Equilibrium Practice

Take 5 chapter-wise practice tests of 25 questions each on Equilibrium for NEET with +4/-1 scoring, answer review, and concise explanations.

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NEET Chemistry - Chapter 5

Equilibrium Practice Tests

5 original practice tests, 25 questions each, NEET 4/-1 marking, and answer review after submission.

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Basic idea
1. A system at equilibrium has:
- A. No net macroscopic change (Correct)
- B. No reaction at all
- C. Only products
- D. Only reactants
Explanation: Forward and reverse changes balance macroscopically.
Dynamic nature
2. Chemical equilibrium is:
- A. Dynamic (Correct)
- B. Static
- C. Stopped
- D. One-directional
Explanation: Both forward and reverse reactions continue at equal rates.
Rate relation
3. At equilibrium, forward and reverse reaction rates are:
- A. Equal (Correct)
- B. Zero
- C. Maximum
- D. Minimum
Explanation: This equality sustains constant concentrations.
Law of mass action
4. For $aA+bB\rightleftharpoons cC+dD$, equilibrium constant in concentration form is:
- A. $K_c=\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ (Correct)
- B. $K_c=\dfrac{[A]^a[B]^b}{[C]^c[D]^d}$
- C. $K_c=[A]+[B]+[C]+[D]$
- D. $K_c=RT$
Explanation: Products appear in numerator and reactants in denominator.
Kc meaning
5. A large value of $K_c$ indicates equilibrium mixture favors:
- A. Products (Correct)
- B. Reactants
- C. Neither side
- D. Catalysts
Explanation: Products dominate at equilibrium when $K_c$ is large.
Kc meaning
6. A very small value of $K_c$ indicates equilibrium mixture favors:
- A. Reactants (Correct)
- B. Products
- C. Both equally
- D. Only catalyst
Explanation: Little product is present at equilibrium.
Reaction quotient
7. If $Q_c
- A. Forward (Correct)
- B. Backward
- C. Not at all
- D. Randomly
Explanation: It must form more products to reach equilibrium.
Reaction quotient
8. If $Q_c > K_c$, the reaction will proceed:
- A. Backward (Correct)
- B. Forward
- C. Remain unchanged
- D. Explosively
Explanation: It must consume excess products to reach equilibrium.
Gaseous equilibrium
9. $K_p$ is used when equilibrium involves:
- A. Partial pressures (Correct)
- B. Only solids
- C. Only liquids
- D. Only ionic strength
Explanation: It is convenient for gaseous systems.
Relation
10. The relation between $K_p$ and $K_c$ is:
- A. $K_p=K_c(RT)^{\Delta n}$ (Correct)
- B. $K_p=K_c/RT$
- C. $K_p=K_c+RT$
- D. $K_p=\Delta nRT$
Explanation: Here $\Delta n$ means gaseous product moles minus reactant moles.
Heterogeneous equilibrium
11. Pure solids and pure liquids are omitted from equilibrium constant because their:
- A. Active masses are taken as unity (Correct)
- B. Concentrations are always zero
- C. Pressures are undefined
- D. Masses cancel by chance
Explanation: Their effective activity is treated as constant.
Reversing reaction
12. If a reaction is reversed, the equilibrium constant becomes:
- A. Reciprocal (Correct)
- B. Double
- C. Unchanged
- D. Zero
Explanation: Forward and reverse constants are inverses.
Scaling reaction
13. If all coefficients in a reaction are doubled, the new equilibrium constant becomes:
- A. $K^2$ (Correct)
- B. $2K$
- C. $K/2$
- D. Unchanged
Explanation: Multiplying the equation multiplies logarithmic extent and raises K to that power.
Combining reactions
14. When two reactions are added, the overall equilibrium constant is:
- A. Product of individual constants (Correct)
- B. Sum of constants
- C. Difference of constants
- D. Always 1
Explanation: This parallels Hess-like algebra for equilibrium constants.
Homogeneous equilibrium
15. All reactants and products in the same phase is called:
- A. Homogeneous equilibrium (Correct)
- B. Heterogeneous equilibrium
- C. Ionic equilibrium only
- D. Thermal equilibrium
Explanation: All species share a common phase.
