The Henderson’s equation used to calculate the ${\rm{pOH}}$ of basic buffer is:

${\rm{pOH}} = {\rm{p}}{K_b} + \log \frac{{\left[ {{\rm{Conjugate\; acid}}} \right]}}{{\left[ {{\rm{\;Base}}} \right]}}$

${\rm{pOH}} = {\rm{p}}{K_b} - \log \frac{{\left[ {{\rm{Conjugate\; acid}}} \right]}}{{\left[ {{\rm{\;Base}}} \right]}}$

${\rm{pOH}} = {\rm{p}}{K_b} + \log \frac{{\left[ {{\rm{Base}}} \right]}}{{\left[ {{\rm{Conjugate\; acid}}} \right]}}$

${\rm{pOH}} = {\rm{p}}{K_b} - \log \frac{{\left[ {{\rm{Base}}} \right]}}{{\left[ {{\rm{Conjugate\; acid}}} \right]}}$

Equilibrium MCQs for NEET UG CHEMISTRY — Easy, Page 2

31 Easy

31 of 144

Application +4 | -1

The reaction,

2{\mathrm{A}_{\left( \mathrm{g} \right)}} + {\mathrm{B}_{\left( \mathrm{g} \right)}}\leftrightharpoons3{\mathrm{C}_{\left( \mathrm{g} \right)}} + {\mathrm{D}_{\left( \mathrm{g} \right)}}

is begun with the concentration of

A

and

B

both at an initial value of

1.00\;M.

When equilibrium is reached, the concentration of

D

is measured and found to be

0.25\;M

. The value for the equilibrium constant for this reaction is given by the expression :

View answer and explanation

Correct option: B

\left[ {{{\left( {0.75} \right)}^3}\left( {0.25} \right)} \right] \div \left[ {{{\left( {0.50} \right)}^2}\left( {0.75} \right)} \right]

Option B is correct because it accurately represents the equilibrium constant expression based on the changes in concentration given the initial concentrations and the equilibrium concentration of D.

Solution approach

Correct Answer + Core Reason: $\left[ {{{\left( {0.75} \right)}^3}\left( {0.25} \right)} \right] \div \left[ {{{\left( {0.50} \right)}^2}\left( {0.75} \right)} \right]$ . The equilibrium constant is calculated based on the change in concentrations.

Step-by-Step Solution:

Step 1: Write the reaction: $2A(g) + B(g) \rightleftharpoons 3C(g) + D(g)$

Step 2: Set up an ICE table:\nInitial: 1.00 1.00 0 0\nChange: -2x -x +3x +x\nEquilibrium: 1-2x 1-x 3x x

Step 3: Given that the equilibrium concentration of D is 0.25 M, then x = 0.25.

Step 4: Calculate the equilibrium concentrations of A, B, and C:
[A] = 1 - 2x = 1 - 2(0.25) = 1 - 0.50 = 0.50 M
[B] = 1 - x = 1 - 0.25 = 0.75 M
[C] = 3x = 3(0.25) = 0.75 M

Step 5: Write the equilibrium constant expression:
$K = \frac{{[C]^3[D]}}{{[A]^2[B]}} = \frac{{(0.75)^3(0.25)}}{{(0.50)^2(0.75)}}$

Why other options are wrong:
- Option A: This option incorrectly uses the initial concentrations of A and B in the denominator instead of the equilibrium concentrations.
- Option C: This option incorrectly uses 0.25 for the concentration of B in the denominator.
- Option D: This option incorrectly uses 0.75 and 0.25 for the concentrations of A and B in the denominator.

Key Concepts:
- Equilibrium constant expression
- ICE tables

Memory Aids:
- Equilibrium concentrations are used in the equilibrium constant expression.

Quick Tips:
- Always use an ICE table to track changes in concentration.

32 Easy

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Application +4 | -1

At 3000 K, the equilibrium pressure of

\mathrm{C}{\mathrm{O}_2},\mathrm{CO}

and

{\mathrm{O}_2}

are 0.6, 0.4 and 0.2 atm respectively.

{K_p}

for the reaction

2\mathrm{C}{\mathrm{O}_2}\leftrightharpoons2\mathrm{CO} + {\mathrm{O}_2}

is

View answer and explanation

Correct option: A

0.089

Option A (

0.089

) is correct because it is the correct calculation of Kp using the given partial pressures.

Solution approach

Correct Answer + Core Reason: $0.089$ . Kp is calculated using the partial pressures of the reactants and products at equilibrium.

Step-by-Step Solution:

Step 1: Write the balanced chemical equation: $2\mathrm{C}{\mathrm{O}_2}\leftrightharpoons2\mathrm{CO} + {\mathrm{O}_2}$

Step 2: Write the expression for Kp:
${K_p} = \frac{{P_{CO}^2 \times P_{O_2}}}{{P_{CO_2}^2}}$

Step 3: Substitute the given partial pressures:
${K_p} = \frac{{(0.4)^2 \times (0.2)}}{(0.6)^2}$

Step 4: Calculate Kp:
${K_p} = \frac{{0.16 \times 0.2}}{{0.36}}$
${K_p} = \frac{{0.032}}{{0.36}}$
${K_p} = 0.08888... \approx 0.089$

Why other options are wrong:
- Option B: $0.098$ : This is incorrect due to a calculation error.
- Option C: $0.189$ : This is incorrect due to a calculation error.
- Option D: $0.198$ : This is incorrect due to a calculation error.

Key Concepts:
- Equilibrium constant (Kp)
- Partial pressures

Memory Aids:
- Remember that Kp is the ratio of partial pressures of products to reactants, each raised to the power of their stoichiometric coefficients.

Quick Tips:
- Always write the balanced chemical equation and the Kp expression before substituting values.

33 Easy

33 of 144

Remembering +4 | -1

The indicators used in the titration of iodine against sodium thiosulphate is

View answer and explanation

Correct option: A

Starch

Option A (Starch) is correct because starch acts as a specific indicator for iodine in iodometric titrations, forming a deep blue complex.

Solution approach

Correct Answer + Core Reason: Starch. Starch specifically interacts with iodine to form a colored complex, making it suitable for detecting the endpoint.

Step-by-Step Solution:

Step 1: Understand the titration process. Iodine is being titrated against sodium thiosulphate.

Step 2: Consider the role of an indicator. An indicator is used to visually signal the endpoint of the titration.

Step 3: Recall the specific indicator used for iodine titrations. Starch is used because it forms a deep blue complex with iodine, which disappears when all the iodine has reacted with the thiosulphate.

Step 4: Therefore, the indicator used is starch.

Why other options are wrong:
- Option B: ${\mathrm{K}_3}\mathrm{Fe}{\left( {\mathrm{CN}} \right)_6}$ : This is potassium ferricyanide, which is not used as an indicator in iodine titrations.
- Option C: ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$ : This is potassium chromate, which is used as an indicator in precipitation titrations, not iodine titrations.
- Option D: Potassium: Potassium itself is not an indicator.

Key Concepts:
- Iodometric titration
- Indicators in titrations
- Starch-iodine complex

Memory Aids:
- Starch turns blue/black in the presence of iodine.

Quick Tips:
- Remember the specific indicators used for different types of titrations.

