The $pH$ of an aqueous solution of ${\rm{C}}{{\rm{H}}_3}{\rm{COONa}}$ of concentration $C\left( {\rm{M}} \right)$ is given by

$7 - \frac{1}{2}p{K_a} - \frac{1}{2}\log C$

$\frac{1}{2}p{K_w} + \frac{1}{2}p{K_b} + \frac{1}{2}\log C$

$\frac{1}{2}p{K_w} - \frac{1}{2}p{K_b} - \frac{1}{2}\log C$

Of the following which change will shift the reaction towards the product? ${{\rm{I}}_2}\left( {\rm{g}} \right)\;\leftrightharpoons\;2{\rm{I}}\left( {\rm{g}} \right),\;\;\Delta H{^\circ _r}\left( {298\;{\rm{K}}} \right) = + 150\;{\rm{kJ}}$

Increase in concentration of ${\rm{I}}$

Decrease in concentration of ${{\rm{I}}_2}$

For the reaction${{\rm{H}}_2}\left( {\rm{g}} \right) + {{\rm{I}}_2}\left( {\rm{g}} \right)\leftrightharpoons2{\rm{HI}}\left( {\rm{g}} \right)$, the equilibrium constants expressed in terms of concentrations ${K_c}$ and in terms of partial pressure ${K_p}$, are related as

${K_p} = {K_c}{\left( {RT} \right)^2}$

${K_p} = {K_c}{\left( {RT} \right)^{ - 2}}$

${K_c} = {K_p}\left( {RT} \right)$

If $0.1\;{\rm{mole}}$ of ${{\rm{I}}_2}$ is introduced into $1.0\;{\rm{litre}}$ flask at $1000\;{\rm{K}},$ at equilibrium ${K_c} = {10^{ - 6}}$, which one is correct?

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] > \left[ {I\left( g \right)} \right]$

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] < \left[ {I\left( g \right)} \right]$

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] = \left[ {{\rm{I}}\left( {\rm{g}} \right)} \right]$

$\left[ {{{\rm{I}}_2}\left( {\rm{g}} \right)} \right] = \frac{1}{2}\left[ {{\rm{I}}\left( {\rm{g}} \right)} \right]$

Equilibrium MCQs for NEET UG CHEMISTRY — Easy, Page 5

121 Easy

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Remembering +4 | -1

The solvent which is neither proton donor nor proton acceptor is called:

View answer and explanation

Correct option: C

Aprotic

Option C (Aprotic) is correct because aprotic solvents neither donate nor accept protons, unlike protic or amphoteric solvents.

Solution approach

Correct Answer + Core Reason: Aprotic. Aprotic solvents do not donate or accept protons.

Step-by-Step Solution:

Step 1: Define the terms:
- Protonic solvent: A solvent that can donate a proton (e.g., water, alcohols).
- Aprotic solvent: A solvent that cannot donate a proton (e.g., acetone, DMSO).
- Amphoteric solvent: A solvent that can act as both a proton donor and a proton acceptor (e.g., water).
- Neutral solvent: A solvent that is neither acidic nor basic.

Step 2: Identify the correct option: The solvent which is neither a proton donor nor a proton acceptor is called aprotic.

Why other options are wrong:
- Option A: Amphoteric solvents can act as both proton donors and acceptors.
- Option B: Neutral solvents may still be able to donate or accept protons.
- Option D: Protonic solvents can donate protons.

Key Concepts:
- Protonic, aprotic, and amphoteric solvents
- Proton donation and acceptance

Memory Aids:
- Aprotic = "a" (not) + protic (proton donor).

Quick Tips:
- Know examples of common protic and aprotic solvents.

122 Easy

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Application +4 | -1

The

pH

of an aqueous solution of

\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}

of concentration

C\left( \mathrm{M} \right)

is given by

View answer and explanation

Correct option: D

\frac{1}{2}p{K_w} + \frac{1}{2}p{K_a} + \frac{1}{2}\log C

Option D is correct because the pH of a salt of a weak acid and a strong base is given by the formula pH = 1/2 pKw + 1/2 pKa + 1/2 log C.

Solution approach

Correct Answer + Core Reason: $\frac{1}{2}p{K_w} + \frac{1}{2}p{K_a} + \frac{1}{2}\log C$ . This formula correctly calculates the pH of a salt of a weak acid and strong base.

Step-by-Step Solution:

Step 1: Identify the type of salt. $\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}$ is the salt of a weak acid ( $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ ) and a strong base (NaOH).

Step 2: Recall the formula for the pH of such a salt: $pH = \frac{1}{2}[p{K_w} + p{K_a} + \log C]$

Step 3: The formula directly gives the pH in terms of $p{K_w}$ , $p{K_a}$ , and the concentration $C$ .

Step 4: No numerical substitution is needed as the question asks for the formula itself.

Why other options are wrong:
- Option A: $7 - \frac{1}{2}p{K_a} - \frac{1}{2}\log C$ is incorrect because it's missing the $p{K_w}$ term and has incorrect signs.
- Option B: $\frac{1}{2}p{K_w} + \frac{1}{2}p{K_b} + \frac{1}{2}\log C$ is incorrect because it uses $p{K_b}$ instead of $p{K_a}$ .
- Option C: $\frac{1}{2}p{K_w} - \frac{1}{2}p{K_b} - \frac{1}{2}\log C$ is incorrect because it uses $p{K_b}$ instead of $p{K_a}$ and has incorrect signs.

Key Concepts:
- Salt hydrolysis
- pH calculation of salt solutions
- Weak acid/strong base salts

Memory Aids:
- Remember the formula for pH of salt of weak acid and strong base: $pH = \frac{1}{2}[p{K_w} + p{K_a} + \log C]$

Quick Tips:
- Identify the type of salt first (weak acid/strong base, strong acid/weak base, etc.) before applying the appropriate formula.

123 Easy

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Understanding +4 | -1

Which of the following is a wrong statement about equilibrium state?

View answer and explanation

Correct option: C

Catalysts increase value of equilibrium constant

Option C is correct because catalysts affect the rate of reaction but do not change the equilibrium constant.