Le Chatelier
16. Le Chatelier's principle predicts how an equilibrium responds to:
- A. A disturbance (Correct)
- B. Atomic number
- C. Catalyst color
- D. Magnetic field only
Explanation: The system shifts to oppose the imposed change.
Concentration effect
17. Adding a reactant generally shifts equilibrium toward:
- A. Products (Correct)
- B. Reactants
- C. No change ever
- D. Both equally
Explanation: The system consumes part of the added reactant.
Product removal
18. Removing a product shifts equilibrium toward:
- A. Products (Correct)
- B. Reactants
- C. No change
- D. Catalyst
Explanation: The system forms more product to compensate.
Catalyst effect
19. A catalyst affects equilibrium by:
- A. Reaching it faster without changing its position (Correct)
- B. Shifting it to products
- C. Shifting it to reactants
- D. Changing K
Explanation: Catalysts accelerate both directions similarly.
Temperature effect
20. Changing temperature can change equilibrium constant because temperature changes:
- A. Relative energetic favorability (Correct)
- B. Stoichiometric coefficients
- C. Activities of solids only
- D. Only color
Explanation: K depends on temperature alone for a given reaction.
Endothermic shift
21. For an endothermic reaction, increasing temperature shifts equilibrium toward:
- A. Products (Correct)
- B. Reactants
- C. No change
- D. Catalyst side
Explanation: Heat behaves like a reactant in an endothermic direction.
Exothermic shift
22. For an exothermic reaction, increasing temperature shifts equilibrium toward:
- A. Reactants (Correct)
- B. Products
- C. No change
- D. Solids
Explanation: Added heat favors the reverse endothermic direction.
Pressure effect
23. Increasing pressure affects gaseous equilibrium significantly when total moles of gas:
- A. Differ on two sides (Correct)
- B. Are equal on both sides
- C. Contain a catalyst
- D. Include solids
Explanation: Equilibrium shifts toward fewer gaseous moles.
Inert gas
24. Adding inert gas at constant volume to a gaseous equilibrium causes:
- A. No shift in equilibrium position (Correct)
- B. Shift to products
- C. Shift to reactants
- D. K to change
Explanation: Partial pressures of reacting gases remain unchanged at constant volume.
Chapter logic
25. The cleanest equilibrium workflow is to write the expression, compare Q with K if needed, and then apply:
- A. Le Chatelier or ICE-table logic (Correct)
- B. Only memorised trends
- C. Only catalyst rule
- D. Only oxidation state
Explanation: That handles both conceptual and numerical equilibrium problems.
Acid base
26. An acid according to Bronsted-Lowry is a proton:
- A. Donor (Correct)
- B. Acceptor
- C. Carrier only
- D. Neutral species
Explanation: Bronsted acid donates H$^+$.
Acid base
27. A base according to Bronsted-Lowry is a proton:
- A. Acceptor (Correct)
- B. Donor
- C. Oxidising agent
- D. Reducing agent
Explanation: Bronsted base accepts H$^+$.
Conjugate pair
28. When NH$_3$ accepts a proton, it forms its conjugate acid:
- A. NH$_4^+$ (Correct)
- B. NH$_2^-$
- C. NH$_4$
- D. NH
Explanation: Addition of one proton gives ammonium ion.
Conjugate pair
29. The conjugate base of $H_2CO_3$ is:
- A. $HCO_3^-$ (Correct)
- B. $CO_3^{2-}$
- C. $H_3O^+$
- D. $OH^-$
Explanation: Loss of one proton gives bicarbonate.
Strong acid
30. A strong acid in water is one that:
- A. Ionises almost completely (Correct)
- B. Ionises partially
- C. Does not donate proton
- D. Always has high molar mass
Explanation: Strong acids dissociate nearly fully in dilute solution.
Weak acid
31. A weak acid in water is one that:
- A. Ionises partially (Correct)
- B. Ionises completely
- C. Never ionises
- D. Has no hydrogen
Explanation: Weak acids establish an equilibrium with water.
Ka
32. Acid dissociation constant for HA is:
- A. $K_a=\dfrac{[H^+][A^-]}{[HA]}$ (Correct)
- B. $K_a=\dfrac{[HA]}{[H^+][A^-]}$
- C. $K_a=[H^+][HA]$
- D. $K_a=[A^-]/[H^+]$
Explanation: This expression measures acid strength.