34 Easy

34 of 144

Understanding +4 | -1

Which of the following statements is correct for a reversible process in a state of equilibrium

View answer and explanation

Correct option: A

\Delta {G^{^0}} = - 2.303\,RT\,\log \,K

Option A (

\Delta {G^{^0}} = - 2.303\,RT\,\log \,K

) is correct because it represents the relationship between the standard free energy change and the equilibrium constant for a reversible process at equilibrium.

Solution approach

Correct Answer + Core Reason: $\Delta {G^{^0}} = - 2.303\,RT\,\log \,K$ . This equation directly relates the standard Gibbs free energy change to the equilibrium constant.

Step-by-Step Solution:

Step 1: Define the context. We are looking for the correct equation that relates the standard Gibbs free energy change ( $\Delta G^0$ ) to the equilibrium constant (K) for a reversible process at equilibrium.

Step 2: Recall the relevant equation. The relationship is given by: $\Delta G^0 = -RT \ln K$

Step 3: Convert the natural logarithm to base-10 logarithm. Since $\ln x = 2.303 \log x$ , we can rewrite the equation as: $\Delta G^0 = -2.303 RT \log K$

Step 4: Final Answer: $\Delta G^0 = -2.303 RT \log K$

Why other options are wrong:
- Option B: This option has the wrong sign. The relationship is negative, not positive.
- Option C: This option uses $\Delta G$ instead of $\Delta G^0$ . $\Delta G = -2.303RT \log Q$ , where Q is the reaction quotient, not the equilibrium constant K. This is only true at equilibrium when $\Delta G = 0$ and Q = K, but the equation provided is for standard conditions.
- Option D: This option has the wrong sign and uses $\Delta G$ instead of $\Delta G^0$ .

Key Concepts:
- Gibbs Free Energy
- Equilibrium Constant
- Relationship between Gibbs Free Energy and Equilibrium Constant

Memory Aids:
- Remember that a negative $\Delta G^0$ indicates a spontaneous reaction (K > 1).

Quick Tips:
- Always remember the negative sign in the equation relating $\Delta G^0$ and K.
- Distinguish between $\Delta G$ and $\Delta G^0$ . $\Delta G^0$ refers to standard conditions.

35 Easy

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Remembering +4 | -1

2 \mathrm{SO_2(g})+\mathrm{O}_2(\mathrm{g}) \xrightarrow{\text { Catalyst }} 2 \mathrm{SO}_3(\mathrm{g})

Identify the catalyst from the options given below:

View answer and explanation

Correct option: A

\mathrm{V}_2 \mathrm{O}_5

Option A (

\mathrm{V}_2 \mathrm{O}_5

) is correct because Vanadium(V) oxide (

\mathrm{V}_2 \mathrm{O}_5

) is the catalyst used in the Contact process for the production of

\mathrm{SO}_3

.

Solution approach

Correct Answer + Core Reason: $\mathrm{V}_2 \mathrm{O}_5$ . Vanadium(V) oxide is the catalyst used in the Contact process for the production of $\mathrm{SO}_3$ .

Step-by-Step Solution:

Step 1: Recall the Contact process for the production of sulfuric acid.

Step 2: Identify the catalyst used in the oxidation of $\mathrm{SO}_2$ to $\mathrm{SO}_3$ .\nThe catalyst used in this step is Vanadium(V) oxide ( $\mathrm{V}_2 \mathrm{O}_5$ ).

Step 3: Write the reaction.
$2 \mathrm{SO_2(g)} + \mathrm{O_2(g)} \xrightarrow{\mathrm{V_2O_5}} 2 \mathrm{SO_3(g)}$

Why other options are wrong:
- Option B: $\mathrm{CuCl}_2$ is not the catalyst used in the Contact process. It is used in Deacon's process for the production of chlorine.
- Option C: $\mathrm{MnO}_2$ is not the catalyst used in the Contact process. It is used as a catalyst in the decomposition of potassium chlorate.
- Option D: Incorrect because only $\mathrm{V}_2 \mathrm{O}_5$ is the correct catalyst.

Key Concepts:
- Contact Process
- Catalysis

Memory Aids:
- Remember that $\mathrm{V}_2 \mathrm{O}_5$ is used in the Contact process.

Quick Tips:
- Memorize the catalysts used in important industrial processes.
- The Contact process is used for the production of sulfuric acid, and $\mathrm{V}_2 \mathrm{O}_5$ is the catalyst for the oxidation of $\mathrm{SO}_2$ to $\mathrm{SO}_3$ .

36 Easy

36 of 144

Analyzing +4 | -1

Which one of the following is least likely to act as a Lewis base?

View answer and explanation

Correct option: A

{\mathrm{I}^ + }

Option A (

{\mathrm{I}^ + }

) is correct because it has a positive charge and a deficiency of electrons, making it an electron acceptor rather than a donor.

Solution approach

Correct Answer + Core Reason: ${\mathrm{I}^ + }$ is least likely to act as a Lewis base because it has a positive charge and is electron deficient.

Step-by-Step Solution:

Step 1: Define a Lewis base. A Lewis base is a species that donates an electron pair.

Step 2: Analyze each option:
- ${\mathrm{I}^ + }$ : Iodine with a positive charge has lost an electron and has a strong tendency to accept electrons, not donate them. Therefore, it is not a Lewis base.
- $\mathrm{I}$ : Neutral iodine has lone pairs of electrons and can donate them, acting as a Lewis base.
- $\mathrm{SC}{\mathrm{l}_2}$ : Sulfur has lone pairs and can donate them, acting as a Lewis base.
- $\mathrm{PC}{\mathrm{l}_3}$ : Phosphorus has a lone pair and can donate it, acting as a Lewis base.

Step 3: Conclusion: ${\mathrm{I}^ + }$ is the least likely to act as a Lewis base.

Why other options are wrong:
- Option B: $\mathrm{I}$ : Neutral iodine has lone pairs and can act as a Lewis base by donating these electrons.
- Option C: $\mathrm{SC}{\mathrm{l}_2}$ : Sulfur in $\mathrm{SC}{\mathrm{l}_2}$ has lone pairs and can act as a Lewis base.
- Option D: $\mathrm{PC}{\mathrm{l}_3}$ : Phosphorus in $\mathrm{PC}{\mathrm{l}_3}$ has a lone pair and can act as a Lewis base.

Key Concepts:
- Lewis base: Electron pair donor
- Lewis acid: Electron pair acceptor

Memory Aids:
- Lewis Acid Accepts, Lewis Base Donates (LA LABD)

Quick Tips:
- Look for species with lone pairs of electrons to identify Lewis bases.
- Positively charged species are usually Lewis acids.

37 Easy

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Application +4 | -1

{K_C}

for

{H_2} + {I_2}\leftrightharpoons2HI

is "

K

" then for

HI\leftrightharpoons\frac{1}{2}\,{H_2} + \frac{1}{2}{I_2},

it is

View answer and explanation

Correct option: C

\frac{1}{{\sqrt K }}

Option C (

\frac{1}{{\sqrt K }}

) is correct because reversing a reaction and halving the stoichiometric coefficients results in taking the inverse square root of the original equilibrium constant.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{\sqrt K }}$ . Reversing the reaction and halving the coefficients results in the inverse square root of the equilibrium constant.

Step-by-Step Solution:

Step 1: Given reaction: ${H_2} + {I_2}\leftrightharpoons2HI$ with equilibrium constant ${K_C} = K$

Step 2: The reverse reaction is: $2HI\leftrightharpoons{H_2} + {I_2}$ with equilibrium constant ${K'_C} = \frac{1}{K}$

Step 3: Now, consider the reaction: $HI\leftrightharpoons\frac{1}{2}{H_2} + \frac{1}{2}{I_2}$ \nThis is half of the reverse reaction.

