Ionic Equilibrium
What to remember
- pH = −log [H⁺] and pOH = −log [OH⁻]. At 298 K, Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴, so pH + pOH = 14. For a weak acid, [H⁺] = √(Ka C) and α = √(Ka/C). Ka × Kb = Kw for a conjugate pair.
- A buffer resists a change in pH. Henderson-Hasselbalch equation: pH = pKa + log ([salt]/[acid]); for a basic buffer, pOH = pKb + log ([salt]/[base]).
- A salt of a strong base and a weak acid gives pH > 7; a salt of a strong acid and a weak base gives pH < 7; a salt of a strong acid and a strong base gives pH = 7.
- Solubility product: for AB, Ksp = s²; for AB₂ or A₂B, Ksp = 4s³. Precipitation occurs when the ionic product exceeds Ksp. A common ion lowers solubility.
Electrolytes and the concepts of acids and bases
An electrolyte gives ions in solution (or in the molten state) and conducts electricity. A strong electrolyte ionises almost completely (HCl, NaOH, NaCl). A weak electrolyte ionises only partly, and an equilibrium exists between the ions and the unionised molecules (CH₃COOH, NH₃, HCN).
| Concept | Acid | Base |
|---|---|---|
| Arrhenius | Gives H⁺ in water | Gives OH⁻ in water |
| Brønsted-Lowry | Proton (H⁺) donor | Proton acceptor |
| Lewis | Electron pair acceptor | Electron pair donor |
- In the Brønsted-Lowry idea, an acid and the base formed when it loses H⁺ are a conjugate acid-base pair. A strong acid has a weak conjugate base, and a weak acid has a strong conjugate base.
- A substance that can act as both an acid and a base is amphoteric, for example H₂O, HCO₃⁻, HSO₄⁻ and NH₃.
- Lewis acids: BF₃, AlCl₃, H⁺, Ag⁺. Lewis bases: NH₃, H₂O, OH⁻, Cl⁻. Neutralisation in the Lewis idea is the formation of a coordinate bond.
| Acid | Conjugate base | Base | Conjugate acid |
|---|---|---|---|
| HCl | Cl⁻ | NH₃ | NH₄⁺ |
| CH₃COOH | CH₃COO⁻ | H₂O | H₃O⁺ |
| H₂O | OH⁻ | CO₃²⁻ | HCO₃⁻ |
| H₂SO₄ | HSO₄⁻ | OH⁻ | H₂O |
Ionisation of water and the pH scale
Water ionises very slightly: 2H₂O ⇌ H₃O⁺ + OH⁻. The ionic product of water is
Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K
In pure water [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol L⁻¹. Kw increases with temperature, so the pH of pure water falls below 7 at higher temperature, but the water stays neutral because [H⁺] = [OH⁻].
pH = −log [H⁺]; pOH = −log [OH⁻]; pH + pOH = pKw = 14 (at 298 K).
| Solution | [H⁺] | pH at 298 K |
|---|---|---|
| Acidic | > 10⁻⁷ | < 7 |
| Neutral | = 10⁻⁷ | = 7 |
| Basic | < 10⁻⁷ | > 7 |
A change of one pH unit means a tenfold change in [H⁺].
pH of strong acids and bases. A strong acid and a strong base are fully ionised, so [H⁺] (or [OH⁻]) equals the concentration multiplied by the number of H⁺ (or OH⁻) given per formula unit.
Worked example 1. 0.01 M HCl: [H⁺] = 10⁻² M, pH = 2. 0.001 M NaOH: [OH⁻] = 10⁻³ M, pOH = 3, pH = 14 − 3 = 11. 0.005 M H₂SO₄ (both H⁺ assumed released): [H⁺] = 0.01 M, pH = 2.
Weak acids and weak bases
For a weak acid HA ⇌ H⁺ + A⁻, with initial concentration C and degree of ionisation α:
| Quantity | Expression |
|---|---|
| Ka | Cα² / (1 − α) |
| Ka (when α is small) | Cα² |
| α | √(Ka / C) |
| [H⁺] | Cα = √(Ka C) |
| pH | ½ (pKa − log C) |
This relation α = √(Ka/C) is Ostwald's dilution law: the degree of ionisation increases on dilution (it is inversely proportional to the square root of concentration). Ka itself does not change with dilution; it changes only with temperature.