Kb
33. Base dissociation constant for BOH-type or ammonia-type bases measures:
- A. Strength of a base (Correct)
- B. Strength of an acid
- C. Only pH of water
- D. Only solubility
Explanation: Larger Kb means stronger base.
Water ionisation
34. The ionic product of water at 25$^\circ$C is:
- A. $10^{-14}$ (Correct)
- B. $10^{-7}$
- C. 14
- D. 1
Explanation: $K_w=[H^+][OH^-]=10^{-14}$ at 25$^\circ$C.
pH
35. pH is defined as:
- A. $-\log[H^+]$ (Correct)
- B. $\log[H^+]$
- C. $-\log[OH^-]$
- D. $1/[H^+]$
Explanation: It is the negative logarithm of hydrogen ion concentration.
pOH
36. pOH is defined as:
- A. $-\log[OH^-]$ (Correct)
- B. $\log[OH^-]$
- C. $-\log[H^+]$
- D. $1/[OH^-]$
Explanation: It is the hydroxide-ion analogue of pH.
Relation
37. At 25$^\circ$C, pH + pOH =
- A. 14 (Correct)
- B. 7
- C. 1
- D. 0
Explanation: This comes directly from $K_w=10^{-14}$.
Neutral solution
38. At 25$^\circ$C, a neutral solution has pH:
- A. 7 (Correct)
- B. 0
- C. 14
- D. 1
Explanation: Hydrogen and hydroxide concentrations are equal at $10^{-7}$ M.
Acidic solution
39. A solution with pH 3 has hydrogen ion concentration:
- A. $10^{-3}$ M (Correct)
- B. $10^{-11}$ M
- C. 3 M
- D. $3\times10^{-7}$ M
Explanation: pH is the negative logarithm of [H$^+$].
Basic solution
40. A solution with pOH 2 has hydroxide ion concentration:
- A. $10^{-2}$ M (Correct)
- B. $10^{-12}$ M
- C. 2 M
- D. $5\times10^{-2}$ M
Explanation: By definition, pOH = -log[OH$^-$].
pH of strong acid
41. pH of 0.01 M HCl is:
- A. 2 (Correct)
- B. 1
- C. 12
- D. 7
Explanation: Strong acid dissociates almost fully, so [H$^+$] = $10^{-2}$ M.
pH of strong base
42. pH of 0.001 M NaOH is:
- A. 11 (Correct)
- B. 3
- C. 10
- D. 13
Explanation: pOH = 3, so pH = 11.
Weak acid
43. For a weak acid, increasing dilution generally:
- A. Increases degree of ionisation (Correct)
- B. Decreases degree of ionisation
- C. Keeps ionisation fixed
- D. Makes Ka zero
Explanation: Ostwald's dilution law captures this trend.
Strength relation
44. If Ka of acid A is greater than Ka of acid B, acid A is:
- A. Stronger (Correct)
- B. Weaker
- C. Equally strong always
- D. More basic
Explanation: Greater dissociation constant means greater ionisation tendency.
Strength relation
45. If Kb of base X is greater than Kb of base Y, base X is:
- A. Stronger (Correct)
- B. Weaker
- C. Neutral
- D. Amphoteric only
Explanation: Larger Kb indicates stronger proton-accepting tendency.
Conjugate concept
46. A strong acid has a conjugate base that is generally:
- A. Very weak (Correct)
- B. Very strong
- C. Always strong
- D. Neutral only
Explanation: Acid-base strengths are inversely related in conjugate pairs.
Conjugate concept
47. A strong base has a conjugate acid that is generally:
- A. Very weak (Correct)
- B. Very strong
- C. Neutral only
- D. Always acidic
Explanation: The conjugate pair shows inverse strength relation.
Amphiprotic species
48. $HCO_3^-$ can act as both acid and base, so it is:
- A. Amphiprotic (Correct)
- B. Only acidic
- C. Only basic
- D. Non-electrolyte
Explanation: It can donate or accept a proton.
Ka Kb relation
49. For a conjugate acid-base pair at 25$^\circ$C:
- A. $K_aK_b=K_w$ (Correct)
- B. $K_a+K_b=K_w$
- C. $K_a=K_b$ always
- D. $K_a/K_b=14$
Explanation: This relation connects conjugate strengths.
