Ionic Equilibrium – 40 MCQs Practice

Ionic Equilibrium – 40 MCQs for NEET & JEE Main

This section provides 40 multiple-choice questions covering key concepts from Ionic Equilibrium, essential for your NEET and JEE Main preparation. Each question has four options, with only one correct answer.

Section 1: Multiple Choice Questions (MCQs)

Instructions: Choose the single best answer for each question.

  1. Which of the following is a strong electrolyte? A) Acetic acid (CH3COOH) B) Ammonia (NH3) C) Sodium Chloride (NaCl) D) Water (H2O)
  2. According to the Arrhenius concept, an acid is a substance that: A) Donates a proton in solution. B) Accepts an electron pair. C) Dissociates in water to give H+ ions. D) Dissociates in water to give OH- ions.
  3. The conjugate base of H2SO4 is: A) H2SO4+ B) SO4^2- C) HSO4- D) SO3
  4. Which of the following is an amphiprotic species? A) Cl- B) H2O C) NH4+ D) CH3COOH
  5. A Lewis acid is a species that can: A) Donate a proton. B) Accept a proton. C) Donate an electron pair. D) Accept an electron pair.
  6. Which of the following is a Lewis base? A) BF3 B) AlCl3 C) H+ D) NH3
  7. For a weak acid, HA, if its concentration is C and degree of ionization is α, the expression for Ka (acid dissociation constant) is: A) Cα2 B) Cα2/(1−α) C) C(1−α)/α2 D) α/C
  8. According to Ostwald’s Dilution Law, the degree of ionization (α) of a weak electrolyte: A) Decreases with dilution. B) Increases with dilution. C) Remains constant with dilution. D) Is independent of temperature.
  9. If the pH of a solution is 3 at 25°C, its pOH will be: A) 3 B) 7 C) 11 D) 14
  10. The ionic product of water (Kw) at 25°C is: A) 1.0×10−7 B) 1.0×10−14 C) 1.0×1014 D) 1.0×107
  11. If the hydrogen ion concentration ([H+]) of a solution is 1.0×10−5 M, the solution is: A) Acidic B) Basic C) Neutral D) Amphoteric
  12. The pH of a 0.001 M HCl solution at 25°C is: A) 1 B) 2 C) 3 D) 11
  13. The pH of a 0.005 M NaOH solution at 25°C is: A) 2.3 B) 11.7 C) 7 D) 9.5
  14. For a weak acid, if its concentration (C) is increased, its hydrogen ion concentration ([H+]) will: A) Increase B) Decrease C) Remain constant D) Cannot be determined
  15. Which of the following statements is true for a basic solution at 25°C? A) [H+]>[OH−] B) pH<7 C) pOH<7 D) [H+]=10−7 M
  16. A salt of a strong acid and strong base, when dissolved in water, will form a solution that is: A) Acidic B) Basic C) Neutral D) Amphoteric
  17. The hydrolysis of which type of salt will result in an acidic solution? A) Salt of strong acid and strong base B) Salt of weak acid and strong base C) Salt of strong acid and weak base D) Salt of weak acid and weak base (where Ka = Kb)
  18. For the hydrolysis of a salt of a weak acid and strong base, the hydrolysis constant (Kh) is given by: A) Ka​×Kb​ B) Kw​/Ka​ C) Kw​/Kb​ D) Ka​/Kb​
  19. The pH of a solution of CH3COONH4 (salt of weak acid and weak base) will be approximately 7 if: A) Ka > Kb B) Ka < Kb C) Ka = Kb D) It is impossible for it to be 7
  20. A buffer solution is a mixture that: A) Completely neutralizes any added acid or base. B) Resists change in its pH upon addition of small amounts of strong acid or base. C) Has a pH of exactly 7. D) Contains only strong acid and strong base.
