2\mathrm{HI}\left( \mathrm{g} \right)\;\leftrightharpoons\;{\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)

At room temperature is 2.85 and that at

698\;K

, it is

1.4 \times {10^{ - 2}}

. This implies that

View answer and explanation

Correct option: D

\mathrm{HI}

is less stable than

{\mathrm{H}_2}

and

{\mathrm{I}_2}

Option D is correct because the equilibrium constant decreases with increasing temperature, indicating that the reverse reaction (formation of HI) is exothermic, meaning HI is less stable than H2 and I2.

Solution approach

Correct Answer + Core Reason: $\mathrm{HI}$ is less stable than ${\mathrm{H}_2}$ and ${\mathrm{I}_2}$ . A decreasing equilibrium constant with increasing temperature indicates an exothermic reverse reaction, implying HI is less stable.

Step-by-Step Solution:

Step 1: Analyze the given information. The equilibrium constant, $K$ , decreases as the temperature increases.

Step 2: Recall the relationship between the equilibrium constant and temperature. If $K$ decreases with increasing temperature, the reverse reaction is favored by higher temperatures. This means the reverse reaction is exothermic.

Step 3: Relate the exothermic reverse reaction to the stability of HI. Since the reverse reaction is exothermic (releases heat), it means that HI has higher energy compared to H2 and I2. Therefore, HI is less stable than H2 and I2.

Why other options are wrong:
- Option A: $\mathrm{HI}$ is resonance stabilised: Resonance stabilization would generally increase stability, which contradicts the observed temperature dependence of the equilibrium constant.
- Option B: $\mathrm{HI}$ is exothermic compound: The formation of HI from H2 and I2 is endothermic, not exothermic. The reverse reaction is exothermic.
- Option C: $\mathrm{HI}$ is stable at room temperature: The change in K with temperature indicates that HI is not particularly stable, especially at higher temperatures.

Key Concepts:
- Equilibrium constant and temperature relationship
- Exothermic and endothermic reactions
- Stability and energy relationship

Memory Aids:
- Le Chatelier's principle: A system at equilibrium will shift to relieve stress (in this case, temperature change).
- Exothermic reactions release heat; endothermic reactions absorb heat.

Quick Tips:
- When the equilibrium constant decreases with increasing temperature, the reverse reaction is exothermic.

85 Easy

85 of 144

Understanding +4 | -1

Which of the following molecules acts as a Lewis acid?

View answer and explanation

Correct option: B

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{B}

Option B (

( {\mathrm{C}{\mathrm{H}_3}} )_3\mathrm{B}

) is correct because boron in trimethylborane has an incomplete octet and can accept a pair of electrons, thus acting as a Lewis acid.

Solution approach

Correct Answer + Core Reason: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{B}$ . Boron has an incomplete octet and accepts electron pairs.

Step-by-Step Solution:

Step 1: Recall the definition of a Lewis acid.

Step 2: A Lewis acid is a substance that can accept an electron pair.

Step 3: Analyze each option:
- ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{N}$ has a lone pair on nitrogen and acts as a Lewis base.
- ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{B}$ has an incomplete octet on boron and can accept an electron pair, acting as a Lewis acid.
- ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{O}$ has lone pairs on oxygen and acts as a Lewis base.
- ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{P}$ has a lone pair on phosphorus and acts as a Lewis base.

Step 4: Therefore, ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{B}$ acts as a Lewis acid.

Why other options are wrong:
- Option A: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{N}$ acts as a Lewis base because nitrogen has a lone pair of electrons.
- Option B: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{B}$ is the correct answer because boron has an incomplete octet and can accept an electron pair.
- Option C: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{O}$ acts as a Lewis base because oxygen has lone pairs of electrons.
- Option D: ${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{P}$ acts as a Lewis base because phosphorus has a lone pair of electrons.