Chapter logic
50. Ionic-equilibrium questions become far easier once you translate acid or base strength into one of four anchors: Ka, Kb, pH, or:
- A. Degree of ionisation (Correct)
- B. Boiling point
- C. Orbital count
- D. Molar mass only
Explanation: These anchors keep the chapter quantitative and organized.
Solubility product
51. The equilibrium constant for dissolution of a sparingly soluble salt is:
- A. $K_{sp}$ (Correct)
- B. $K_a$
- C. $K_b$
- D. $K_w$
Explanation: Solubility product applies to saturated sparingly soluble salts.
Solubility product
52. For $AB(s)\rightleftharpoons A^+ + B^-$, $K_{sp}$ equals:
- A. $[A^+][B^-]$ (Correct)
- B. $[AB]$
- C. $[A^+]+[B^-]$
- D. $[A^+]^2[B^-]^2$
Explanation: The solid is omitted from the expression.
Solubility product
53. For $A_2B(s)\rightleftharpoons2A^+ + B^{2-}$, $K_{sp}$ is:
- A. $[A^+]^2[B^{2-}]$ (Correct)
- B. $[A^+][B^{2-}]$
- C. $2[A^+][B^{2-}]$
- D. $[A_2B]$
Explanation: Stoichiometric coefficients become powers in the expression.
Ion product
54. If ionic product exceeds $K_{sp}$, then:
- A. Precipitation occurs (Correct)
- B. More salt dissolves
- C. Nothing happens
- D. pH becomes 7
Explanation: The solution is supersaturated.
Ion product
55. If ionic product is less than $K_{sp}$, the solution is:
- A. Unsaturated (Correct)
- B. Supersaturated
- C. Exactly saturated
- D. Always neutral
Explanation: More solute can still dissolve.
Common ion effect
56. Solubility of a sparingly soluble salt generally decreases in presence of:
- A. A common ion (Correct)
- B. A catalyst
- C. Inert gas
- D. Indicator
Explanation: Equilibrium shifts left to oppose added common ion.
Buffer
57. A buffer solution resists change in:
- A. pH (Correct)
- B. Temperature
- C. Volume
- D. Color
Explanation: It resists pH change on adding small acid or base amounts.
Buffer type
58. A basic buffer can be made from:
- A. Weak base and its salt with strong acid (Correct)
- B. Strong acid and strong base
- C. Weak acid alone
- D. Strong base alone
Explanation: Example: NH$_4$OH/NH$_4$Cl.
Buffer type
59. An acidic buffer can be made from:
- A. Weak acid and its salt with strong base (Correct)
- B. Strong acid and strong base
- C. Weak base alone
- D. Strong acid alone
Explanation: Example: CH$_3$COOH/CH$_3$COONa.
Buffer action
60. A buffer works because one component consumes added acid and the other consumes:
- A. Added base (Correct)
- B. Water
- C. Salt
- D. Heat
Explanation: This paired action stabilizes pH.
Hydrolysis
61. A salt of strong acid and strong base gives solution that is generally:
- A. Neutral (Correct)
- B. Acidic
- C. Basic
- D. Always buffer
Explanation: Neither ion hydrolyses appreciably.
Hydrolysis
62. A salt of weak acid and strong base generally gives:
- A. Basic solution (Correct)
- B. Acidic solution
- C. Neutral solution
- D. Buffer always
Explanation: The anion hydrolyses to produce OH$^-$.
Hydrolysis
63. A salt of strong acid and weak base generally gives:
- A. Acidic solution (Correct)
- B. Basic solution
- C. Neutral solution
- D. Only buffer
Explanation: The cation hydrolyses to produce H$^+$.
Indicator
64. An acid-base indicator changes color because its two forms have different:
- A. Colors in different protonation states (Correct)
- B. Molar masses
- C. Atomic numbers
- D. Crystal defects
Explanation: Indicator equilibrium shifts with pH.
Salt hydrolysis
65. Sodium acetate solution is expected to be:
- A. Basic (Correct)
- B. Acidic
- C. Neutral
- D. Strongly acidic
Explanation: Acetate ion is conjugate base of weak acid acetic acid.