For a weak base B + H₂O ⇌ BH⁺ + OH⁻: Kb = Cα² (small α), [OH⁻] = √(Kb C), pOH = ½ (pKb − log C), pKb = −log Kb.
Conjugate pair relation: Ka × Kb = Kw, so pKa + pKb = 14 at 298 K. A stronger acid has a larger Ka (smaller pKa). The conjugate base of a stronger acid is weaker.
Worked example 2. A weak acid HA has Ka = 1.0 × 10⁻⁵ and C = 0.1 M. α = √(10⁻⁵/10⁻¹) = 10⁻² (1%). [H⁺] = Cα = 10⁻³ M, pH = 3.
Worked example 3. A 0.01 M weak acid is 2% ionised. [H⁺] = 0.01 × 0.02 = 2 × 10⁻⁴ M. pH = 4 − log 2 = 4 − 0.30 = 3.70 (log 2 = 0.30). Ka = Cα² = 0.01 × (0.02)² = 4 × 10⁻⁶.
Polybasic acids and acid strength
A polybasic acid ionises in steps, and each step has its own constant. The first constant is always the largest, because it is harder to remove H⁺ from a negatively charged ion.
| Acid | Ka₁ | Ka₂ | Ka₃ |
|---|---|---|---|
| H₂CO₃ | 4.3 × 10⁻⁷ | 5.6 × 10⁻¹¹ | – |
| H₂S | 9.1 × 10⁻⁸ | 1.2 × 10⁻¹³ | – |
| H₃PO₄ | 7.5 × 10⁻³ | 6.2 × 10⁻⁸ | 4.2 × 10⁻¹³ |
Ka₁ > Ka₂ > Ka₃. In a solution of a polybasic acid, nearly all the H⁺ comes from the first step.
Factors affecting acid strength (for H–A):
- Bond strength: in a group, the weaker the H–A bond, the stronger the acid. HF < HCl < HBr < HI in acid strength in water, although electronegativity decreases down the group.
- Electronegativity of A in a period: acid strength increases as A becomes more electronegative. CH₄ < NH₃ < H₂O < HF.
- Greater stability of the conjugate base A⁻ means a stronger acid.
Common ion effect
When a salt that has an ion in common is added to a weak electrolyte solution, ionisation of the weak electrolyte decreases. Example: adding CH₃COONa to CH₃COOH gives more CH₃COO⁻, so the equilibrium CH₃COOH ⇌ CH₃COO⁻ + H⁺ shifts backward. [H⁺] falls and pH rises. Likewise, NH₄Cl added to NH₃ solution lowers [OH⁻].
Hydrolysis of salts
Hydrolysis is the reaction of the ions of a salt with water that changes the pH of the solution.
| Salt of | Example | Ion hydrolysed | Nature | pH at 298 K | Standard formula |
|---|---|---|---|---|---|
| Strong acid + strong base | NaCl, KNO₃ | None | Neutral | 7 | pH = 7 |
| Weak acid + strong base | CH₃COONa, Na₂CO₃ | Anion | Basic | > 7 | pH = 7 + ½ (pKa + log C) |
| Strong acid + weak base | NH₄Cl, CuSO₄ | Cation | Acidic | < 7 | pH = 7 − ½ (pKb + log C) |
| Weak acid + weak base | CH₃COONH₄ | Both | Depends on Ka and Kb | about 7 | pH = 7 + ½ (pKa − pKb) |
- For the anion of a weak acid: Kh = Kw / Ka and h = √(Kh/C). For the cation of a weak base: Kh = Kw / Kb.
- For a salt of a weak acid and a weak base, the pH does not depend on concentration. For CH₃COONH₄, pKa = pKb, so pH = 7 (neutral).
Worked example 4. Find the pH of 0.1 M CH₃COONa if Ka (acetic acid) = 1.0 × 10⁻⁵ (pKa = 5).
pH = 7 + ½ (5 + log 0.1) = 7 + ½ (5 − 1) = 9.