  21. Which of the following combinations forms an acidic buffer solution? A) HCl + NaCl B) CH3COOH + CH3COONa C) NH4OH + NH4Cl D) NaOH + NaCl
  22. The pH of an acidic buffer solution can be calculated using the Henderson-Hasselbalch equation: A) pH=pKa​+log([Acid]/[Salt]) B) pH=pKa​+log([Salt]/[Acid]) C) pOH=pKb​+log([Salt]/[Base]) D) pH=−log[H+]
  23. A buffer solution shows maximum buffer capacity when: A) [Salt] = [Acid] B) [Salt] = 2[Acid] C) [Acid] = 2[Salt] D) pH = 7
  24. The solubility product expression for Silver Chromate (Ag2CrO4) is: A) Ksp​=[Ag+][CrO42−] B) Ksp​=[Ag+]2[CrO42−] C) Ksp​=[Ag+][CrO42−]2 D) Ksp​=[Ag2CrO4]
  25. If the solubility of Calcium Fluoride (CaF2) is ‘s’ mol/L, its solubility product (Ksp​) will be: A) s2 B) 4s2 C) 2s3 D) 4s3
  26. The solubility of a sparingly soluble salt decreases significantly in the presence of a: A) Non-common ion B) Common ion C) Acid D) Base
  27. If the ionic product (Q) of a solution is greater than the solubility product (Ksp​), then: A) The solution is unsaturated. B) No precipitation will occur. C) Precipitation will occur. D) The solution is at equilibrium.
  28. Which of the following has the highest pH at 25°C? A) 0.1 M HCl B) 0.1 M CH3COOH (Ka = 1.8 x 10^-5) C) 0.1 M NaOH D) 0.1 M NH4OH (Kb = 1.8 x 10^-5)
  29. If the pH of a solution changes from 2 to 5, the hydrogen ion concentration ([H+]) changes by a factor of: A) 2.5 B) 3 C) 100 D) 1000
  30. The relationship between Ka and Kb for a conjugate acid-base pair is: A) Ka​/Kb​=Kw​ B) Ka​×Kb​=Kw​ C) Ka​+Kb​=Kw​ D) Ka​−Kb​=Kw​
  31. Which of the following is a non-electrolyte? A) Potassium Nitrate (KNO3) B) Glucose (C6H12O6) C) Hydrochloric Acid (HCl) D) Sodium Hydroxide (NaOH)
  32. What is the pH of a 1.0 M solution of a strong acid like HNO3? A) 0 B) 1 C) 7 D) 14
  33. When a small amount of strong acid is added to a buffer solution, the pH change is minimal because: A) The added acid is completely absorbed by water. B) The acid component of the buffer neutralizes the added acid. C) The base component of the buffer neutralizes the added acid. D) The salt in the buffer precipitates out.
  34. For a sparingly soluble salt PbCl2, if Ksp​=1.6×10−5, its solubility ‘s’ is: A) 1.6×10−5 M B) 4.0×10−3 M C) 1.6×10−2 M D) 1.6×10−4 M
  35. Which of the following salts will produce a neutral solution upon hydrolysis? A) CH3COONa B) NH4Cl C) KNO3 D) (NH4)2SO4
  36. The pKa of acetic acid is 4.74. A buffer solution containing 0.1 M CH3COOH and 0.1 M CH3COONa will have a pH of: A) 2.37 B) 4.74 C) 7.00 D) 9.26
  37. If a solution has a pOH of 5 at 25°C, its [H+] concentration is: A) 1.0×10−5 M B) 1.0×10−9 M C) 1.0×10−7 M D) 1.0×10−14 M
  38. The degree of hydrolysis (h) for a salt of weak acid and weak base is (approximately) independent of concentration when: A) The acid is very weak. B) The base is very weak. C) Both the acid and base are weak, and the ions are monovalent. D) The salt is very soluble.
  39. Adding Sodium Acetate (CH3COONa) to an acetic acid (CH3COOH) solution will: A) Increase the pH. B) Decrease the pH. C) Not change the pH. D) Make the solution strongly acidic.
  40. The solubility of Calcium Hydroxide (Ca(OH)2) in water will decrease if we add: A) HCl B) NaOH C) H2O D) NH4Cl