Salt hydrolysis
66. Ammonium chloride solution is expected to be:
- A. Acidic (Correct)
- B. Basic
- C. Neutral
- D. Strongly basic
Explanation: Ammonium ion is conjugate acid of weak base ammonia.
Precipitation
67. Mixing solutions of $AgNO_3$ and NaCl forms precipitate because:
- A. $AgCl$ has low solubility product (Correct)
- B. NaNO$_3$ is colored
- C. Ag$^+$ is neutral
- D. Cl$^-$ is acidic
Explanation: Ionic product easily exceeds the small $K_{sp}$ of silver chloride.
Buffer relation
68. For an acidic buffer, Henderson equation is:
- A. $pH=pK_a+\log\dfrac{[salt]}{[acid]}$ (Correct)
- B. $pH=pK_a-\log\dfrac{[salt]}{[acid]}$
- C. $pH=pOH+14$
- D. $pH=\log K_a$
Explanation: This is the Henderson-Hasselbalch form.
Buffer relation
69. For a basic buffer, pOH is given by:
- A. $pOH=pK_b+\log\dfrac{[salt]}{[base]}$ (Correct)
- B. $pOH=pK_b-\log\dfrac{[salt]}{[base]}$
- C. $pOH=14-pK_b$
- D. $pOH=\log K_b$
Explanation: This is the analogous expression for basic buffers.
Degree of ionization
70. Degree of ionization of a weak electrolyte generally:
- A. Increases on dilution (Correct)
- B. Decreases on dilution
- C. Remains constant
- D. Becomes zero
Explanation: Less concentrated solutions dissociate to a greater extent fractionally.
Ostwald law
71. Ostwald's dilution law is best applied to:
- A. Weak electrolytes (Correct)
- B. Strong electrolytes
- C. Only strong acids
- D. Only salts
Explanation: Strong electrolytes are too extensively dissociated for the approximation.
Solubility
72. Molar solubility of a 1:1 salt AB in pure water is s, so $K_{sp}$ equals:
- A. $s^2$ (Correct)
- B. $2s$
- C. $s$
- D. $4s^2$
Explanation: At equilibrium [A$^+$] = [B$^-$] = s.
Solubility
73. Molar solubility of $A_2B$ in pure water is s, so $K_{sp}$ equals:
- A. $4s^3$ (Correct)
- B. $s^2$
- C. $2s^2$
- D. $s^3$
Explanation: [A$^+$] = 2s and [B$^{2-}$] = s, so product is $(2s)^2(s)$.
Chapter logic
74. Most precipitation questions reduce to comparing:
- A. Ionic product with $K_{sp}$ (Correct)
- B. Ka with Kb
- C. pH with pOH
- D. Kp with Kc
Explanation: That tells whether the solution is unsaturated or supersaturated.
Chapter logic
75. Most buffer questions become routine once the weak acid/base pair and its salt are identified, then linked to:
- A. Henderson-type logarithmic form (Correct)
- B. Gas law
- C. Bond enthalpy
- D. Orbital filling
Explanation: That is the standard shortcut used in NEET.
Numerical pH
76. pH of $10^{-4}$ M HCl is:
- A. 4 (Correct)
- B. 10
- C. 3
- D. 6
Explanation: Strong acid concentration directly gives [H$^+$].
Numerical pH
77. pOH of $10^{-3}$ M KOH is:
- A. 3 (Correct)
- B. 11
- C. 10
- D. 1
Explanation: Strong base concentration directly gives [OH$^-$].
Numerical pH
78. pH of $10^{-2}$ M NaOH is:
- A. 12 (Correct)
- B. 2
- C. 10
- D. 11
Explanation: pOH = 2, so pH = 12.
Numerical pH
79. If [H$^+$] = $10^{-5}$ M, then pOH is:
- A. 9 (Correct)
- B. 5
- C. 14
- D. 10
Explanation: pH = 5 and pH + pOH = 14.
Weak acid numeric
80. For a weak acid HA of concentration C and small ionisation, [H$^+$] is approximately:
- A. $\sqrt{K_aC}$ (Correct)
- B. $K_a/C$
- C. $K_a+C$
- D. $C^2/K_a$
Explanation: This common approximation is valid when ionisation is small.