Step 4: The equilibrium constant for this reaction, ${K''_C}$ , is the square root of the equilibrium constant of the reaction $2HI\leftrightharpoons{H_2} + {I_2}$ .
${K''_C} = \sqrt{{K'_C}} = \sqrt{\frac{1}{K}} = \frac{1}{{\sqrt K }}$

Why other options are wrong:
- Option A: Squaring the equilibrium constant is incorrect as it corresponds to doubling the stoichiometric coefficients of the original reaction.
- Option B: Taking the square root of the equilibrium constant is incorrect as it corresponds to halving the stoichiometric coefficients of the original reaction, not reversing it.
- Option D: Taking the inverse square of the equilibrium constant is incorrect. It would correspond to reversing the reaction and then doubling the coefficients of the reversed reaction.

Key Concepts:
- Equilibrium Constant
- Reversing a reaction
- Changing stoichiometric coefficients

Memory Aids:
- Reversing a reaction inverts the equilibrium constant.
- Multiplying coefficients by n raises the equilibrium constant to the power of n.

Quick Tips:
- Remember how the equilibrium constant changes when the reaction is reversed or when the coefficients are multiplied by a factor.

38 Easy

38 of 144

Remembering +4 | -1

Which of the following mixture is a buffer solution?

View answer and explanation

Correct option: D

\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}

Option D (

\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}

) is correct because a buffer solution consists of a weak acid/base and its salt with a strong base/acid, and this option contains a weak base (ammonium hydroxide) and its salt (ammonium chloride).

Solution approach

Correct Answer + Core Reason: $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$ . This mixture contains a weak base and its salt, which is a buffer solution.

Step-by-Step Solution:

Step 1: Define a buffer solution.

Step 2: A buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.

Step 3: Analyze the given options:
- $\mathrm{NaCl}+\mathrm{NaOH}$ : Strong base and its salt (not a buffer).
- $\mathrm{HCl}+\mathrm{NH}_4 \mathrm{Cl}$ : Strong acid and its salt (not a buffer).
- $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{HCl}$ : Weak acid and strong acid (not a buffer).
- $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$ : Weak base and its salt (a buffer).

Step 4: Therefore, $\mathrm{NH}_4 \mathrm{OH}+\mathrm{NH}_4 \mathrm{Cl}$ is a buffer solution.

Why other options are wrong:
- Option A: $\mathrm{NaCl}+\mathrm{NaOH}$ : This is a mixture of a strong base and its salt, not a buffer.
- Option B: $\mathrm{HCl}+\mathrm{NH}_4 \mathrm{Cl}$ : This is a mixture of a strong acid and its salt, not a buffer.
- Option C: $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{HCl}$ : This is a mixture of a weak acid and a strong acid, not a buffer.

Key Concepts:
- Buffer solution
- Weak acid/base
- Conjugate acid/base

Memory Aids:
- Buffer = Weak + Salt

Quick Tips:
- Identify weak acids/bases and their salts to find buffer solutions.

39 Easy

39 of 144

Remembering +4 | -1

Isoelectric point is defined as the pH at which:

View answer and explanation

Correct option: D

\mathrm{Zwitter}

ion has zero charge

Option D (Zwitter ion has zero charge) is correct because the isoelectric point is the pH at which the zwitterion form of an amino acid has no net electrical charge.

Solution approach

Correct Answer + Core Reason: $\mathrm{Zwitter}$ ion has zero charge. The isoelectric point is the pH at which a molecule carries no net electrical charge.

Step-by-Step Solution:

Step 1: Define isoelectric point.\nThe isoelectric point (pI) is the pH at which a molecule, such as an amino acid or protein, carries no net electrical charge. This occurs when the number of positive and negative charges are equal.

Step 2: Explain the zwitterion.\nAt the isoelectric point, the amino acid exists primarily as a zwitterion, which is a dipolar ion with both positive and negative charges, but the net charge is zero.

Step 3: Relate isoelectric point to zwitterion charge.\nTherefore, the isoelectric point is defined as the pH at which the zwitterion has zero charge.

Why other options are wrong:
- Option A: An amino acid becomes acidic at a pH lower than its pI.
- Option B: An amino acid becomes basic at a pH higher than its pI.
- Option C: At pH values other than the isoelectric point, the zwitterion can have a positive or negative charge.

Key Concepts:
- Isoelectric point
- Zwitterion
- Amino acids

Memory Aids:
- Isoelectric point = "iso" (equal) + electric (charge) = equal positive and negative charges.

Quick Tips:
- The isoelectric point is a characteristic property of amino acids and proteins.

40 Easy

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Remembering +4 | -1

The reaction quotient

\left( Q \right)

at equilibrium is:

View answer and explanation

Correct option: B

= K

Option B (= K) is correct because at equilibrium, the reaction quotient (Q) is equal to the equilibrium constant (K).

Solution approach

Correct Answer + Core Reason: = K. At equilibrium, the reaction quotient equals the equilibrium constant.

Step-by-Step Solution:

Step 1: Define the reaction quotient (Q): It is a measure of the relative amounts of products and reactants present in a reaction at any given time. It indicates whether a reaction will favor product or reactant formation to reach equilibrium.

Step 2: Define the equilibrium constant (K): It is the value of the reaction quotient when the reaction is at equilibrium.

Step 3: At equilibrium, the forward and reverse reaction rates are equal, and the net change in concentrations of reactants and products is zero. Therefore, the reaction quotient (Q) is equal to the equilibrium constant (K).

Why other options are wrong:
- Option A: = 1 is incorrect because the equilibrium constant can be any value, not just 1.
- Option C: > K is incorrect because if Q > K, the reaction will shift towards reactants to reach equilibrium.
- Option D: < K is incorrect because if Q < K, the reaction will shift towards products to reach equilibrium.

Key Concepts:
- Reaction quotient (Q)
- Equilibrium constant (K)
- Chemical equilibrium

Memory Aids:
- At equilibrium, Q = K.

Quick Tips:
- Understand the relationship between Q and K to predict the direction of a reaction.

41 Easy

41 of 144

Remembering +4 | -1

Acidosis is diagnosed when blood

\mathrm{pH}

:

View answer and explanation

Correct option: A

Falls below

7.35

Option A (Falls below

7.35

) is correct because acidosis is a condition where the blood pH falls below the normal range of 7.35-7.45.

Solution approach

Correct Answer + Core Reason: Falls below $7.35$ . Acidosis is a condition where the blood pH is lower than the normal range.

Step-by-Step Solution:

Step 1: Define normal blood pH.\nThe normal blood pH range is between 7.35 and 7.45.

Step 2: Define acidosis.\nAcidosis is a condition in which the blood pH falls below 7.35.

Step 3: Define alkalosis.\nAlkalosis is a condition in which the blood pH rises above 7.45.

Step 4: Determine the correct answer.\nAcidosis is diagnosed when blood pH falls below 7.35.

Why other options are wrong:
- Option B: Rises above $7.45$ indicates alkalosis, not acidosis.
- Option C: Both (a) and (b) is incorrect because only (a) is the condition for acidosis.
- Option D: Option A is correct, so this option is incorrect.