Section 2: Answer Key

  1. C
  2. C
  3. C
  4. B
  5. D
  6. D
  7. B
  8. B
  9. C
  10. B
  11. A
  12. C
  13. B
  14. A
  15. C
  16. C
  17. C
  18. B
  19. C
  20. B
  21. B
  22. B
  23. A
  24. B
  25. D
  26. B
  27. C
  28. C
  29. D
  30. B
  31. B
  32. A
  33. C
  34. C
  35. C
  36. B
  37. B
  38. C
  39. A
  40. B

Section 3: Detailed Explanations

  1. C) Sodium Chloride (NaCl)
    • Explanation: Strong electrolytes ionize almost completely in solution. HCl, H2SO4, NaOH, KOH, and most salts like NaCl are strong electrolytes. Acetic acid and ammonia are weak electrolytes, and water is a very weak electrolyte.
  2. C) Dissociates in water to give H+ ions.
    • Explanation: According to the Arrhenius concept, an acid is a substance that produces H+ ions (or H3O+ ions) when dissolved in water. A base produces OH- ions.
  3. C) HSO4-
    • Explanation: A conjugate base is formed when an acid donates a proton (H+). H2SO4 (acid) – H+ = HSO4- (conjugate base).
  4. B) H2O
    • Explanation: An amphiprotic (or amphoteric) substance can act as both a Bronsted acid (proton donor) and a Bronsted base (proton acceptor). Water can donate a proton to form OH- (acid) or accept a proton to form H3O+ (base). HCO3- and HSO4- are also common examples.
  5. D) Accept an electron pair.
    • Explanation: According to the Lewis concept, a Lewis acid is an electron-pair acceptor (often has an empty orbital). A Lewis base is an electron-pair donor (often has a lone pair).
  6. D) NH3
    • Explanation: NH3 (ammonia) has a lone pair of electrons on the nitrogen atom, which it can donate to form a coordinate bond. Therefore, NH3 is a Lewis base. BF3, AlCl3, and H+ are electron-deficient and act as Lewis acids.
  7. B) Cα2/(1−α)
    • Explanation: For a weak acid HA <=> H+ + A-, if initial concentration is C and degree of ionization is α: At equilibrium: [HA]=C(1−α), [H+]=Cα, [A−]=Cα. Ka=[H+][A−]/[HA]=(Cα)(Cα)/C(1−α)=Cα2/(1−α).
  8. B) Increases with dilution.
    • Explanation: Ostwald’s Dilution Law states that for a weak electrolyte, the degree of ionization (α) is inversely proportional to the square root of its concentration (if α is small). So, as concentration decreases (dilution increases), α increases.
  9. C) 11
    • Explanation: At 25°C, pH + pOH = 14. If pH = 3, then pOH = 14 – 3 = 11.
  10. B) 1.0×10−14
    • Explanation: At 25°C, the ionic product of water (Kw​) is 1.0×10−14. This value increases with increasing temperature.
  11. A) Acidic
    • Explanation: At 25°C, a solution is acidic if [H+]>1.0×10−7 M. Since 1.0×10−5 M is greater than 1.0×10−7 M, the solution is acidic. (pH = 5).
  12. C) 3
    • Explanation: HCl is a strong acid, so it ionizes completely. [H+]=concentration of HCl=0.001 M=10−3 M. pH=−log[H+]=−log(10−3)=3.
  13. B) 11.7
    • Explanation: NaOH is a strong base, so it ionizes completely. [OH−]=concentration of NaOH=0.005 M=5×10−3 M. pOH=−log(5×10−3)=−log5+3=3−0.7=2.3. pH=14−pOH=14−2.3=11.7.
  14. A) Increase
    • Explanation: For a weak acid, [H+]=Ka​C​ (approximation). As the concentration (C) increases, the hydrogen ion concentration ([H+]) will increase, although the degree of ionization (α) will decrease.
  15. C) pOH<7
    • Explanation: For a basic solution at 25°C, pH > 7. Since pH + pOH = 14, this implies pOH < 7. Also, in a basic solution, [OH−]>[H+].
  16. C) Neutral