Weak base numeric
81. For a weak base BOH-type or NH$_3$ type of concentration C and small ionisation, [OH$^-$] is approximately:
- A. $\sqrt{K_bC}$ (Correct)
- B. $K_b/C$
- C. $K_b+C$
- D. $C^2/K_b$
Explanation: This is the analogous weak-base approximation.
Degree ionisation
82. If a weak acid has Ka = $10^{-5}$ and concentration 0.1 M, [H$^+$] is approximately:
- A. $10^{-3}$ M (Correct)
- B. $10^{-5}$ M
- C. $10^{-2}$ M
- D. $10^{-4}$ M
Explanation: $\sqrt{10^{-5}\times10^{-1}}=10^{-3}$.
Degree ionisation
83. For the same weak acid, degree of ionisation is:
- A. 1% (Correct)
- B. 10%
- C. 0.1%
- D. 100%
Explanation: $\alpha=[H^+]/C=10^{-3}/10^{-1}=10^{-2}=1\%$.
Buffer numeric
84. A buffer has equal concentrations of weak acid and its salt. Then pH is:
- A. Equal to pKa (Correct)
- B. Greater than pKa by 1
- C. Less than pKa by 1
- D. Always 7
Explanation: Log term becomes zero in Henderson equation.
Buffer numeric
85. If [salt]/[acid] = 10 in an acidic buffer, then pH is:
- A. pKa + 1 (Correct)
- B. pKa - 1
- C. 14 - pKa
- D. pKa only
Explanation: Log 10 = 1.
Buffer numeric
86. If [salt]/[base] = 100 in a basic buffer, then pOH is:
- A. pKb + 2 (Correct)
- B. pKb - 2
- C. pKb only
- D. 14 - pKb
Explanation: Log 100 = 2.
Ksp numeric
87. For AB with $K_{sp}=10^{-10}$, molar solubility in pure water is:
- A. $10^{-5}$ M (Correct)
- B. $10^{-10}$ M
- C. $10^{-2}$ M
- D. $5\times10^{-5}$ M
Explanation: For a 1:1 salt, $s=\sqrt{K_{sp}}$.
Ksp numeric
88. For $CaF_2$ with molar solubility s, $K_{sp}$ equals:
- A. $4s^3$ (Correct)
- B. $s^2$
- C. $2s^2$
- D. $s^3$
Explanation: [Ca$^{2+}$] = s and [F$^-$] = 2s.
Ksp numeric
89. If $K_{sp}$ of $AgCl$ is $10^{-10}$, molar solubility is:
- A. $10^{-5}$ M (Correct)
- B. $10^{-10}$ M
- C. $10^{-2}$ M
- D. $2\times10^{-5}$ M
Explanation: Again, for 1:1 salts $s=\sqrt{K_{sp}}$.
Common ion numeric
90. Solubility of AgCl in 0.1 M NaCl is much less than in pure water because of:
- A. Common ion effect from Cl$^-$ (Correct)
- B. Catalyst effect
- C. Temperature rise
- D. Hydrogen bonding
Explanation: Added chloride suppresses dissolution equilibrium.
Hydrolysis numeric
91. A salt from strong acid and weak base gives pH:
- A. Less than 7 (Correct)
- B. Greater than 7
- C. Exactly 7
- D. Always 14
Explanation: Cation hydrolysis releases H$^+$.
Hydrolysis numeric
92. A salt from weak acid and strong base gives pH:
- A. Greater than 7 (Correct)
- B. Less than 7
- C. Exactly 7
- D. Always 0
Explanation: Anion hydrolysis generates OH$^-$.
Approximation
93. The approximation $x\ll C$ in weak-electrolyte problems is used to:
- A. Simplify equilibrium expression (Correct)
- B. Change Ka value
- C. Make the acid strong
- D. Ignore temperature
Explanation: It converts quadratic forms into simpler square-root relations.
Approach
94. The best setup for many equilibrium numericals is:
- A. ICE table (Correct)
- B. Bohr radius table
- C. Periodic table only
- D. Density table
Explanation: Initial-change-equilibrium organization prevents algebra mistakes.
Approach
95. If initial concentration of product is already present, the easiest way to handle equilibrium numerically is:
- A. Include it in the ICE table from the start (Correct)
- B. Ignore it
- C. Assume K changes
- D. Treat it as catalyst
Explanation: Initial products alter the reaction quotient and equilibrium shift.