Key Concepts:
- Blood pH
- Acidosis
- Alkalosis

Memory Aids:
- Acidosis: pH is acidic (lower than normal).
- Alkalosis: pH is alkaline (higher than normal).

Quick Tips:
- Remember the normal blood pH range and the conditions for acidosis and alkalosis.

42 Easy

42 of 144

Understanding +4 | -1

Which of the following will supress the ionisation of acetic acid is aqueous solution?

View answer and explanation

Correct option: B

\mathrm{HCl}

Option B (

\mathrm{HCl}

) is correct because it provides a common ion (H+) which suppresses the ionization of acetic acid due to the common ion effect.

Solution approach

Correct Answer + Core Reason: $\mathrm{HCl}$ . HCl provides a common ion (H+) which suppresses the ionization of acetic acid.

Step-by-Step Solution:

Step 1: Acetic acid ( $\mathrm{CH_3COOH}$ ) is a weak acid that partially ionizes in water: $\mathrm{CH_3COOH(aq) \leftrightharpoons CH_3COO^-(aq) + H^+(aq)}$

Step 2: According to Le Chatelier's principle, adding a common ion to the solution will shift the equilibrium to the left, suppressing the ionization of the weak acid.

Step 3: HCl is a strong acid that completely ionizes in water: $\mathrm{HCl(aq) \rightarrow H^+(aq) + Cl^-(aq)}$

Step 4: The H+ ions from HCl will act as a common ion, suppressing the ionization of acetic acid.

Why other options are wrong:
- Option A: $\mathrm{NaCl}$ : NaCl is a salt of a strong acid and a strong base. It does not provide a common ion that would suppress the ionization of acetic acid.
- Option C: $\mathrm{KCl}$ : KCl is a salt of a strong acid and a strong base. It does not provide a common ion that would suppress the ionization of acetic acid.
- Option D: Unpredictable: The effect is predictable based on the common ion effect.

Key Concepts:
- Common ion effect
- Le Chatelier's principle
- Weak acid ionization

Memory Aids:
- Common ion effect: Adding a common ion suppresses ionization.

Quick Tips:
- Identify the weak acid/base and look for a strong electrolyte that provides a common ion.

43 Easy

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Analyzing +4 | -1

Although

\mathrm{CO}

is neutral but it shows acidic nature on reaction with high

P

and

T

:

View answer and explanation

Correct option: B

\mathrm{NaOH}

The correct answer is B (NaOH) because CO reacts with strong bases like NaOH under high pressure and temperature to form sodium formate, exhibiting acidic behavior.

Solution approach

Correct Answer + Core Reason: $\mathrm{NaOH}$ . CO reacts with strong bases like NaOH under high pressure and temperature.

Step-by-Step Solution:

Step 1: CO is a neutral oxide but can act as a weak Lewis acid under extreme conditions.

Step 2: Strong bases like NaOH are more likely to react with CO under high pressure and temperature.

Step 3: The reaction forms sodium formate: $\mathrm{CO + NaOH} \to \mathrm{HCOONa}$

Step 4: Weak bases like $\mathrm{Ca(OH)}_2$ , $\mathrm{Mg(OH)}_2$ , and $\mathrm{LiOH}$ are less likely to react effectively with CO under these conditions.

Why other options are wrong:
- Option A: $\mathrm{Ca(OH)}_2$ is a weaker base compared to NaOH and less likely to react with CO under high pressure and temperature.
- Option C: $\mathrm{Mg(OH)}_2$ is a weaker base compared to NaOH and less likely to react with CO under high pressure and temperature.
- Option D: $\mathrm{LiOH}$ is a strong base, but less reactive than NaOH under these specific conditions due to other factors like solubility or reaction kinetics.

Key Concepts:
- Acid-base reactions
- Lewis acids and bases
- Reactivity of oxides

Memory Aids:
- Strong bases react more readily with acidic oxides.

Quick Tips:
- Consider the strength of the base when predicting reactions with oxides.

44 Easy

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Remembering +4 | -1

The Henderson’s equation used to calculate the

\mathrm{pOH}

of basic buffer is:

View answer and explanation

Correct option: A

\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}

Option A (

\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}

) is correct because it represents the Henderson-Hasselbalch equation for a basic buffer, relating pOH to pKb and the ratio of conjugate acid to base concentrations.

Solution approach

Correct Answer + Core Reason: $\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}$ . This equation correctly represents the Henderson-Hasselbalch equation for calculating the pOH of a basic buffer.

Step-by-Step Solution:

Step 1: Recall the Henderson-Hasselbalch equation for a basic buffer.

Step 2: The Henderson-Hasselbalch equation relates the pOH of a basic buffer to the $pK_b$ of the base and the ratio of the concentrations of the conjugate acid and the base.

Step 3: The correct equation is $\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}$

Why other options are wrong:
- Option B: $\mathrm{pOH} = \mathrm{p}{K_b} - \log \frac{{\left[ {\mathrm{Conjugate\; acid}} \right]}}{{\left[ {\mathrm{\;Base}} \right]}}$ : This equation is incorrect because it has a negative sign before the logarithm term, which is not the correct form of the Henderson-Hasselbalch equation for a basic buffer.
- Option C: $\mathrm{pOH} = \mathrm{p}{K_b} + \log \frac{{\left[ {\mathrm{Base}} \right]}}{{\left[ {\mathrm{Conjugate\; acid}} \right]}}$ : This equation is incorrect because the ratio of concentrations is inverted. The correct ratio is [Conjugate acid]/[Base].
- Option D: $\mathrm{pOH} = \mathrm{p}{K_b} - \log \frac{{\left[ {\mathrm{Base}} \right]}}{{\left[ {\mathrm{Conjugate\; acid}} \right]}}$ : This equation is incorrect because it has a negative sign before the logarithm term and the ratio of concentrations is inverted.

Key Concepts:
- Henderson-Hasselbalch equation
- Basic buffer
- $pK_b$
- Conjugate acid-base pairs

Memory Aids:
- Remember that for a basic buffer, the equation involves pOH and $pK_b$ .
- The ratio is always [Conjugate acid]/[Base].

Quick Tips:
- Know the difference between the Henderson-Hasselbalch equation for acidic and basic buffers.

45 Easy

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Understanding +4 | -1

Acidity of

B{F_3}

can be explained on the basis of which of the following concept

View answer and explanation

Correct option: C

Lewis concept

Option C (Lewis concept) is correct because

BF_3

is an electron-deficient molecule that accepts a pair of electrons, fitting the definition of a Lewis acid.

Solution approach

Correct Answer + Core Reason: Lewis concept. $BF_3$ accepts electron pairs.

Step-by-Step Solution:

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

Step 2: Arrhenius acids produce $H^+$ ions in water.

Step 3: Bronsted-Lowry acids donate protons.

Step 4: Lewis acids accept electron pairs.

Step 5: $BF_3$ has an incomplete octet and can accept a pair of electrons from a Lewis base.

Step 6: Therefore, the acidity of $BF_3$ can be explained by the Lewis concept.

Why other options are wrong:
- Option A: Arrhenius concept is incorrect because $BF_3$ does not produce $H^+$ ions in water.
- Option B: Bronsted Lowry concept is incorrect because $BF_3$ does not donate protons.
- Option C: Lewis concept is the correct answer because $BF_3$ accepts electron pairs.
- Option D: 2 and 3 both is incorrect because only the Lewis concept applies.