    • Explanation: A salt formed from a strong acid and a strong base (e.g., NaCl, KNO3) consists of ions that are very weak conjugate acid/base and do not undergo significant hydrolysis in water. Thus, the solution remains neutral with pH = 7.
  17. C) Salt of strong acid and weak base
    • Explanation: In a salt of a strong acid and a weak base (e.g., NH4Cl), the cation (NH4+) is the conjugate acid of a weak base and hydrolyzes to produce H+ ions, while the anion (Cl-) does not hydrolyze. NH4+ + H2O <=> NH4OH + H+ This makes the solution acidic.
  18. B) Kw​/Ka​
    • Explanation: For a salt of a weak acid and strong base (e.g., A-Na+), the anion A- undergoes hydrolysis: A- + H2O <=> HA + OH-. The hydrolysis constant Kh​=Kw​/Ka​.
  19. C) Ka = Kb
    • Explanation: For a salt of a weak acid and weak base, the pH of the solution is given by pH=7+(1/2)pKa​−(1/2)pKb​. If Ka​=Kb​, then pKa​=pKb​, which makes the term (1/2)pKa​−(1/2)pKb​=0. So, pH = 7.
  20. B) Resists change in its pH upon addition of small amounts of strong acid or base.
    • Explanation: This is the defining characteristic of a buffer solution. It contains a weak acid and its conjugate base, or a weak base and its conjugate acid, which can neutralize small added amounts of H+ or OH- ions.
  21. B) CH3COOH + CH3COONa
    • Explanation: An acidic buffer solution is a mixture of a weak acid and its salt with a strong base. CH3COOH is a weak acid, and CH3COONa is its salt with strong base (NaOH).
  22. B) pH=pKa​+log([Salt]/[Acid])
    • Explanation: This is the correct Henderson-Hasselbalch equation for an acidic buffer.
  23. A) [Salt] = [Acid]
    • Explanation: Buffer capacity is maximum when the concentration of the weak acid (or weak base) is equal to the concentration of its conjugate salt. At this point, pH = pKa (or pOH = pKb).
  24. B) Ksp​=[Ag+]2[CrO42−]
    • Explanation: For a sparingly soluble salt Ag2CrO4, the dissociation equilibrium is: Ag2CrO4 (s) <=> 2Ag+(aq) + CrO4^2-(aq). The solubility product Ksp​ is the product of ion concentrations raised to their stoichiometric coefficients: Ksp​=[Ag+]2[CrO42−].
  25. D) 4s3
    • Explanation: For CaF2 (s) <=> Ca2+(aq) + 2F-(aq). If solubility is ‘s’, then [Ca2+]=s and [F−]=2s. Ksp​=[Ca2+][F−]2=(s)(2s)2=s×4s2=4s3.
  26. B) Common ion
    • Explanation: According to the Common Ion Effect (Le Chatelier’s Principle), the solubility of a sparingly soluble salt decreases when a soluble salt containing a common ion is added to the solution. This shifts the dissolution equilibrium of the sparingly soluble salt to the left.
  27. C) Precipitation will occur.
    • Explanation:
      • If Q (Ionic Product) > Ksp (Solubility Product): The solution is supersaturated, and precipitation will occur until Q becomes equal to Ksp.
      • If Q < Ksp: The solution is unsaturated, and no precipitation occurs (more salt can dissolve).
      • If Q = Ksp: The solution is saturated, and the system is at equilibrium.
  28. C) 0.1 M NaOH
    • Explanation:
      • 0.1 M HCl: Strong acid, pH = 1.
      • 0.1 M CH3COOH: Weak acid, pH will be between 1 and 7 (e.g., approx 2.87 for Ka = 1.8×10^-5).
      • 0.1 M NaOH: Strong base, pOH = 1, pH = 13.
      • 0.1 M NH4OH: Weak base, pOH will be between 7 and 14 (e.g., approx 2.87 for Kb = 1.8×10^-5, so pH approx 11.13). Therefore, 0.1 M NaOH has the highest pH (most basic).
  29. D) 1000