Key Concepts:
- Lewis acids and bases
- Electron-pair acceptance
- Incomplete octet

Memory Aids:
- Boron is electron deficient, so it's a Lewis acid.

Quick Tips:
- Look for molecules with incomplete octets or positive charges to identify Lewis acids.

86 Easy

86 of 144

Analyzing +4 | -1

Aqueous solution of which salt has the lowest

pH

?

View answer and explanation

Correct option: B

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

Option B (

\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

) is correct because it is a salt of a weak base and a strong acid, leading to hydrolysis and a lower pH.

Solution approach

Correct Answer + Core Reason: $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ . Salts of weak bases and strong acids produce acidic solutions.

Step-by-Step Solution:

Step 1: Identify the nature of each salt.

Step 2: NaOH is a strong base, so its aqueous solution will have a high pH.

Step 3: $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ is a salt of a weak base ( $\mathrm{N}{\mathrm{H}_4}\mathrm{OH}$ ) and a strong acid (HCl). It undergoes hydrolysis to produce $\mathrm{H}^+$ ions, lowering the pH.

Step 4: $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$ is a salt of a strong base (NaOH) and a weak acid ( $\mathrm{H}_2\mathrm{CO}_3$ ). It undergoes hydrolysis to produce $\mathrm{OH}^-$ ions, increasing the pH.

Step 5: NaCl is a salt of a strong acid (HCl) and a strong base (NaOH). It does not undergo hydrolysis, so its pH is approximately 7.

Step 6: Compare the pH values. Strong bases have the highest pH, followed by salts of strong bases and weak acids. Salts of weak bases and strong acids have the lowest pH, and neutral salts have a pH around 7.

Step 7: Conclusion: $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ will have the lowest pH.

Why other options are wrong:
- Option A: NaOH is a strong base and will have a high pH.
- Option C: $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$ is a salt of a strong base and a weak acid and will have a pH greater than 7.
- Option D: NaCl is a salt of a strong acid and a strong base and will have a pH around 7.

Key Concepts:
- Salt hydrolysis
- Acidic, basic, and neutral salts
- pH scale

Memory Aids:
- Strong Acid + Strong Base = Neutral Salt
- Strong Acid + Weak Base = Acidic Salt
- Weak Acid + Strong Base = Basic Salt

Quick Tips:
- Remember the rules for salt hydrolysis to quickly determine the pH of salt solutions.

87 Easy

87 of 144

Analyzing +4 | -1

STATEMENT-1 : The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature.

STATEMENT-2 : When a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of change

View answer and explanation

Correct option: D

If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Option D (If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE) is correct. Endothermic reactions are favored at higher temperatures, and exothermic reactions are favored at lower temperatures. STATEMENT-2 correctly describes Le Chatelier's principle.

Solution approach

Correct Answer + Core Reason: If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE. Endothermic favored at high T, exothermic at low T. Statement 2 is Le Chatelier's.

Step-by-Step Solution:

Step 1: STATEMENT-1: The endothermic reactions are favored at lower temperature and the exothermic reactions are favored at higher temperature. This is incorrect. Endothermic reactions are favored at higher temperatures, and exothermic reactions are favored at lower temperatures.

Step 2: STATEMENT-2: When a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of change. This is a statement of Le Chatelier's principle, which is correct.

Step 3: Therefore, STATEMENT-1 is FALSE, and STATEMENT-2 is TRUE.

Why other options are wrong:
- Option A: STATEMENT-1 is false.
- Option B: STATEMENT-1 is false.
- Option C: STATEMENT-1 is false, and STATEMENT-2 is true, but it is not the correct explanation.

Key Concepts:
- Le Chatelier's principle
- Effect of temperature on equilibrium
- Endothermic and exothermic reactions

Memory Aids:
- Remember that endothermic reactions absorb heat and are favored at high temperatures, while exothermic reactions release heat and are favored at low temperatures.