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

Memory Aids:
- $BF_3$ is electron deficient, so it's a Lewis acid.

Quick Tips:
- Remember that molecules with incomplete octets often act as Lewis acids.

46 Easy

46 of 144

Understanding +4 | -1

Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest

pH

?

View answer and explanation

Correct option: A

\mathrm{BaC}{\mathrm{l}_2}

Option A (

\mathrm{BaC}{\mathrm{l}_2}

) is correct because barium hydroxide, formed by the hydrolysis of BaCl2, is the strongest base among the hydroxides formed by the other alkaline earth metals listed, leading to the highest pH.

Solution approach

Correct Answer + Core Reason: $\mathrm{BaC}{\mathrm{l}_2}$ . $\mathrm{BaC}{\mathrm{l}_2}$ will produce the highest pH because it forms the strongest base upon hydrolysis.

Step-by-Step Solution:

Step 1: Explain hydrolysis of salts.\nSalts formed from strong acids and strong bases result in neutral solutions. Salts formed from strong bases and weak acids result in basic solutions. Salts formed from weak bases and strong acids result in acidic solutions.

Step 2: Analyze the hydrolysis of each salt.\nAll the given salts are chlorides of group 2 elements (alkaline earth metals). They are formed from the reaction of a strong acid (HCl) and a base of the form $M(OH)_2$ , where M is the metal.
$MCl_2(aq) + 2H_2O(l) \rightleftharpoons M(OH)_2(aq) + 2HCl(aq)$

Step 3: Determine the strength of the bases.\nThe strength of the bases $M(OH)_2$ increases down the group. Therefore, $Ba(OH)_2$ is the strongest base among the given options, followed by $Sr(OH)_2$ , $Ca(OH)_2$ , and $Mg(OH)_2$ .

Step 4: Relate base strength to pH.\nThe stronger the base, the higher the pH of the solution.

Step 5: Determine the correct answer.\nTherefore, $\mathrm{BaC}{\mathrm{l}_2}$ will produce the highest pH.

Why other options are wrong:
- Option B: $\mathrm{MgC}{\mathrm{l}_2}$ will produce the lowest pH among the given options because $Mg(OH)_2$ is the weakest base.
- Option C: $\mathrm{CaC}{\mathrm{l}_2}$ will produce a pH lower than $\mathrm{BaC}{\mathrm{l}_2}$ because $Ca(OH)_2$ is a weaker base than $Ba(OH)_2$ .
- Option D: $\mathrm{SrC}{\mathrm{l}_2}$ will produce a pH lower than $\mathrm{BaC}{\mathrm{l}_2}$ because $Sr(OH)_2$ is a weaker base than $Ba(OH)_2$ .

Key Concepts:
- Hydrolysis of salts
- Strength of bases
- pH
- Group 2 elements

Memory Aids:
- Base strength increases down group 2.

Quick Tips:
- Remember the trends in base strength for group 2 hydroxides.

47 Easy

47 of 144

Remembering +4 | -1

According to Arrhenius concept, base is a substance that:

View answer and explanation

Correct option: D

Gives

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

ions in aqueous solution

Option D (Gives

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

ions in aqueous solution) is correct because, according to the Arrhenius concept, a base is a substance that produces hydroxide ions in water.

Solution approach

Correct Answer + Core Reason: Gives $\mathrm{O}{\mathrm{H}^ - }$ ions in aqueous solution. Arrhenius bases produce hydroxide ions in water.

Step-by-Step Solution:

Step 1: Recall the Arrhenius definition of a base.

Step 2: According to Arrhenius, a base is a substance that increases the concentration of hydroxide ions ( $\mathrm{O}{\mathrm{H}^ - }$ ) in aqueous solution.

Step 3: Therefore, the correct answer is that a base gives $\mathrm{O}{\mathrm{H}^ - }$ ions in aqueous solution.

Why other options are wrong:
- Option A: Gives a pair of protons is incorrect because this describes an acid according to some definitions, but not Arrhenius.
- Option B: Donates a proton is incorrect because this describes a Bronsted-Lowry acid.
- Option C: Accepts an electron pair is incorrect because this describes a Lewis acid.
- Option D: Gives $\mathrm{O}{\mathrm{H}^ - }$ ions in aqueous solution is the correct definition of an Arrhenius base.

Key Concepts:
- Arrhenius definition of acids and bases
- Bronsted-Lowry definition of acids and bases
- Lewis definition of acids and bases

Memory Aids:
- Arrhenius: Acids release H+, Bases release OH-

Quick Tips:
- Remember the different definitions of acids and bases (Arrhenius, Bronsted-Lowry, Lewis).

48 Easy

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Understanding +4 | -1

A higher value for equilibrium constant,

K

shows that:

View answer and explanation

Correct option: A

The reaction has gone to near completion towards right

Option A (The reaction has gone to near completion towards right) is correct because a higher value of the equilibrium constant K indicates that the concentration of products is much greater than the concentration of reactants at equilibrium.

Solution approach

Correct Answer + Core Reason: The reaction has gone to near completion towards right. A high K indicates a large product to reactant ratio.

Step-by-Step Solution:

Step 1: Define the equilibrium constant, K:
$K = \frac{{\text{{[Products]}}}}{{\text{{[Reactants]}}}}$

Step 2: Interpret a high value of K: A high value of K means that the numerator (concentration of products) is much larger than the denominator (concentration of reactants) at equilibrium.

Step 3: Conclude: This indicates that the reaction has proceeded nearly to completion towards the right, forming a large amount of products.

Why other options are wrong:
- Option B: The reaction has not yet started: If the reaction hasn't started, there would be no products, and K would be very small or zero.
- Option C: The reaction has gone to near completion towards left: This would mean that the reactants are in much higher concentration than the products, resulting in a very small value of K.
- Option D: None of the above: Option A is correct.

Key Concepts:
- Equilibrium constant (K)
- Interpretation of K values
- Product-to-reactant ratio

Memory Aids:
- High K: Products favored. Low K: Reactants favored.

Quick Tips:
- Remember that K is the ratio of products to reactants at equilibrium.

49 Easy

49 of 144

Analyzing +4 | -1

Assertion :

S O_2(g)+\frac{1}{2} O_2(g) \rightleftharpoons S O_3(g)+

heat.
Forward reaction is favoured at high temperature and low pressure.

Reason : Reaction is exothermic.

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. The forward reaction is favored at low temperature and high pressure because it is exothermic and has fewer moles of gas on the product side.

Solution approach

Correct Answer + Core Reason: If assertion is false but reason is true. Exothermic reactions are favored at low temperatures and high pressure favors fewer gas moles.

Step-by-Step Solution:

Step 1: The given reaction is $S O_2(g)+\frac{1}{2} O_2(g) \rightleftharpoons S O_3(g) + \text{heat}$ .

Step 2: The reaction is exothermic, as heat is released.

Step 3: According to Le Chatelier's principle, exothermic reactions are favored at low temperatures.

Step 4: The number of moles of gas on the reactant side is $1 + \frac{1}{2} = \frac{3}{2}$ , and on the product side is 1. Therefore, the forward reaction is favored at high pressure.

Step 5: The assertion states that the forward reaction is favored at high temperature and low pressure, which is incorrect.

Step 6: The reason states that the reaction is exothermic, which is correct.

Why other options are wrong:
- Option A: The assertion is false.
- Option B: The assertion is false.
- Option C: The assertion is false.