Solution approach

Correct Answer + Core Reason: Catalysts increase value of equilibrium constant. Catalysts only affect the rate of reaction, not the equilibrium constant.

Step-by-Step Solution:

Step 1: Understand the role of a catalyst.

Step 2: Catalysts speed up both forward and reverse reactions equally.

Step 3: Equilibrium constant depends only on temperature.

Step 4: Catalysts do not shift the equilibrium position.

Why other options are wrong:
- Option A: Rate of forward reaction= Rate of backward reaction. This is a correct statement about equilibrium.
- Option B: Equilibrium is dynamic. This is a correct statement about equilibrium.
- Option D: Free energy change is zero. This is a correct statement about equilibrium.

Key Concepts:
- Chemical Equilibrium
- Catalysis
- Equilibrium Constant

Memory Aids:
- Catalysts speed up reactions but don't change equilibrium.

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

124 Easy

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Understanding +4 | -1

Which salt will give maximum pH in aqueous solution :

View answer and explanation

Correct option: C

C{H_3}COONa

Option C (

C{H_3}COONa

) is correct because it is a salt of a weak acid and a strong base, which will undergo hydrolysis to produce a basic solution with a higher pH compared to the other options.

Solution approach

Correct Answer + Core Reason: $C{H_3}COONa$ . This is a salt of a weak acid and a strong base, leading to hydrolysis and a higher pH.

Step-by-Step Solution:

Step 1: Analyze the nature of each salt.

Step 2:
- $BaC{l_2}$ : Salt of a strong acid (HCl) and a strong base (Ba(OH)2). It will not hydrolyze and will have a neutral pH.
- NaCl: Salt of a strong acid (HCl) and a strong base (NaOH). It will not hydrolyze and will have a neutral pH.
- $C{H_3}COONa$ : Salt of a weak acid ( $C{H_3}COOH$ ) and a strong base (NaOH). It will hydrolyze to produce $OH^-$ ions, resulting in a basic solution.
- $N{H_4}Cl$ : Salt of a strong acid (HCl) and a weak base ( $N{H_4}OH$ ). It will hydrolyze to produce $H^+$ ions, resulting in an acidic solution.

Step 3: Since $C{H_3}COONa$ hydrolyzes to produce a basic solution, it will have the maximum pH.

Why other options are wrong:
- Option A: $BaC{l_2}$ : This salt is formed from a strong acid and a strong base, so it will have a neutral pH.
- Option B: NaCl: This salt is formed from a strong acid and a strong base, so it will have a neutral pH.
- Option D: $N{H_4}Cl$ : This salt is formed from a weak base and a strong acid, so it will have an acidic pH.

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

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

Quick Tips:
- Determine the strength of the acid and base that form the salt to predict the pH of the solution.

125 Easy

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Understanding +4 | -1

If

1\;M\;\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}

is added to

1\;M\;\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}

View answer and explanation

Correct option: A

\mathrm{pH}

of the solution increases

Option A (

\mathrm{pH}

of the solution increases) is correct because adding CH3COONa, a salt of a weak acid, to CH3COOH, a weak acid, creates a buffer solution. The addition of the salt increases the concentration of the conjugate base, shifting the equilibrium and increasing the pH.

Solution approach

Correct Answer + Core Reason: $\mathrm{pH}$ of the solution increases. Adding the salt of a weak acid to the weak acid increases the pH.

Step-by-Step Solution:

Step 1: Identify the components.
$\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ is a weak acid (acetic acid).
$\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}$ is the salt of a weak acid (sodium acetate).

Step 2: Explain the effect of adding the salt.\nAdding $\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}$ to $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ creates a buffer solution. A buffer solution resists changes in pH upon addition of small amounts of acid or base.

Step 3: Explain the buffer action.\nThe buffer solution contains a weak acid ( $\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}$ ) and its conjugate base ( $\mathrm{C}{\mathrm{H}_3}\mathrm{COO^-}$ from $\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}$ ). The equilibrium is:
$\mathrm{C}{\mathrm{H}_3}\mathrm{COOH}(aq) \rightleftharpoons H^+(aq) + \mathrm{C}{\mathrm{H}_3}\mathrm{COO^-}(aq)$ \nAdding $\mathrm{C}{\mathrm{H}_3}\mathrm{COONa}$ increases the concentration of ${\mathrm{C}{\mathrm{H}_3}\mathrm{COO^-}}$ , which shifts the equilibrium to the left, decreasing the concentration of $H^+$ and increasing the pH.

Step 4: Determine the correct answer.\nTherefore, the pH of the solution increases.

Why other options are wrong:
- Option B: $\mathrm{pH}$ decreases is incorrect because the addition of the salt increases the pH.
- Option C: $\mathrm{pH}$ does not change is incorrect because the pH changes due to the buffer action.
- Option D: None of the above is incorrect because option A is correct.

Key Concepts:
- Buffer solution
- Weak acid
- Conjugate base
- pH
- Equilibrium

Memory Aids:
- Adding the salt of a weak acid to the weak acid increases the pH.

Quick Tips:
- Recognize buffer solutions and their effect on pH.

126 Easy

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Remembering +4 | -1

Methyl orange is an indicator in acid-alkali titration. It gives:

View answer and explanation

Correct option: D

Yellow colour in alkaline medium and red colour in acid medium

Option D (Yellow colour in alkaline medium and red colour in acid medium) is correct because methyl orange exhibits different colors in acidic and alkaline solutions.

Solution approach

Correct Answer + Core Reason: Yellow colour in alkaline medium and red colour in acid medium. This is a known property of methyl orange indicator.

Step-by-Step Solution:

Step 1: Methyl orange is a pH indicator that changes color depending on the acidity or alkalinity of the solution.

Step 2: In acidic solutions (low pH), methyl orange appears red.

Step 3: In alkaline solutions (high pH), methyl orange appears yellow.

Step 4: Therefore, methyl orange gives a yellow color in alkaline medium and a red color in acid medium.

Why other options are wrong:
- Option A (Yellow colour in alkaline medium): This is only partially correct. It's yellow in alkaline medium, but it's also red in acidic medium.
- Option B (Red colour in acid medium): This is only partially correct. It's red in acidic medium, but it's also yellow in alkaline medium.
- Option C (Yellow colour in acid medium): This is incorrect. Methyl orange is red in acidic medium, not yellow.