Quick Tips:
- Always remember the relationship between temperature and the type of reaction (endothermic or exothermic) when applying Le Chatelier's principle.

88 Easy

88 of 144

Analyzing +4 | -1

The equilibrium constant for the reaction

\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})

is

K_1

and the equilibrium constant for the reaction

\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g})

is

K_2

both at the same temperature. The value of

K_1

and

K_2

are related as

View answer and explanation

Correct option: A

K_1=\left(\frac{1}{K_2}\right)^2

Option A (

K_1=\left(\frac{1}{K_2}\right)^2

) is correct because the second reaction is the reverse of the first reaction divided by 2, so their equilibrium constants are related by the inverse square.

Solution approach

Correct Answer + Core Reason: $K_1=\left(\frac{1}{K_2}\right)^2$ . The equilibrium constants are related because the second reaction is the reverse of the first reaction divided by 2.

Step-by-Step Solution:

Step 1: Write the equilibrium constant for the first reaction: $\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$
$K_1 = \frac{[NO]^2}{[N_2][O_2]}$

Step 2: Write the equilibrium constant for the second reaction: $\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{N}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g})$
$K_2 = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}$

Step 3: Relate $K_1$ and $K_2$ : Notice that $K_2 = \frac{1}{\sqrt{K_1}}$ . Squaring both sides, we get $K_2^2 = \frac{1}{K_1}$ . Therefore, $K_1 = \frac{1}{K_2^2} = \left(\frac{1}{K_2}\right)^2$ .

Why other options are wrong:
- Option B: $K_1=K_2^2$ : This is incorrect because the relationship is inverse and squared.
- Option C: $K_2=\left(\frac{1}{K_1}\right)^2$ : This is incorrect because the powers are on the wrong side.
- Option D: $K_2=K_1^2$ : This is incorrect because the relationship is inverse and squared.

Key Concepts:
- Equilibrium Constant
- Relationship between equilibrium constants for forward and reverse reactions

Memory Aids:
- When a reaction is reversed, the new equilibrium constant is the inverse of the original. When a reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power of n.

Quick Tips:
- If a reaction is reversed, the new equilibrium constant is the inverse of the original. If a reaction is multiplied by a factor n, the new equilibrium constant is the original raised to the power of n.

89 Easy

89 of 144

Application +4 | -1

In the equilibrium,

AB\;\leftrightharpoons\;A + B,

if the equilibrium concentration of

A

is doubled, the equilibrium concentration of

B

would become

View answer and explanation

Correct option: A

Half

Option A (Half) is correct because, according to the equilibrium constant expression, if [A] is doubled, [B] must be halved to maintain the constant value of K.

Solution approach

Correct Answer + Core Reason: Half. The equilibrium constant expression dictates the relationship between the concentrations of A and B.

Step-by-Step Solution:

Step 1: Write the equilibrium constant expression: $K = \frac{{[A][B]}}{{[AB]}}$

Step 2: Let the initial equilibrium concentrations be $[A]_1$ , $[B]_1$ , and $[AB]_1$ .\nSo, $K = \frac{{[A]_1[B]_1}}{{[AB]_1}}$

Step 3: Now, the equilibrium concentration of A is doubled, so $[A]_2 = 2[A]_1$ . Let the new equilibrium concentration of B be $[B]_2$ .\nThen, $K = \frac{{[A]_2[B]_2}}{{[AB]_2}} = \frac{{2[A]_1[B]_2}}{{[AB]_2}}$

Step 4: Since K is constant, we can equate the two expressions:
$\frac{{[A]_1[B]_1}}{{[AB]_1}} = \frac{{2[A]_1[B]_2}}{{[AB]_2}}$

Step 5: Since the amount of AB will also change to re-establish equilibrium, we can consider a simplified scenario where the change in AB is negligible for approximation. In this case, to maintain K constant, if [A] is doubled, [B] must be halved.
$[B]_2 = \frac{{[B]_1}}{2}$

Why other options are wrong:
- Option B: If [B] were doubled, the equilibrium constant would not remain constant when [A] is also doubled.
- Option C: If [B] became 1/4th, the equilibrium constant would change significantly when [A] is doubled.
- Option D: If [B] became 1/8th, the equilibrium constant would change even more significantly when [A] is doubled.