Key Concepts:
- Le Chatelier's principle
- Exothermic and endothermic reactions
- Effect of temperature and pressure on equilibrium

Memory Aids:
- Remember that exothermic reactions are favored at low temperatures and endothermic reactions are favored at high temperatures.
- High pressure favors the side with fewer moles of gas.

Quick Tips:
- Always consider the effect of temperature and pressure separately when applying Le Chatelier's principle.

50 Easy

50 of 144

Application +4 | -1

The

\mathrm{pH}

of a

\mathrm{0}\mathrm{.02}\;\mathrm{M}

solution of hydrochloric acid is:

View answer and explanation

Correct option: B

\mathrm{1}\mathrm{.7}

Option B (

\mathrm{1}\mathrm{.7}

) is correct because the

pH

of a 0.02 M solution of hydrochloric acid, a strong acid, is calculated using the formula

pH = -log[H^+]

.

Solution approach

Correct Answer + Core Reason: $\mathrm{1}\mathrm{.7}$ . The $pH$ is calculated using the formula $pH = -log[H^+]$ .

Step-by-Step Solution:

Step 1: Hydrochloric acid (HCl) is a strong acid, so it completely dissociates in water: $\mathrm{HCl} \to {\mathrm{H}^ + } + \mathrm{Cl}^-$

Step 2: The concentration of $\mathrm{H}^+$ ions is equal to the concentration of HCl, which is 0.02 M.

Step 3: $pH = -log[{\mathrm{H}^ + }] = -log(0.02)$

Step 4: $pH = -log(2 \times 10^{-2}) = -(log(2) + log(10^{-2})) = -(0.3010 - 2) = -(-1.699) = 1.699 \approx 1.7$

Why other options are wrong:
- Option A: This is incorrect. The correct $pH$ is approximately 1.7.
- Option C: This is incorrect. The correct $pH$ is approximately 1.7.
- Option D: This is incorrect. The correct $pH$ is approximately 1.7.

Key Concepts:
- $pH$ calculation
- Strong acids
- Dissociation of acids

Memory Aids:
- $pH = -log[{\mathrm{H}^ + }]$

Quick Tips:
- Remember that strong acids completely dissociate.

51 Easy

51 of 144

Analyzing +4 | -1

Which one of the following statement is not true?

View answer and explanation

Correct option: C

\mathrm{The}\;\mathrm{ pH}\;\mathrm{ of}\;1 \times {10^{ - 8}}\;\mathrm{M}\;\mathrm{ HCl}\;\mathrm{ is}\;8

Option C is correct because the pH of a very dilute strong acid like

1 \times 10^{-8} M \;HCl

cannot be 8. It must be less than 7, considering the contribution of water's autoionization.

Solution approach

Correct Answer + Core Reason: $\mathrm{The}\;\mathrm{ pH}\;\mathrm{ of}\;1 \times {10^{ - 8}}\;\mathrm{M}\;\mathrm{ HCl}\;\mathrm{ is}\;8$ . The pH of a dilute acid cannot be greater than 7.

Step-by-Step Solution:

Step 1: Recognize that HCl is a strong acid and completely dissociates in water.

Step 2: The concentration of $H^+$ ions from HCl is $1 \times 10^{-8} M$ .

Step 3: Consider the autoionization of water: ${H_2}O \rightleftharpoons {H^ + } + O{H^ - }$ , where $[H^+] = [OH^-] = 1 \times 10^{-7} M$ .

Step 4: The total $[H^+]$ is the sum of $[H^+]$ from HCl and water: $1 \times 10^{-8} + 1 \times 10^{-7} = 1.1 \times 10^{-7} M$ .

Step 5: Calculate the pH: $pH = -log[H^+] = -log(1.1 \times 10^{-7}) \approx 6.96$ .

Step 6: The pH should be less than 7, not 8.

Why other options are wrong:
- Option A: The conjugate base of ${\mathrm{H}_2}\mathrm{PO}_4^ - \;\mathrm{is}\;\mathrm{ HPO}_4^{2 - }$ is correct. ${\mathrm{HPO}_4^{2 - }}$ is formed by removing a proton from ${\mathrm{H}_2}\mathrm{PO}_4^ -$ .
- Option B: $\mathrm{pH} + \mathrm{pOH} = 14$ for all aqueous solutions is correct at 25°C.
- Option D: 96500 C of electricity when passed through a $\mathrm{CuS}{\mathrm{O}_4}$ solution deposit 1 g equivalent of copper at the cathode is correct based on Faraday's laws of electrolysis.

Key Concepts:
- pH calculation
- Strong acids
- Autoionization of water
- Conjugate acid-base pairs
- Faraday's laws of electrolysis

Memory Aids:
- pH = -log[H+]
- Strong acids completely dissociate

Quick Tips:
- Always consider the autoionization of water when dealing with very dilute acid or base solutions.
- The pH of an acidic solution must be less than 7.

52 Easy

52 of 144

Understanding +4 | -1

Which is a basic salt?

View answer and explanation

Correct option: D

2\mathrm{PbC}{\mathrm{O}_3}\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2}

Option D (

2\mathrm{PbC}{\mathrm{O}_3}\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2}

) is a basic salt because it contains hydroxide ions in its formula.

Solution approach

Correct Answer + Core Reason: $2\mathrm{PbC}{\mathrm{O}_3}\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2}$ is a basic salt because it contains hydroxide ions.

Step-by-Step Solution:

Step 1: Define a basic salt.\nA basic salt is a salt that contains hydroxide ions in addition to the cation and anion of the salt.

Step 2: Examine each option to determine if it contains hydroxide ions.
$\mathrm{PbS}$ is a normal salt.
$\mathrm{PbC}{\mathrm{O}_3}$ is a normal salt.
$\mathrm{PbS}{\mathrm{O}_4}$ is a normal salt.
$2\mathrm{PbC}{\mathrm{O}_3}\mathrm{Pb}{\left( {\mathrm{OH}} \right)_2}$ contains hydroxide ions.

Why other options are wrong:
- Option A: $\mathrm{PbS}$ is a normal salt formed from a weak acid and a weak base. It does not contain hydroxide ions.
- Option B: $\mathrm{PbC}{\mathrm{O}_3}$ is a normal salt formed from a weak acid and a weak base. It does not contain hydroxide ions.
- Option C: $\mathrm{PbS}{\mathrm{O}_4}$ is a normal salt formed from a strong acid and a weak base. It does not contain hydroxide ions.

Key Concepts:
- Basic salts
- Salt hydrolysis

Memory Aids:
- Basic salts contain hydroxide ions.

Quick Tips:
- Look for hydroxide ions in the formula of the salt to identify basic salts.

53 Easy

53 of 144

Application +4 | -1

The pH of millimolar

\mathrm{HCl}

is

View answer and explanation

Correct option: B

3

Option B (3) is correct because millimolar HCl means 10^-3 M, and pH is the negative logarithm of the H+ concentration.

Solution approach

Correct Answer + Core Reason: 3. pH is the negative logarithm of the H+ concentration.

Step-by-Step Solution:

Step 1: Convert millimolar to molar.\nMillimolar means $10^{-3}$ M, so the concentration of HCl is $10^{-3}$ M.