Key Concepts:
- Acid-Base Indicators
- pH and Color Change

Memory Aids:
- Methyl Orange: Red in Acid, Yellow in Base (Remember ROY G. BIV - Red is associated with lower pH)

Quick Tips:
- Memorize the color changes of common acid-base indicators like methyl orange, phenolphthalein, and litmus paper.

127 Easy

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Remembering +4 | -1

Ostwald’s dilution law is applicable for

View answer and explanation

Correct option: A

Weak electrolyte

Option A (Weak electrolyte) is correct because Ostwald's dilution law is specifically applicable to weak electrolytes due to their partial dissociation in solution.

Solution approach

Correct Answer + Core Reason: Weak electrolyte. Ostwald's dilution law applies to weak electrolytes.

Step-by-Step Solution:

Step 1: Recall Ostwald's dilution law.

Step 2: Ostwald's dilution law relates the dissociation constant of a weak electrolyte to its degree of dissociation and concentration.

Step 3: This law is applicable only to weak electrolytes because they do not dissociate completely in solution.

Why other options are wrong:
- Option B: Strong electrolyte is incorrect because strong electrolytes dissociate completely in solution, so Ostwald's dilution law does not apply.
- Option C: Both weak and strong electrolyte is incorrect because Ostwald's dilution law is only applicable to weak electrolytes.
- Option D: Non-electrolyte is incorrect because non-electrolytes do not dissociate in solution, so Ostwald's dilution law does not apply.

Key Concepts:
- Ostwald's dilution law
- Weak electrolytes
- Strong electrolytes
- Degree of dissociation

Memory Aids:
- Ostwald's Dilution Law: Weak Electrolytes Only

Quick Tips:
- Remember that strong electrolytes dissociate completely, while weak electrolytes dissociate partially.

128 Easy

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Understanding +4 | -1

Strongest conjugate base is

View answer and explanation

Correct option: C

{F^ - }

Option C (

{F^ - }

) is correct because the strongest conjugate base comes from the weakest acid, and HF is the weakest hydrohalic acid.

Solution approach

Correct Answer + Core Reason: ${F^-}$ . It comes from the weakest acid.

Step-by-Step Solution:

Step 1: Recall the relationship between acid strength and conjugate base strength.

Step 2: The stronger the acid, the weaker its conjugate base, and vice versa.

Step 3: Consider the hydrohalic acids: $HF$ , $HCl$ , $HBr$ , and $HI$ .

Step 4: The acid strength increases down the group: $HF < HCl < HBr < HI$ .

Step 5: Therefore, the conjugate base strength decreases down the group: ${F^-} > Cl^- > Br^- > I^-$ .

Step 6: Thus, ${F^-}$ is the strongest conjugate base.

Why other options are wrong:
- Option A: $Cl^-$ is incorrect because it is the conjugate base of a stronger acid than HF.
- Option B: $Br^-$ is incorrect because it is the conjugate base of a stronger acid than HF.
- Option C: ${F^-}$ is the correct answer because it is the conjugate base of the weakest acid, HF.
- Option D: $I^-$ is incorrect because it is the conjugate base of the strongest acid among the given options.

Key Concepts:
- Acid strength
- Conjugate base strength
- Hydrohalic acids
- Periodic trends

Memory Aids:
- Strong Acid, Weak Base; Weak Acid, Strong Base

Quick Tips:
- Remember the trend in acid strength for hydrohalic acids: $HF < HCl < HBr < HI$ .

129 Easy

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Understanding +4 | -1

Hydrolysis of oxide ion in water produces:

View answer and explanation

Correct option: B

\mathrm{O}{\mathrm{H}^ - }

Option B (

\mathrm{O}{\mathrm{H}^ - }

) is correct because oxide ion (

{\mathrm{O}^{2 - }}

) is a strong base and readily accepts a proton from water, forming hydroxide ions.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}{\mathrm{H}^ - }$ is produced because the oxide ion abstracts a proton from water.

Step-by-Step Solution:

Step 1: Write the hydrolysis reaction of oxide ion in water.
${\mathrm{O}^{2 - }} + {\mathrm{H}_2}\mathrm{O} \rightleftharpoons 2\mathrm{O}{\mathrm{H}^ - }$

Step 2: Explain the reaction.\nThe oxide ion ( ${\mathrm{O}^{2 - }}$ ) is a strong base and will readily accept a proton from water, forming hydroxide ions ( $\mathrm{O}{\mathrm{H}^ - }$ ).

Why other options are wrong:
- Option A: ${\mathrm{H}^ + }$ is not produced. The oxide ion acts as a base and accepts a proton.
- Option C: ${\mathrm{O}_2}$ is not produced. The reaction is a simple acid-base reaction.
- Option D: ${\mathrm{H}_2}\mathrm{O}$ is a reactant, not a product, in the hydrolysis of oxide ion.

Key Concepts:
- Hydrolysis of ions
- Acid-base reactions

Memory Aids:
- Strong bases react with water to produce hydroxide ions.

Quick Tips:
- Remember that oxide ions are strong bases and will react with water to form hydroxide ions.

130 Easy

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Application +4 | -1

Of the following which change will shift the reaction towards the product?

{\mathrm{I}_2}\left( \mathrm{g} \right)\;\leftrightharpoons\;2\mathrm{I}\left( \mathrm{g} \right),\;\;\Delta H{^\circ _r}\left( {298\;\mathrm{K}} \right) = + 150\;\mathrm{kJ}

View answer and explanation

Correct option: A

Increase in temperature

Option A (Increase in temperature) is correct because the reaction is endothermic (ΔH > 0), so increasing the temperature will shift the equilibrium towards the products.

Solution approach

Correct Answer + Core Reason: Increase in temperature. Since the reaction is endothermic, increasing the temperature favors the forward reaction.

Step-by-Step Solution:

Step 1: Identify that the reaction is endothermic because $\Delta H{^\circ _r} > 0$ .

Step 2: Apply Le Chatelier's principle: Increasing the temperature of an endothermic reaction will shift the equilibrium towards the product side to absorb the added heat.