Buffer solutions
A buffer is a solution whose pH changes very little when a small amount of acid or base is added, or when it is diluted.
| Type | Composition | Example | Equation |
|---|---|---|---|
| Acidic buffer | Weak acid + its salt with a strong base | CH₃COOH + CH₃COONa | pH = pKa + log ([salt]/[acid]) |
| Basic buffer | Weak base + its salt with a strong acid | NH₃ + NH₄Cl | pOH = pKb + log ([salt]/[base]) |
- The first equation is the Henderson-Hasselbalch equation. When [salt] = [acid], pH = pKa.
- Buffer action: in CH₃COOH/CH₃COONa, added H⁺ is taken up by CH₃COO⁻; added OH⁻ is neutralised by CH₃COOH.
- The buffer is most effective when the ratio [salt]/[acid] is close to 1 (usually between 0.1 and 10), so a suitable buffer has pH within one unit of its pKa.
- Buffer capacity is greater when the concentrations of the components are larger.
Worked example 5. A buffer has 0.2 M CH₃COONa and 0.1 M CH₃COOH, pKa = 4.75. pH = 4.75 + log (0.2/0.1) = 4.75 + 0.30 = 5.05 (log 2 = 0.30).
Solubility equilibria
For a sparingly soluble salt in water, the dissolved ions are in equilibrium with the solid. The solubility product Ksp is the product of the ion concentrations (each raised to its coefficient) in a saturated solution. The solid is omitted from the expression.
If the molar solubility is s mol L⁻¹:
| Type | Example | Dissolution | Ksp | s |
|---|---|---|---|---|
| AB | AgCl, BaSO₄ | AB ⇌ A⁺ + B⁻ | s² | √Ksp |
| AB₂ | CaF₂, Mg(OH)₂ | AB₂ ⇌ A²⁺ + 2B⁻ | 4s³ | (Ksp/4)^(1/3) |
| A₂B | Ag₂CrO₄ | A₂B ⇌ 2A⁺ + B²⁻ | 4s³ | (Ksp/4)^(1/3) |
Ionic product and precipitation. The ionic product Q has the form of Ksp but uses the actual concentrations.
- Q < Ksp: unsaturated; no precipitate forms.
- Q = Ksp: saturated; equilibrium.
- Q > Ksp: precipitation occurs until Q = Ksp.
Common ion effect on solubility: the solubility of a sparingly soluble salt decreases in a solution containing a common ion. In 0.1 M NaCl, the Cl⁻ from NaCl is much larger than that from AgCl, so [Ag⁺] = Ksp / [Cl⁻].
Worked example 6. AgCl has s = 1.0 × 10⁻⁵ mol L⁻¹ in water, so Ksp = s² = 1.0 × 10⁻¹⁰. In 0.1 M NaCl: s = Ksp/[Cl⁻] = 10⁻¹⁰/10⁻¹ = 10⁻⁹ mol L⁻¹.
Worked example 7. For Mg(OH)₂, Ksp = 4 × 10⁻¹². Ksp = 4s³, so s³ = 10⁻¹² and s = 10⁻⁴ mol L⁻¹. Then [OH⁻] = 2s = 2 × 10⁻⁴ M.
NEET traps
- 1. pH of pure water at a temperature above 298 K is below 7, but the water is still neutral.
- 2. A strong acid has a weak conjugate base. A weak acid has a strong conjugate base.
- 3. Ka depends only on temperature. The degree of ionisation α changes with dilution but Ka does not.
- 4. For polybasic acids Ka₁ > Ka₂ > Ka₃. Do not use Ka₂ for the first ionisation.
- 5. HI is a stronger acid than HF in water, because the H–I bond is weaker, even though F is more electronegative.
- 6. Ammonium acetate is a salt of a weak acid and a weak base, yet its solution is nearly neutral because Ka ≈ Kb.
- 7. Sodium carbonate solution is basic. Ammonium chloride solution is acidic. Sodium chloride solution is neutral.
- 8. A buffer has the highest capacity when [salt] = [acid]. At that point pH = pKa.
- 9. For AB₂ and A₂B salts Ksp = 4s³, but for AB it is s². Do not mix them. For comparing solubility, use s and not Ksp, unless the salts are of the same type.
- 10. Precipitation needs Q > Ksp. A common ion lowers solubility but does not change Ksp.