Step 2: Calculate the pH.\nSince HCl is a strong acid, it completely dissociates in water, so $[H^+] = [HCl] = 10^{-3} M$ .
$pH = -log[H^+] = -log(10^{-3}) = 3$

Step 3: Final Answer
$pH = 3$

Why other options are wrong:
- Option A: This is incorrect because a pH of 1 corresponds to a 0.1 M solution of HCl, not a millimolar solution.
- Option C: This is incorrect because a pH of 2 corresponds to a 0.01 M solution of HCl, not a millimolar solution.
- Option D: This is incorrect because a pH of 4 corresponds to a $10^{-4}$ M solution of HCl, not a millimolar solution.

Key Concepts:
- pH calculation
- Strong acid dissociation
- Molarity and millimolarity

Memory Aids:
- pH = -log[H+]
- Strong acids completely dissociate

Quick Tips:
- Remember that strong acids completely dissociate in water.
- Pay attention to the units (millimolar vs. molar).

54 Easy

54 of 144

Understanding +4 | -1

If the value of equilibrium constant for a particular reaction is

1.6 \times {10^{12}}

, then at equilibrium the
system will contain

View answer and explanation

Correct option: B

mostly products

Option B (mostly products) is correct because a very large equilibrium constant indicates that the reaction strongly favors the formation of products at equilibrium.

Solution approach

Correct Answer + Core Reason: mostly products. A large equilibrium constant indicates that the reaction favors product formation.

Step-by-Step Solution:

Step 1: Understand the meaning of the equilibrium constant (K). K is the ratio of products to reactants at equilibrium.

Step 2: A large value of K (e.g., $1.6 \times 10^{12}$ ) indicates that the concentration of products is much greater than the concentration of reactants at equilibrium.

Step 3: Therefore, at equilibrium, the system will contain mostly products.

Why other options are wrong:
- Option A: mostly reactants This is incorrect because a large K value indicates the opposite.
- Option C: similar amounts of reactants and products This is incorrect because a very large K value indicates a significant difference in concentrations.
- Option D: all reactants This is incorrect because a large K value indicates that most of the reactants have been converted to products.

Key Concepts:
- Equilibrium constant (K)
- Product-favored vs. Reactant-favored reactions

Memory Aids:
- Large K = mostly products; Small K = mostly reactants

Quick Tips:
- A very large K value (>>1) indicates that the reaction goes almost to completion, forming mostly products.

55 Easy

55 of 144

Application +4 | -1

If equilibrium constants of reactions

AB\,and\,\,CD\,are\,{K_1}\,and\,\,{K_2}

respectively, then the equilibrium constant for the reaction

D + BA + C

will be

View answer and explanation

Correct option: B

\frac{1}{{{K_1}{K_2}}}

Option B is correct because reversing and adding reactions results in multiplying or inverting the equilibrium constants.

Solution approach

Correct Answer + Core Reason: $\frac{1}{{{K_1}{K_2}}}$ . Reversing and adding reactions involves inverting and multiplying equilibrium constants.

Step-by-Step Solution:

Step 1: Write the given reactions and their equilibrium constants:
$A + B \rightleftharpoons AB$ , $K_1$
$C + D \rightleftharpoons CD$ , $K_2$

Step 2: Write the target reaction: $D + BA \rightleftharpoons A + C + D$ \nThis can be obtained by reversing the first two reactions and adding them:
$AB \rightleftharpoons A + B$ , $1/K_1$
$CD \rightleftharpoons C + D$ , $1/K_2$ \nAdding these two: $AB + CD \rightleftharpoons A + B + C + D$

Step 3: The target reaction is $D + BA \rightleftharpoons A + C + D$ . This is obtained by reversing the first reaction and adding it to the reverse of the second reaction. The equilibrium constant is then:
$K = \frac{1}{K_1} \times \frac{1}{K_2} = \frac{1}{K_1K_2}$

Why other options are wrong:
- Option A: This is incorrect because equilibrium constants are multiplied when reactions are added, and inverted when reactions are reversed, not added in the denominator.
- Option C: This is incorrect because it only considers the ratio of the original equilibrium constants, not the effect of reversing the reactions.
- Option D: This is incorrect because it would be the equilibrium constant for adding the two original reactions, not for the target reaction.

Key Concepts:
- Manipulating equilibrium reactions
- Combining equilibrium constants

Memory Aids:
- Reversing a reaction inverts K.
- Multiplying reactions multiplies K.

Quick Tips:
- Always write out the reactions and their corresponding equilibrium constants before manipulating them.

56 Easy

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Remembering +4 | -1

For a homogeneous chemical reaction,

K_p=K_C

when

View answer and explanation

Correct option: A

\Delta n=0

Option A (

\Delta n=0

) is correct because

K_p = K_c(RT)^{\Delta n}

, so when

\Delta n = 0

,

K_p = K_c

.

Solution approach

Correct Answer + Core Reason: $\Delta n=0$ . $K_p = K_c$ when the change in the number of moles of gaseous reactants and products is zero.

Step-by-Step Solution:

Step 1: Recall the relationship between $K_p$ and $K_c$ .
$K_p = K_c(RT)^{\Delta n}$ , where $\Delta n = \text{moles of gaseous products} - \text{moles of gaseous reactants}$ .

Step 2: Determine the condition for $K_p = K_c$ .\nFor $K_p$ to be equal to $K_c$ , the term $(RT)^{\Delta n}$ must be equal to 1.

Step 3: Analyze the term $(RT)^{\Delta n} = 1$ .\nThis is only true if $\Delta n = 0$ , because $(RT)^0 = 1$ .

Step 4: Conclude the condition.\nTherefore, $K_p = K_c$ when $\Delta n = 0$ .

Why other options are wrong:
- Option B: Incorrect because if $\Delta n = 1$ , then $K_p = K_c(RT)$ , which means $K_p \neq K_c$ .
- Option C: Incorrect because if $\Delta n = 2$ , then $K_p = K_c(RT)^2$ , which means $K_p \neq K_c$ .
- Option D: Incorrect because if $\Delta n = \infty$ , then $K_p$ would be infinitely larger than $K_c$ , which means $K_p \neq K_c$ .

Key Concepts:
- Equilibrium Constants $K_p$ and $K_c$
- Relationship between $K_p$ and $K_c$
- Change in the number of moles of gaseous species ( $\Delta n$ )

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

Quick Tips:
- Always calculate $\Delta n$ carefully by subtracting the moles of gaseous reactants from the moles of gaseous products.
- If $\Delta n = 0$ , $K_p = K_c$ regardless of the temperature.

57 Easy

57 of 144

Application +4 | -1

Calculate the

pH

of a solution in which hydrogen ion concentration is

0.005\mathrm{ }\;g - equi/L?

View answer and explanation

Correct option: A

2.3

Option A (2.3) is correct because pH is calculated as the negative logarithm of the hydrogen ion concentration, and in this case, it's -log(0.005) which is approximately 2.3.

Solution approach

Correct Answer + Core Reason: 2.3. pH is the negative logarithm of the hydrogen ion concentration.

Step-by-Step Solution:

Step 1: Understand the definition of pH: $pH = -log[H^+]$ .

Step 2: The hydrogen ion concentration is given as 0.005 g-equiv/L. Since $H^+$ has a molar mass of approximately 1 g/mol, 0.005 g-equiv/L is approximately equal to 0.005 M.

Step 3: Calculate the pH: $pH = -log(0.005)$ .

Step 4: $pH = -log(5 \times 10^{-3})$ .

Step 5: $pH = - (log(5) + log(10^{-3}))$ .