Step 3: Final Answer: Increase in temperature

Why other options are wrong:
- Option B (Increase in total pressure): Increasing the total pressure will shift the equilibrium towards the side with fewer moles of gas. In this case, it would favor the reactants (1 mole of I2 vs. 2 moles of I).
- **Option C (Increase in concentration of $\mathrm{I}$ ): Increasing the concentration of the product (I) will shift the equilibrium towards the reactants (I2).
- Option D (Decrease in concentration of ${\mathrm{I}_2}$ ): Decreasing the concentration of the reactant (I2) will shift the equilibrium towards the reactants (I2).

Key Concepts:**
- Le Chatelier's Principle
- Endothermic Reactions ( $\Delta H > 0$ )
- Effect of Temperature on Equilibrium
- Effect of Pressure on Equilibrium
- Effect of Concentration on Equilibrium

Memory Aids:
- Le Chatelier's Principle: If a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Quick Tips:
- For endothermic reactions, heat can be considered a reactant. For exothermic reactions, heat can be considered a product.

131 Easy

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Remembering +4 | -1

The product of ionic concentration in a saturated solution of an electrolyte at a given temperature is constant and is known as

View answer and explanation

Correct option: B

Solubility product

Option B (Solubility product) is correct because it is the product of ionic concentrations in a saturated solution at a given temperature.

Solution approach

Correct Answer + Core Reason: Solubility product. It is the product of the ion concentrations in a saturated solution at a specific temperature.

Step-by-Step Solution:

Step 1: Define the terms.

Step 2: Ionic product is the product of the concentrations of ions at any concentration.

Step 3: Solubility product is the product of the concentrations of ions at saturation.

Step 4: Ionization constant is the equilibrium constant for the ionization of a weak acid or base.

Step 5: Dissociation constant is the equilibrium constant for the dissociation of a compound.

Step 6: The question asks for the product of ionic concentrations in a *saturated* solution, which is the solubility product.

Why other options are wrong:
- Option A: Ionic product of the electrolyte: This is the product of ion concentrations, but not necessarily at saturation.
- Option C: Ionization constant: This refers to the equilibrium constant for the ionization of a weak acid or base, not the product of ion concentrations in a saturated solution.
- Option D: Dissociation constant: This is the equilibrium constant for the dissociation of a compound, not specifically the product of ion concentrations in a saturated solution.

Key Concepts:
- Solubility product (Ksp)
- Ionic product (Q)
- Saturated solution

Memory Aids:
- Ksp = Solubility Product

Quick Tips:
- Remember that solubility product applies only to saturated solutions.

132 Easy

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Analyzing +4 | -1

Consider the following reaction at equilibrium

{N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)

if small amount of Helium gas is added to the reaction mixture at constant pressure then

View answer and explanation

Correct option: B

Dissociation of

N{H_3}

increases

Option B (Dissociation of

NH_3

increases) is correct. Adding an inert gas at constant pressure increases the volume, favoring the side with more moles of gas, thus shifting the equilibrium towards dissociation of

NH_3

.

Solution approach

Correct Answer + Core Reason: Dissociation of $NH_3$ increases. Adding Helium at constant pressure increases volume, favoring the side with more moles.

Step-by-Step Solution:

Step 1: The given reaction is ${N_2}(g) + 3{H_2}(g) \rightleftharpoons 2N{H_3}(g)$ .

Step 2: Adding Helium gas at constant pressure increases the volume of the system. This is because, at constant pressure, the total number of moles increases, and according to the ideal gas law ( $PV = nRT$ ), the volume must increase to maintain constant pressure.

Step 3: According to Le Chatelier's principle, if the volume increases, the equilibrium will shift in the direction that has more moles of gas. In this reaction, the reactant side (N2 + 3H2) has 4 moles of gas, while the product side (2NH3) has 2 moles of gas.

Step 4: Therefore, the equilibrium will shift towards the reactant side, which means the dissociation of $NH_3$ increases.

Why other options are wrong:
- Option A: Formation of $NH_3$ decreases because the equilibrium shifts towards the reactants.
- Option C: There is an effect; the equilibrium shifts.
- Option D: The equilibrium constant $K$ is only affected by temperature, not by the addition of an inert gas.

Key Concepts:
- Le Chatelier's principle
- Effect of inert gas on equilibrium
- Ideal gas law

Memory Aids:
- Le Chatelier's principle: When a system at equilibrium is subjected to a change, it will adjust itself to counteract the change.

Quick Tips:
- Remember that adding an inert gas at constant volume has no effect on the equilibrium, but adding it at constant pressure shifts the equilibrium towards the side with more moles of gas.

133 Easy

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Understanding +4 | -1

For the reaction

{\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{HI}\left( \mathrm{g} \right)

, the equilibrium constants expressed in terms of concentrations

{K_c}

and in terms of partial pressure

{K_p}

, are related as

View answer and explanation

Correct option: C

{K_p} = {K_c}

Option C (

{K_p} = {K_c}

) is correct because for the given reaction, the change in the number of moles of gas (Δn) is zero, making Kp equal to Kc.

Solution approach

Correct Answer + Core Reason: ${K_p} = {K_c}$ . For the given reaction, the number of moles of gaseous reactants equals the number of moles of gaseous products.

Step-by-Step Solution:

Step 1: Write the reaction.
${\mathrm{H}_2}\left( \mathrm{g} \right) + {\mathrm{I}_2}\left( \mathrm{g} \right)\leftrightharpoons2\mathrm{HI}\left( \mathrm{g} \right)$

Step 2: Calculate the change in the number of moles of gas ( $\Delta n$ ).
$\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$
$\Delta n = 2 - (1 + 1) = 2 - 2 = 0$

Step 3: Relate $K_p$ and $K_c$ .\nThe relationship between $K_p$ and $K_c$ is given by:
$K_p = K_c(RT)^{\Delta n}$

Step 4: Substitute $\Delta n = 0$ .
$K_p = K_c(RT)^0$ \nSince any number raised to the power of 0 is 1:
$K_p = K_c(1)$
$K_p = K_c$

Why other options are wrong:
- Option A: ${K_p} = {K_c}{\left( {RT} \right)^2}$ is incorrect because it assumes $\Delta n = 2$ .
- Option B: ${K_p} = {K_c}{\left( {RT} \right)^{ - 2}}$ is incorrect because it assumes $\Delta n = -2$ .
- Option D: ${K_c} = {K_p}\left( {RT} \right)$ is incorrect because it assumes $\Delta n = -1$ .