Temperature and K
96. For a given reaction, equilibrium constant changes with:
- A. Temperature (Correct)
- B. Pressure only
- C. Catalyst only
- D. Initial concentration only
Explanation: K is temperature dependent.
Pressure and K
97. Changing pressure does not change the value of equilibrium constant because:
- A. K depends only on temperature (Correct)
- B. Pressure never affects equilibrium
- C. Moles stay same always
- D. Catalysts cancel pressure
Explanation: Pressure may shift equilibrium position, not K itself.
Catalyst and K
98. A catalyst does not change K because it:
- A. Speeds both forward and reverse reactions equally (Correct)
- B. Only speeds reverse reaction
- C. Changes reaction enthalpy
- D. Changes temperature
Explanation: Equilibrium is reached sooner, but its composition stays the same.
Shortcut
99. For strong monoprotic acid, pH calculation is usually direct because:
- A. Ionisation is essentially complete (Correct)
- B. Ka is zero
- C. Water ionises fully
- D. No equilibrium exists
Explanation: [H$^+$] is taken equal to acid concentration.
Chapter logic
100. Equilibrium numericals become manageable once you identify whether the problem belongs to K-expression, pH, buffer, or:
- A. Solubility-product branch (Correct)
- B. Atomic structure branch
- C. Thermochemistry branch
- D. Kinetics branch
Explanation: That classification avoids mixing unrelated formulas.
Le Chatelier mix
101. For $N_2 + 3H_2 \rightleftharpoons 2NH_3$, increasing pressure shifts equilibrium toward:
- A. Ammonia (Correct)
- B. Nitrogen and hydrogen
- C. No shift
- D. Catalyst
Explanation: Equilibrium favors the side with fewer moles of gas.
Le Chatelier mix
102. For the Haber process, removing ammonia from the reaction mixture will:
- A. Shift equilibrium to the right (Correct)
- B. Shift equilibrium to the left
- C. Have no effect
- D. Decrease K
Explanation: Product removal drives further product formation.
Le Chatelier mix
103. If the forward Haber reaction is exothermic, increasing temperature will:
- A. Reduce ammonia yield at equilibrium (Correct)
- B. Increase ammonia yield at equilibrium
- C. Leave yield unchanged
- D. Change catalyst only
Explanation: Higher temperature favors the reverse endothermic direction.
Reaction quotient
104. For a system initially containing only reactants, the value of Q at start is:
- A. 0 (Correct)
- B. 1
- C. K
- D. Infinity
Explanation: No products are present initially, so numerator is zero.
Reaction quotient
105. For a system initially containing excess products and almost no reactants, Q is likely:
- A. Greater than K for many product-rich cases (Correct)
- B. Always zero
- C. Always equal to K
- D. Always less than K
Explanation: Product-heavy mixtures often have Q above K and shift backward.
Acid strength
106. Among acids with Ka values $10^{-2}$ and $10^{-5}$, the stronger acid is the one with:
- A. $10^{-2}$ (Correct)
- B. $10^{-5}$
- C. Both same
- D. Cannot compare
Explanation: Larger Ka means greater ionisation.
Base strength
107. Among bases with Kb values $10^{-4}$ and $10^{-7}$, the stronger base is the one with:
- A. $10^{-4}$ (Correct)
- B. $10^{-7}$
- C. Both same
- D. Cannot compare
Explanation: Again, larger dissociation constant means greater strength.
Buffer choice
108. A suitable acidic buffer pair is:
- A. Acetic acid and sodium acetate (Correct)
- B. HCl and NaCl
- C. NaOH and KOH
- D. HNO$_3$ and KNO$_3$
Explanation: Weak acid plus salt of its conjugate base gives acidic buffer.
Buffer choice
109. A suitable basic buffer pair is:
- A. Ammonia and ammonium chloride (Correct)
- B. HCl and NH$_4$Cl
- C. NaOH and NaCl
- D. Acetic acid and HCl
Explanation: Weak base plus salt of its conjugate acid gives basic buffer.
pH comparison
110. The solution with higher pH is:
- A. More basic (Correct)
- B. More acidic
- C. More neutral
- D. Always more concentrated
Explanation: Higher pH means lower hydrogen-ion concentration.
pH comparison
111. The solution with lower pOH is:
- A. More basic (Correct)
- B. More acidic
- C. Always neutral
- D. Always dilute
Explanation: Lower pOH means larger hydroxide-ion concentration.