Step 6: $pH = - (0.699 - 3)$ .

Step 7: $pH = - (-2.301)$ .

Step 8: $pH = 2.301 \approx 2.3$ .

Why other options are wrong:
- Option B (2.8): This is not the correct calculation of -log(0.005).
- Option C (2.9): This is not the correct calculation of -log(0.005).
- Option D (2.6): This is not the correct calculation of -log(0.005).

Key Concepts:
- pH definition
- Logarithms

Memory Aids:
- $pH = -log[H^+]$

Quick Tips:
- Remember the definition of pH and how to calculate it using logarithms.

58 Easy

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Remembering +4 | -1

In the titration of

\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}\;\mathrm{and}\;\mathrm{ HCl}

, the indicator used is

View answer and explanation

Correct option: A

Methyl orange

Option A (Methyl orange) is correct because methyl orange is a suitable indicator for titrations involving a strong acid (HCl) and a weak base (Na2CO3), as its color change occurs in the acidic pH range.

Solution approach

Correct Answer + Core Reason: Methyl orange. Methyl orange is a suitable indicator for the titration of a strong acid and a weak base.

Step-by-Step Solution:

Step 1: Identify the type of titration.\nThe titration involves a strong acid (HCl) and a weak base ( $\mathrm{N}{\mathrm{a}_2}\mathrm{C}{\mathrm{O}_3}$ ).

Step 2: Determine the appropriate indicator.\nFor titrations involving a strong acid and a weak base, an indicator that changes color in the acidic pH range is required. Methyl orange is a suitable indicator because its color change occurs in the pH range of 3.1 - 4.4.

Step 3: Explain why other indicators are not suitable.
- Phenolphthalein changes color in the pH range of 8.3 - 10.0, which is basic and not suitable for this titration.
- Methylene blue and litmus are not commonly used as indicators in acid-base titrations.

Why other options are wrong:
- Option B: Methylene blue is not a common indicator for acid-base titrations.
- Option C: Phenolphthalein is suitable for titrations involving a strong base and a weak acid.
- Option D: Litmus has a broad pH range and is not precise enough for titrations.

Key Concepts:
- Acid-base titrations
- Indicators
- pH range

Memory Aids:
- Methyl orange is for strong acid-weak base titrations.

Quick Tips:
- Choose an indicator whose pH range corresponds to the expected pH at the equivalence point.

59 Easy

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Remembering +4 | -1

{K_c}

for

{m_1}A + {m_2}B = {n_1}C + {n_2}D

is given by:

View answer and explanation

Correct option: C

{K_c} = \frac{{{{\left[ C \right]}^{{n_1}}}{{\left[ D \right]}^{{n_2}}}}}{{{{\left[ A \right]}^{{m_1}}}{{\left[ B \right]}^{{m_2}}}}}

Option C (

{K_c} = \frac{{{{\left[ C \right]}^{{n_1}}}{{\left[ D \right]}^{{n_2}}}}}}{{{{\left[ A \right]}^{{m_1}}}{{\left[ B \right]}^{{m_2}}}}}}

) is correct because it represents the equilibrium constant expression with products raised to their stoichiometric coefficients divided by reactants raised to their stoichiometric coefficients.

Solution approach

Correct Answer + Core Reason: ${K_c} = \frac{{{{\left[ C \right]}^{{n_1}}}{{\left[ D \right]}^{{n_2}}}}}}{{{{\left[ A \right]}^{{m_1}}}{{\left[ B \right]}^{{m_2}}}}}$ . The equilibrium constant is products over reactants, each raised to the power of their stoichiometric coefficients.

Step-by-Step Solution:

Step 1: Write the general equilibrium reaction: ${m_1}A + {m_2}B \rightleftharpoons {n_1}C + {n_2}D$

Step 2: The equilibrium constant $K_c$ is defined as the ratio of the product of the concentrations of the products, each raised to the power of its stoichiometric coefficient, to the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

Step 3: Therefore, $K_c = \frac{{[C]^{n_1}[D]^{n_2}}}{{[A]^{m_1}[B]^{m_2}}}$

Why other options are wrong:
- Option A: ${K_c} = \frac{{{{\left[ A \right]}^{{m_1}}}{{\left[ B \right]}^{{m_2}}}}}{{\left[ C \right] \times \left[ D \right]}}$ This is incorrect because it has reactants in the numerator and products in the denominator, and it doesn't raise the concentrations to the power of their stoichiometric coefficients.
- Option B: ${K_c} = \frac{{{{\left[ A \right]}^{{n_1}}}{{\left[ B \right]}^{{n_2}}}}}{{{{\left[ C \right]}^{{m_1}}}{{\left[ D \right]}^{{m_2}}}}}$ This is incorrect because it has reactants in the numerator and products in the denominator, and it uses the wrong stoichiometric coefficients.
- Option D: ${K_c} = \frac{{{{\left[ C \right]}^{{m_1}}} \times {{\left[ D \right]}^{{m_2}}}}}{{{{\left[ A \right]}^{{n_1}}} \times {{\left[ B \right]}^{{n_2}}}}}$ This is incorrect because it uses the wrong stoichiometric coefficients.

Key Concepts:
- Equilibrium constant expression
- Stoichiometric coefficients

Memory Aids:
- Products over reactants (always).

Quick Tips:
- Always remember to raise the concentrations to the power of their stoichiometric coefficients in the balanced chemical equation.

60 Easy

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Analyzing +4 | -1

Which of the following is the strongest conjugate base?

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }

The correct answer is B (

\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }

) because it is the conjugate base of a weak acid (acetic acid), making it a stronger base than the conjugate bases of strong acids.

Solution approach

Correct Answer + Core Reason: $\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }$ . It is the conjugate base of a weak acid.

Step-by-Step Solution:

Step 1: Stronger acids have weaker conjugate bases, and weaker acids have stronger conjugate bases.

Step 2: $\mathrm{C}{\mathrm{l}^ - }$ is the conjugate base of a strong acid (HCl), so it is a weak base.

Step 3: $\mathrm{SO}_4^{2 - }$ is the conjugate base of a strong acid ( $\mathrm{H}_2\mathrm{SO}_4$ ), so it is a weak base.

Step 4: $\mathrm{NO}_2^ -$ is the conjugate base of a moderately strong acid ( $\mathrm{HNO}_2$ ), so it is a relatively weak base.

Step 5: $\mathrm{C}{\mathrm{H}_3}\mathrm{CO}{\mathrm{O}^ - }$ is the conjugate base of a weak acid ( $\mathrm{CH}_3\mathrm{COOH}$ ), so it is a strong base.

Why other options are wrong:
- Option A: $\mathrm{C}{\mathrm{l}^ - }$ is the conjugate base of a strong acid (HCl) and is a weak base.
- Option C: $\mathrm{SO}_4^{2 - }$ is the conjugate base of a strong acid ( $\mathrm{H}_2\mathrm{SO}_4$ ) and is a weak base.
- Option D: $\mathrm{NO}_2^ -$ is the conjugate base of a moderately strong acid ( $\mathrm{HNO}_2$ ) and is a relatively weak base.

Key Concepts:
- Conjugate acid-base pairs
- Strength of acids and bases

Memory Aids:
- Strong acid = weak conjugate base, weak acid = strong conjugate base.

Quick Tips:
- Identify the parent acid to determine the strength of the conjugate base.

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