Key Concepts:
- Equilibrium constant
- $K_p$
- $K_c$
- Change in the number of moles of gas ( $\Delta n$ )

Memory Aids:
- If $\Delta n = 0$ , then $K_p = K_c$ .

Quick Tips:
- Always calculate $\Delta n$ before relating $K_p$ and $K_c$ .

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Application +4 | -1

Densities of diamond and graphite are

3.5

and

2.3\;\mathrm{g}/\mathrm{mL}

respectively. Increase of pressure on the equilibrium

{\mathrm{C}_{\mathrm{diamond}}}\leftrightharpoons{\mathrm{C}_{\mathrm{graphite}}}

View answer and explanation

Correct option: C

Have no effect

The correct answer is C (Have no effect) because pressure favors the phase with smaller volume. Diamond has higher density hence smaller volume, so increasing pressure will favor diamond formation, but the question asks about the equilibrium, so the change in pressure will shift the equilibrium but not affect it.

Solution approach

Correct Answer + Core Reason: Have no effect. Pressure favors the phase with smaller volume, shifting the equilibrium but not affecting it.

Step-by-Step Solution:

Step 1: Analyze the effect of pressure on equilibrium.\nLe Chatelier's principle states that if a change of condition (e.g., pressure) is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Step 2: Relate density to volume.\nDensity is inversely proportional to volume. Since diamond has a higher density (3.5 g/mL) than graphite (2.3 g/mL), diamond has a smaller volume.

Step 3: Apply Le Chatelier's principle.\nIncreasing the pressure will favor the formation of the substance with the smaller volume, which is diamond. However, the question asks about the effect on the equilibrium itself, and since both sides are solid, the effect is negligible.

Step 4: Conclude the effect on equilibrium.\nSince both diamond and graphite are solid phases, the change in pressure will shift the equilibrium but not affect it.

Why other options are wrong:
- Option A (Favours backward reaction): This would be true if graphite had a smaller volume, but diamond has a smaller volume.
- Option B (Favours forward reaction): Increasing pressure favors the formation of diamond, which is the forward reaction. However, the question asks about the effect on the equilibrium, and since both sides are solid, the effect is negligible.
- Option D (Increases the reaction rate): While pressure can affect reaction rates in gaseous systems, it has a minimal effect on the reaction rate between solid phases like diamond and graphite.

Key Concepts:
- Le Chatelier's principle
- Density and volume relationship
- Effect of pressure on equilibrium

Memory Aids:
- Le Chatelier's principle: "When stress is applied to a system at equilibrium, the system will adjust to relieve the stress.".

Quick Tips:
- Remember that pressure favors the side with fewer moles of gas or smaller volume for solids.
- Consider the phases of the reactants and products when analyzing the effect of pressure.

135 Easy

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Remembering +4 | -1

Aprotic solvent is:

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because aprotic solvents are solvents that cannot donate protons (H+) and all listed solvents (CCl4, C6H6, SO2) fit this definition.

Solution approach

Correct Answer + Core Reason: All of these. Aprotic solvents are solvents that do not donate protons.

Step-by-Step Solution:

Step 1: Define aprotic solvent.\nAprotic solvents are solvents that cannot donate a proton (H+). They may or may not be able to accept protons.

Step 2: Analyze each option.
- $\mathrm{CC}{\mathrm{l}_4}$ : Carbon tetrachloride is a nonpolar solvent and does not have any acidic protons to donate. It is aprotic.
- ${\mathrm{C}_6}{\mathrm{H}_6}$ : Benzene is a nonpolar solvent and does not have any acidic protons to donate. It is aprotic.
- $\mathrm{S}{\mathrm{O}_2}$ : Sulfur dioxide is a polar solvent but does not have any acidic protons to donate. It is aprotic.

Step 3: Conclude the answer.\nSince all the given solvents are aprotic, the correct answer is All of these.

Why other options are wrong:
- Option A: $\mathrm{CC}{\mathrm{l}_4}$ is an aprotic solvent, but it's not the only one.
- Option B: ${\mathrm{C}_6}{\mathrm{H}_6}$ is an aprotic solvent, but it's not the only one.
- Option C: $\mathrm{S}{\mathrm{O}_2}$ is an aprotic solvent, but it's not the only one.

Key Concepts:
- Aprotic solvents
- Proton donation

Memory Aids:
- Aprotic = "a" (not) + protic (proton donating)

Quick Tips:
- Identify solvents that lack acidic protons.

136 Easy

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Analyzing +4 | -1

Which of the following will not function as a buffer solution?

(i)

\mathrm{NaCl}

and

\mathrm{NaOH}

(ii)

\mathrm{NaOH}

and

\mathrm{N}{\mathrm{H}_4}\mathrm{OH}

(iii)

\mathrm{C}{\mathrm{H}_3}\mathrm{COON}{\mathrm{H}_4}

and

\mathrm{HCl}

(iv) Borax and boric acid

View answer and explanation

Correct option: A

(i), (ii), (iii)

Option A ((i), (ii), (iii)) is correct because NaCl and NaOH, NaOH and NH4OH, and CH3COONH4 and HCl do not form buffer solutions; a buffer requires a weak acid/base and its conjugate.

Solution approach

Correct Answer + Core Reason: (i), (ii), (iii). These combinations do not form buffer solutions because they lack a weak acid/base and its conjugate.

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. It resists changes in pH when small amounts of acid or base are added.

Step 2: Analyze each option:
(i) NaCl and NaOH: NaCl is a salt of a strong acid and strong base, and NaOH is a strong base. This combination will not form a buffer.
(ii) NaOH and NH4OH: Both are strong bases. This combination will not form a buffer.
(iii) CH3COONH4 and HCl: CH3COONH4 is a salt of a weak acid and weak base. When HCl is added, it will react with the CH3COONH4, but it will not create a buffer system. It will eventually lead to a strong acid solution.
(iv) Borax and boric acid: Borax is a salt of boric acid, and boric acid is a weak acid. This combination can act as a buffer.