Solubility comparison
112. If $K_{sp}$ of salt X is smaller than that of salt Y for comparable stoichiometry, salt X is generally:
- A. Less soluble (Correct)
- B. More soluble
- C. Equally soluble
- D. More basic
Explanation: Smaller Ksp usually means lower solubility.
Common ion mix
113. Adding NaF to saturated $CaF_2$ solution will:
- A. Decrease solubility of $CaF_2$ (Correct)
- B. Increase solubility of $CaF_2$
- C. Have no effect
- D. Destroy Ksp
Explanation: Extra fluoride shifts dissolution equilibrium backward.
Salt hydrolysis mix
114. Which salt gives basic solution?
- A. Sodium carbonate (Correct)
- B. Sodium chloride
- C. Potassium nitrate
- D. Barium chloride
Explanation: Carbonate is conjugate base of weak carbonic acid.
Salt hydrolysis mix
115. Which salt gives acidic solution?
- A. Ammonium nitrate (Correct)
- B. Potassium chloride
- C. Sodium sulfate
- D. Calcium nitrate
Explanation: Ammonium ion is acidic in water.
K relation
116. For the reaction $2SO_2 + O_2 \rightleftharpoons 2SO_3$, $\Delta n_g$ equals:
- A. -1 (Correct)
- B. 1
- C. -2
- D. 0
Explanation: Products have 2 gaseous moles and reactants have 3.
K relation
117. For the above reaction, $K_p$ compared to $K_c$ at a given T is:
- A. $K_p=K_c/(RT)$ (Correct)
- B. $K_p=K_c(RT)$
- C. $K_p=K_c$ always
- D. $K_p=K_c(RT)^2$
Explanation: Since $\Delta n=-1$, $K_p=K_c(RT)^{-1}$.
Dissociation
118. The equilibrium constant for decomposition of $N_2O_4\rightleftharpoons2NO_2$ increases with temperature if the forward reaction is:
- A. Endothermic (Correct)
- B. Exothermic
- C. Athermic
- D. Catalysed
Explanation: Higher temperature favors the endothermic direction, increasing K.
Direction prediction
119. If Q equals K, the system is:
- A. Already at equilibrium (Correct)
- B. Shifting right
- C. Shifting left
- D. Supersaturated
Explanation: No net shift is required.
Direction prediction
120. If products are suddenly added to an equilibrium mixture, the immediate value of Q becomes:
- A. Greater than K (Correct)
- B. Less than K
- C. Equal to zero
- D. Unchanged always
Explanation: Adding products enlarges the numerator of Q.
Concept synthesis
121. A catalyst has no effect on pH or K because it does not alter:
- A. Thermodynamic equilibrium condition (Correct)
- B. Reaction pathway speed
- C. Collision frequency
- D. Activation energy
Explanation: It changes kinetics only.
Concept synthesis
122. For weak acids, pKa is related to Ka such that:
- A. Smaller pKa means stronger acid (Correct)
- B. Larger pKa means stronger acid
- C. pKa has no link with strength
- D. pKa is always positive 14
Explanation: pKa = -log Ka, so stronger acids have larger Ka and smaller pKa.
Concept synthesis
123. The most common NEET slip in equilibrium is using concentration changes to claim K changed, even though K changes only with:
- A. Temperature (Correct)
- B. Pressure
- C. Catalyst
- D. Initial volume
Explanation: Pressure or concentration may shift position, not the value of K itself.
Concept synthesis
124. The safest route in mixed equilibrium questions is to separate them into chemical equilibrium, acid-base equilibrium, or:
- A. Solubility equilibrium (Correct)
- B. Redox only
- C. Kinetics only
- D. Quantum equilibrium
Explanation: This prevents formula crossover errors.
Chapter integration
125. The entire equilibrium chapter becomes much more coherent when you view every shift, pH, and precipitation question as a system trying to:
- A. Re-establish equilibrium under a defined constant (Correct)
- B. Increase temperature only
- C. Lower molar mass
- D. Avoid catalysts
Explanation: That restoration idea unites chemical and ionic equilibrium.

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