    • Explanation: pH = −log[H+]. If pH1 = 2, then [H+]1=10−2 M. If pH2 = 5, then [H+]2=10−5 M. Ratio of concentrations = [H+]1/[H+]2=10−2/10−5=103=1000.
  30. B) Ka​×Kb​=Kw​
    • Explanation: This is the fundamental relationship between the acid dissociation constant (Ka​) of a weak acid and the base dissociation constant (Kb​) of its conjugate base (or vice-versa) at a given temperature.
  31. B) Glucose (C6H12O6)
    • Explanation: Non-electrolytes are substances that do not ionize in solution and therefore do not conduct electricity. Glucose, urea, and sucrose are common examples of non-electrolytes. KNO3, HCl, and NaOH are strong electrolytes.
  32. A) 0
    • Explanation: HNO3 is a strong acid, so it completely dissociates. [H+]=concentration of HNO3=1.0 M. pH=−log(1.0)=0.
  33. C) The base component of the buffer neutralizes the added acid.
    • Explanation: In an acidic buffer (Weak Acid + Conjugate Base), the conjugate base (A-) reacts with any added H+ ions (from strong acid) to form the weak acid (HA), thus consuming the added H+ and minimizing pH change. H+ + A- -> HA.
  34. C) 1.6×10−2 M
    • Explanation: For PbCl2(s)<=>Pb2+(aq)+2Cl−(aq). If solubility is ‘s’, then [Pb2+]=s and [Cl−]=2s. Ksp​=[Pb2+][Cl−]2=(s)(2s)2=4s3. Given Ksp​=1.6×10−5. 1.6×10−5=4s3 s3=(1.6×10−5)/4=0.4×10−5=4×10−6 s=34×10−6​=(4)1/3×10−2. Since (13=1,23=8), (4)1/3 is approximately 1.587. So, s≈1.587×10−2 M, which is closest to 1.6×10−2 M.
  35. C) KNO3
    • Explanation: KNO3 is a salt of a strong acid (HNO3) and a strong base (KOH). Its ions (K+, NO3-) do not undergo hydrolysis, so the solution remains neutral (pH=7). CH3COONa is basic, NH4Cl is acidic, and (NH4)2SO4 (salt of weak base and strong acid) is acidic.
  36. B) 4.74
    • Explanation: Use the Henderson-Hasselbalch equation for an acidic buffer: pH=pKa​+log([Salt]/[Acid]). Given pKa​=4.74, [Salt]=0.1 M, [Acid]=0.1 M. pH=4.74+log(0.1/0.1)=4.74+log(1)=4.74+0=4.74.
  37. B) 1.0×10−9 M
    • Explanation: At 25°C, pH + pOH = 14. If pOH = 5, then pH = 14 – 5 = 9. Since pH=−log[H+], then [H+]=10−pH=10−9 M.
  38. C) Both the acid and base are weak, and the ions are monovalent.
    • Explanation: For a salt of a weak acid and a weak base, the degree of hydrolysis (h) is given by h=Kh​​=Kw​/(Ka​×Kb​)​. In this formula, the concentration (C) of the salt does not appear (assuming monovalent ions), meaning ‘h’ is independent of concentration.
  39. A) Increase the pH.
    • Explanation: Adding Sodium Acetate (CH3COONa) provides a common ion (CH3COO-), which is the conjugate base of acetic acid. This common ion shifts the equilibrium of the weak acid dissociation (CH3COOH <=> H+ + CH3COO-) to the left, decreasing the [H+] concentration, and thus increasing the pH. This is the basis of buffer action.
  40. B) NaOH
    • Explanation: Ca(OH)2 (s) <=> Ca2+(aq) + 2OH-(aq). Adding NaOH provides a common ion (OH-). According to Le Chatelier’s Principle, increasing the concentration of a product ion (OH-) will shift the equilibrium to the left, favoring the precipitation of Ca(OH)2 and thus decreasing its solubility. HCl would react with OH-, increasing solubility. H2O would slightly increase solubility by dilution. NH4Cl would produce H+ ions via hydrolysis (NH4+ + H2O <=> NH4OH + H+), which would neutralize OH-, also increasing solubility.

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