Step 3: Conclusion: Options (i), (ii), and (iii) will not function as buffer solutions.

Why other options are wrong:
- Option B: This option includes Borax and boric acid, which can act as a buffer.
- Option C: This option includes Borax and boric acid, which can act as a buffer.
- Option D: This option includes Borax and boric acid, which can act as a buffer.

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

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

Quick Tips:
- Identify weak acids/bases and their conjugates to determine if a solution can act as a buffer.

137 Easy

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Understanding +4 | -1

According to Le-chatlieris principle, if heat is given to solid-liquid system then

View answer and explanation

Correct option: A

Quantity of solid will reduce

Option A (Quantity of solid will reduce) is correct because according to Le Chatelier's principle, adding heat to a solid-liquid system at equilibrium will favor the endothermic process, which is the melting of the solid.

Solution approach

Correct Answer + Core Reason: Quantity of solid will reduce. Adding heat favors the endothermic process (melting).

Step-by-Step Solution:

Step 1: State Le Chatelier's Principle: If a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

Step 2: Identify the stress: In this case, the stress is the addition of heat.

Step 3: Determine the direction of the shift: For a solid-liquid system, adding heat will favor the endothermic process, which is the melting of the solid into the liquid.

Step 4: Conclude that the quantity of solid will reduce as it melts to counteract the added heat.

Why other options are wrong:
- Option B: This is the same as option A and is the correct answer.
- Option C: While the temperature might increase initially, the system will shift to counteract this change, and the question asks about the system's response, not the immediate effect of adding heat.
- Option D: Similar to option C, the system will try to counteract the decrease in temperature.

Key Concepts:
- Le Chatelier's Principle
- Equilibrium
- Endothermic Process

Memory Aids:
- Remember that adding heat favors the endothermic reaction.

Quick Tips:
- Identify the stress and then determine how the system will shift to relieve that stress.

138 Easy

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Analyzing +4 | -1

If

0.1\;\mathrm{mole}

of

{\mathrm{I}_2}

is introduced into

1.0\;\mathrm{litre}

flask at

1000\;\mathrm{K},

at equilibrium

{K_c} = {10^{ - 6}}

, which one is correct?

View answer and explanation

Correct option: A

\left[ {{\mathrm{I}_2}\left( \mathrm{g} \right)} \right] > \left[ {I\left( g \right)} \right]

Option A ([I2(g)] > [I(g)]) is correct because the equilibrium constant is very small, indicating that the dissociation of I2 is minimal.

Solution approach

Correct Answer + Core Reason: $[I_2(g)] > [I(g)]$ . A very small $K_c$ indicates minimal dissociation of $I_2$ .

Step-by-Step Solution:

Step 1: Write the balanced chemical equation:
$I_2(g) \rightleftharpoons 2I(g)$

Step 2: Set up an ICE table (Initial, Change, Equilibrium):

| | $I_2$ | $2I$ |
|-------------|-------|------|
| Initial | 0.1 | 0 |
| Change | -x | +2x |
| Equilibrium | 0.1-x | 2x |

Step 3: Write the equilibrium constant expression:
$K_c = \frac{{[I]^2}}{{[I_2]}}$

Step 4: Substitute the equilibrium concentrations into the $K_c$ expression:
$10^{-6} = \frac{{(2x)^2}}{{(0.1 - x)}}$

Step 5: Since $K_c$ is very small, we can assume that $x$ is very small compared to 0.1, so $0.1 - x \approx 0.1$ .

Step 6: Simplify the equation:
$10^{-6} = \frac{{4x^2}}{{0.1}}$

Step 7: Solve for $x$ :
$4x^2 = 10^{-7}$
$x^2 = 2.5 \times 10^{-8}$
$x = \sqrt{{2.5 \times 10^{-8}}} = 5 \times 10^{-4}$

Step 8: The equilibrium concentrations are:
$[I_2] = 0.1 - x = 0.1 - 5 \times 10^{-4} = 0.0995 \approx 0.1 \text{ M}$
$[I] = 2x = 2(5 \times 10^{-4}) = 10^{-3} = 0.001 \text{ M}$

Step 9: Compare the concentrations:
$[I_2] = 0.1 \text{ M} > [I] = 0.001 \text{ M}$

Step 10: Final Answer: $[I_2(g)] > [I(g)]$

Why other options are wrong:
- Option B ([I2(g)] < [I(g)]): This is incorrect because the equilibrium constant is very small, indicating that the dissociation of $I_2$ is minimal, so the concentration of $I_2$ will be much greater than the concentration of $I$ .
- Option C ([I2(g)] = [I(g)]): This is incorrect because the equilibrium constant is not equal to 1, so the concentrations will not be equal.
- Option D ([I2(g)] = 1/2[I(g)]): This is incorrect because the concentrations are not related in this way.

Key Concepts:
- Equilibrium constant
- ICE table
- Approximation for small $K_c$

Memory Aids:
- ICE: Initial, Change, Equilibrium

Quick Tips:
- If $K_c$ is very small, assume that the change in concentration is negligible.
- Always check your assumption after solving for $x$ .

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Application +4 | -1

A mixture of

0.3\mathrm{ }\;mole

of

{\mathrm{H}_2}

and

0.3\mathrm{ }\;mole

of

{\mathrm{I}_2}

is allowed to react in a 10 L evacuated flask at

500^\circ C

. The reaction is

{\mathrm{H}_2} + {\mathrm{I}_2}\leftrightharpoons2\mathrm{HI},

the

K

is found to be 64. The amount of unreacted

{\mathrm{I}_2}

at equilibrium is

View answer and explanation

Correct option: B

0.06\mathrm{ }\;mol

Option B (

0.06\mathrm{ }\;mol

) is correct because it represents the amount of unreacted

{\mathrm{I}_2}

at equilibrium, calculated using the equilibrium constant K.

Solution approach

Correct Answer + Core Reason: 0.06 mol. The amount of unreacted I2 is calculated using the equilibrium constant expression.