Key Concepts:
- Equilibrium Constant
- Le Chatelier's Principle

Memory Aids:
- The equilibrium constant K remains constant at a given temperature.

Quick Tips:
- When one concentration changes, the others must adjust to keep K constant.

90 Easy

90 of 144

Application +4 | -1

Ionisation constant of

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

is

1.7 \times {10^{ - 5}}

and

\left[ {{\mathrm{H}^ + }} \right]

ions is

3.4 \times {10^{ - 4}}

. Then, initial concentration of

\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

molecules is

View answer and explanation

Correct option: A

6.8 \times {10^{ - 3}}

Option A is correct because the initial concentration of acetic acid can be calculated using the ionization constant and the hydrogen ion concentration.

Solution approach

Correct Answer + Core Reason: $6.8 \times {10^{ - 3}}$ . The initial concentration is calculated using the ionization constant expression.

Step-by-Step Solution:

Step 1: Write the ionization equilibrium for acetic acid: $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH} \rightleftharpoons \mathrm{C}{\mathrm{H}_3}{\mathrm{COO}^ - } + {\mathrm{H}^ + }$

Step 2: Write the expression for the ionization constant, $K_a$ : $K_a = \frac{{\left[ {{\mathrm{H}^ + }} \right]\left[ {\mathrm{C}{\mathrm{H}_3}{\mathrm{COO}^ - }} \right]}}{{\left[ {\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}} \right]}}$

Step 3: Assume that the initial concentration of $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ is $C$ . At equilibrium, $\left[ {{\mathrm{H}^ + }} \right] = \left[ {\mathrm{C}{\mathrm{H}_3}{\mathrm{COO}^ - }} \right] = 3.4 \times {10^{ - 4}}$ .

Step 4: Substitute the values into the $K_a$ expression: $1.7 \times {10^{ - 5}} = \frac{{(3.4 \times {10^{ - 4}})(3.4 \times {10^{ - 4}})}}{{C - 3.4 \times {10^{ - 4}}}}$

Step 5: Since $K_a$ is small, we can approximate $C - 3.4 \times {10^{ - 4}} \approx C$ . Therefore, $1.7 \times {10^{ - 5}} = \frac{{(3.4 \times {10^{ - 4}})^2}}{C}$

Step 6: Solve for $C$ : $C = \frac{{(3.4 \times {10^{ - 4}})^2}}{{1.7 \times {10^{ - 5}}}} = \frac{{11.56 \times {10^{ - 8}}}}{{1.7 \times {10^{ - 5}}}} = 6.8 \times {10^{ - 3}}$

Why other options are wrong:
- Option B: $2.5 \times {10^{ - 4}}$ is incorrect. This value might be obtained from an incorrect calculation or misunderstanding of the equilibrium expression.
- Option C: $3.5 \times {10^{ - 3}}$ is incorrect. This value might be obtained from an incorrect calculation or misunderstanding of the equilibrium expression.
- Option D: $4.5 \times {10^{ - 3}}$ is incorrect. This value might be obtained from an incorrect calculation or misunderstanding of the equilibrium expression.

Key Concepts:
- Weak acid ionization
- Ionization constant ( $K_a$ )
- Equilibrium calculations

Memory Aids:
- $K_a = \frac{{\left[ {{\mathrm{H}^ + }} \right]\left[ {{\mathrm{A}^ - }} \right]}}{{\left[ {\mathrm{HA}} \right]}}$

Quick Tips:
- For weak acids, the change in concentration is often negligible compared to the initial concentration, simplifying the calculation.

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