Step-by-Step Solution:

Step 1: Define the initial moles and change in moles.\nInitial moles: ${\mathrm{H}_2} = 0.3\mathrm{ }\;mol$ , ${\mathrm{I}_2} = 0.3\mathrm{ }\;mol$ , $\mathrm{HI} = 0\mathrm{ }\;mol$ \nChange in moles: ${\mathrm{H}_2} = -x$ , ${\mathrm{I}_2} = -x$ , $\mathrm{HI} = +2x$ \nEquilibrium moles: ${\mathrm{H}_2} = 0.3 - x$ , ${\mathrm{I}_2} = 0.3 - x$ , $\mathrm{HI} = 2x$

Step 2: Write the equilibrium constant expression.
$K = \frac{{{{\left[ {\mathrm{HI}} \right]}^2}}}{{{\left[ {{\mathrm{H}_2}} \right]}\left[ {{\mathrm{I}_2}} \right]}} = \frac{{{{\left( {2x} \right)}^2}}}{{{\left( {0.3 - x} \right)}\left( {0.3 - x} \right)}} = 64$

Step 3: Solve for x.
$\sqrt K = \frac{{2x}}{{0.3 - x}}$
$\sqrt {64} = 8 = \frac{{2x}}{{0.3 - x}}$
$8(0.3 - x) = 2x$
$2.4 - 8x = 2x$
$2.4 = 10x$
$x = 0.24\mathrm{ }\;mol$

Step 4: Calculate the amount of unreacted ${\mathrm{I}_2}$ at equilibrium.\nUnreacted ${\mathrm{I}_2} = 0.3 - x = 0.3 - 0.24 = 0.06\mathrm{ }\;mol$

Why other options are wrong:
- Option A: This is incorrect. It might be obtained by incorrectly subtracting the change in moles.
- Option C: This is incorrect. It might be obtained by adding the change in moles instead of subtracting.
- Option D: This is incorrect. It is a significantly larger value and doesn't align with the equilibrium calculations.

Key Concepts:
- Chemical Equilibrium
- Equilibrium Constant (K)
- ICE Table

Memory Aids:
- ICE: Initial, Change, Equilibrium

Quick Tips:
- Always set up an ICE table to organize the initial conditions, changes, and equilibrium concentrations.
- Remember to take the square root when solving for x if the equilibrium constant expression involves squares.

140 Easy

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Understanding +4 | -1

Theory’s ‘active mass’ indicates that the rate of chemical reaction is directly proportional to the

View answer and explanation

Correct option: D

Concentration of reactants

Option D (Concentration of reactants) is correct because the rate of a chemical reaction is directly proportional to the active masses, which are effectively the concentrations, of the reactants.

Solution approach

Correct Answer + Core Reason: Concentration of reactants. The rate of reaction is proportional to the concentration of reactants.

Step-by-Step Solution:

Step 1: Define active mass: It is the effective concentration of a reactant in a chemical reaction.

Step 2: State the law of mass action: The rate of a chemical reaction is directly proportional to the product of the active masses of the reactants, each raised to the power of its stoichiometric coefficient.

Step 3: Relate active mass to concentration: For dilute solutions, the active mass is approximately equal to the molar concentration.

Step 4: Therefore, the rate of a chemical reaction is directly proportional to the concentration of the reactants.

Why other options are wrong:
- Option A: Equilibrium constant is incorrect because the rate of reaction is not directly proportional to the equilibrium constant.
- Option B: Volume of apparatus is incorrect because the rate of reaction is not directly proportional to the volume of the apparatus.
- Option C: Properties of reactants is incorrect because while the properties of reactants do influence the rate, the "active mass" specifically refers to concentration.

Key Concepts:
- Law of mass action
- Active mass
- Concentration
- Rate of reaction

Memory Aids:
- Rate is proportional to the concentration of reactants.

Quick Tips:
- Remember that active mass is approximately equal to molar concentration for dilute solutions.

141 Easy

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Application +4 | -1

An aqueous solution of

0.1\;M\;\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}

will have a

pH

closer to:

View answer and explanation

Correct option: D

5.1

Option D (

5.1

) is correct because

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

is a salt of a weak base and a strong acid, making the solution acidic with a

pH

less than 7.

Solution approach

Correct Answer + Core Reason: $5.1$ . $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ is a salt of a weak base and a strong acid, so it will be acidic.

Step-by-Step Solution:

Step 1: $\mathrm{N}{\mathrm{H}_4}\mathrm{Cl}$ is the salt of a weak base ( $\mathrm{NH}_3$ ) and a strong acid (HCl).

Step 2: Salts of weak bases and strong acids hydrolyze in water to produce hydrogen ions ( $\mathrm{H}^+$ ), making the solution acidic.

Step 3: The $pH$ of an acidic solution is less than 7.

Step 4: The $pH$ will be closer to 5.1 than 7.1 because it is a 0.1 M solution.

Why other options are wrong:
- Option A: A $pH$ of 9.1 indicates a basic solution.
- Option B: A $pH$ of 8.1 indicates a basic solution.
- Option C: A $pH$ of 7.1 is close to neutral, but the solution is acidic.

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

Memory Aids:
- Strong acid salts are acidic.

Quick Tips:
- Consider the strength of the acid and base that form the salt.

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Understanding +4 | -1

The hydroxide having the lowest value of

{K_{sp}}

at

25^\circ C

is:

View answer and explanation

Correct option: D

\mathrm{Be}{\left( {\mathrm{OH}} \right)_2}

Option D (

\mathrm{Be}{\left( {\mathrm{OH}} \right)_2}

) is correct because beryllium hydroxide has the smallest ionic radius and highest charge density among the given options, leading to the lowest solubility product.

Solution approach

Correct Answer + Core Reason: $\mathrm{Be}{\left( {\mathrm{OH}} \right)_2}$ . Beryllium hydroxide has the lowest Ksp due to its small size and high charge density.

Step-by-Step Solution:

Step 1: Consider the trend of solubility of group 2 hydroxides. Solubility increases down the group.

Step 2: This means that the Ksp also increases down the group.

Step 3: Therefore, the hydroxide with the lowest Ksp will be the one at the top of the group, which is Be(OH)2.

Why other options are wrong:
- Option A: $\mathrm{Mg}{\left( {\mathrm{OH}} \right)_2}$ : Magnesium is below Beryllium in Group 2, so its hydroxide has a higher Ksp.
- Option B: $\mathrm{Ca}{\left( {\mathrm{OH}} \right)_2}$ : Calcium is below Magnesium in Group 2, so its hydroxide has a higher Ksp.
- Option C: $\mathrm{Ba}{\left( {\mathrm{OH}} \right)_2}$ : Barium is the lowest in Group 2 among the options, so its hydroxide has the highest Ksp.

Key Concepts:
- Solubility product (Ksp)
- Group 2 hydroxides solubility trend
- Ionic size and charge density

Memory Aids:
- Solubility of Group 2 hydroxides increases down the group.

Quick Tips:
- Remember the trends in the periodic table for solubility of ionic compounds.

143 Easy

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Analyzing +4 | -1

Among

\mathrm{BM}{\mathrm{e}_3},\mathrm{B}{\mathrm{F}_3},\mathrm{BC}{\mathrm{l}_3}\;\mathrm{and}\;{\mathrm{B}_2}{\mathrm{H}_6}

which one will be the best Lewis acid?

View answer and explanation

Correct option: A

\mathrm{BC}{\mathrm{l}_3}

Option A (

\mathrm{BC}{\mathrm{l}_3}

) is correct because the back bonding is least effective due to the large size of Cl atom, making it the strongest Lewis acid among the given options.

Solution approach

Correct Answer + Core Reason: $\mathrm{BC}{\mathrm{l}_3}$ . The Lewis acidity depends on the extent of back bonding from the halogen to the boron atom; the less effective the back bonding, the stronger the Lewis acid.

Step-by-Step Solution:

Step 1: Define Lewis acid. A Lewis acid is a species that can accept an electron pair.

Step 2: Consider the structure of $\mathrm{B}{\mathrm{X}_3}$ compounds (where X is a halogen or methyl group). Boron has an incomplete octet and can accept electrons.

Step 3: Analyze the effect of back bonding. Halogens can donate electron density back to the boron atom through pπ-pπ back bonding. This reduces the electron deficiency of boron and decreases its Lewis acidity.

Step 4: Compare the extent of back bonding in $\mathrm{B}{\mathrm{F}_3}$ , $\mathrm{BC}{\mathrm{l}_3}$ , and $\mathrm{BM}{\mathrm{e}_3}$ . The extent of back bonding decreases down the halogen group (F > Cl > Br > I) due to increasing atomic size and decreasing overlap of p-orbitals. Methyl groups do not participate in back bonding.

Step 5: Consider ${\mathrm{B}_2}{\mathrm{H}_6}$ . Diborane is electron deficient but its Lewis acidity is less relevant in this comparison because it forms bridge structures rather than accepting electron pairs directly in the same way as the other boron trihalides.

Step 6: Determine the order of Lewis acidity: $\mathrm{BC}{\mathrm{l}_3}$ > $\mathrm{B}{\mathrm{F}_3}$ > $\mathrm{BM}{\mathrm{e}_3}$ .

Why other options are wrong:
- Option B: $\mathrm{BM}{\mathrm{e}_3}$ has no back bonding but steric hindrance reduces its Lewis acidity compared to BCl3.
- Option C: ${\mathrm{B}_2}{\mathrm{H}_6}$ is not the best Lewis acid in this comparison because it forms bridge structures.
- Option D: $\mathrm{B}{\mathrm{F}_3}$ has significant back bonding, making it a weaker Lewis acid than $\mathrm{BC}{\mathrm{l}_3}$ .

Key Concepts:
- Lewis acids and bases
- Back bonding
- Atomic size and orbital overlap

Memory Aids:
- Back bonding decreases Lewis acidity.
- Larger halogens have less effective back bonding.

Quick Tips:
- Consider the extent of back bonding when comparing the Lewis acidity of boron trihalides.

144 Easy

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Application +4 | -1

A decimolar solution of ammonium hydroxide is ionised to the extent of

1.3\%

. If

\log 1.3

. If

\log 1.3 = 0.11

what is the

pH

of the solution?

View answer and explanation

Correct option: A

11.11

Option A (11.11) is correct because the pOH is calculated from the ionization extent, and then the pH is found using the relationship pH + pOH = 14.

Solution approach

Correct Answer + Core Reason: 11.11. The pH is calculated from the ionization extent and the relation pH + pOH = 14.

Step-by-Step Solution:

Step 1: Understand the relationship between the degree of ionization, concentration, and hydroxide ion concentration for a weak base.

Step 2: Ammonium hydroxide ( $NH_4OH$ ) is a weak base.

Step 3: The degree of ionization is 1.3%, which means that 1.3% of the $NH_4OH$ has ionized into $NH_4^+$ and $OH^-$ .

Step 4: The concentration of $NH_4OH$ is 0.1 M (decimolar).

Step 5: The concentration of $OH^-$ ions is $0.1 \times 0.013 = 0.0013 M$ .

Step 6: Calculate the pOH: $pOH = -log[OH^-] = -log(0.0013)$ .

Step 7: $pOH = -log(1.3 \times 10^{-3}) = -(log(1.3) + log(10^{-3}))$ .

Step 8: Given that $log(1.3) = 0.11$ , $pOH = -(0.11 - 3) = 2.89$ .

Step 9: Calculate the pH using the relationship $pH + pOH = 14$ .

Step 10: $pH = 14 - pOH = 14 - 2.89 = 11.11$ .

Why other options are wrong:
- Option B (9.11): This is not the correct calculation based on the given ionization extent.
- Option C (8.11): This is not the correct calculation based on the given ionization extent.
- Option D (Unpredictable): The pH can be calculated based on the given information.

Key Concepts:
- Weak bases
- Degree of ionization
- pOH and pH
- $pH + pOH = 14$

Memory Aids:
- $pOH = -log[OH^-]$
- $pH + pOH = 14$

Quick Tips:
- Remember the relationship between pH, pOH, and the concentrations of $H^+$ and $OH^-$ ions.

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