One-liners
- 1. Kw = 1.0 × 10⁻¹⁴ at 298 K.
- 2. pH + pOH = 14 at 298 K.
- 3. An acid is a proton donor in the Brønsted-Lowry idea, and an electron pair acceptor in the Lewis idea.
- 4. Conjugate base of H₂O is OH⁻, and conjugate acid of H₂O is H₃O⁺.
- 5. For a weak acid [H⁺] = √(Ka C).
- 6. Degree of ionisation α = √(Ka/C) (Ostwald's dilution law).
- 7. Ka × Kb = Kw, and pKa + pKb = 14.
- 8. In H₃PO₄, Ka₁ > Ka₂ > Ka₃.
- 9. Acid strength: HF < HCl < HBr < HI.
- 10. Common ion addition suppresses ionisation of a weak electrolyte.
- 11. CH₃COONa solution is basic and NH₄Cl solution is acidic.
- 12. Henderson-Hasselbalch: pH = pKa + log ([salt]/[acid]).
- 13. CH₃COOH/CH₃COONa and NH₃/NH₄Cl are standard buffers.
- 14. For AgCl, Ksp = s². For Ca(OH)₂, Ksp = 4s³.
- 15. Q > Ksp means precipitation.
Practice questions
The conjugate base of HSO₄⁻ is
- H₃O⁺
- HSO₃⁻
- H₂SO₄
- SO₄²⁻
Answer
D. SO₄²⁻
Loss of H⁺ from HSO₄⁻ gives SO₄²⁻.
The pH of 0.01 M NaOH at 298 K is
- 2
- 12
- 10
- 13
Answer
B. 12
[OH⁻] = 10⁻² M, so pOH = 2 and pH = 14 − 2 = 12.
Assuming complete ionisation to give two H⁺ per molecule, the pH of 0.005 M H₂SO₄ at 298 K is
- 3
- 2.3
- 1.7
- 2
Answer
D. 2
[H⁺] = 2 × 0.005 = 0.01 M, so pH = 2.
The pH of a 0.01 M solution of a weak acid with Ka = 1.0 × 10⁻⁶ is
- 3
- 4
- 5
- 2
Answer
B. 4
[H⁺] = √(Ka C) = √(10⁻⁶ × 10⁻²) = 10⁻⁴ M, so pH = 4.
A 0.01 M solution of a weak acid has Ka = 4 × 10⁻⁶. Its degree of ionisation is
- 0.002
- 0.04
- 0.02
- 0.2
Answer
C. 0.02
α = √(Ka/C) = √(4 × 10⁻⁶/10⁻²) = √(4 × 10⁻⁴) = 0.02.
The Ka of a weak acid is 1.0 × 10⁻⁵ at 298 K. The Kb of its conjugate base at 298 K is
- 1.0 × 10⁻⁹
- 1.0 × 10⁹
- 1.0 × 10⁻⁵
- 1.0 × 10⁻¹⁹
Answer
A. 1.0 × 10⁻⁹
Kb = Kw/Ka = 10⁻¹⁴/10⁻⁵ = 10⁻⁹.
Which of the following is a Lewis acid?
- BF₃
- H₂O
- NH₃
- OH⁻
Answer
A. BF₃
BF₃ has an incomplete octet and accepts an electron pair.
The correct order of increasing acid strength in water is
- HF < HI < HBr < HCl
- HI < HBr < HCl < HF
- HF < HCl < HBr < HI
- HCl < HF < HBr < HI
Answer
C. HF < HCl < HBr < HI
Acid strength rises as the H–X bond becomes weaker down the group.
In an acidic buffer of a weak acid and its salt, when [salt] = [acid] the pH is equal to
- 7
- pKb
- pKw
- pKa
Answer
D. pKa
pH = pKa + log 1 = pKa.
A basic buffer has 0.1 M NH₄Cl and 0.2 M NH₃, with pKb of NH₃ = 4.75. Taking log 2 = 0.30, the pH at 298 K is nearest to
- 9.55
- 8.95
- 4.45
- 5.05
Answer
A. 9.55
pOH = 4.75 + log (0.1/0.2) = 4.75 − 0.30 = 4.45; pH = 14 − 4.45 = 9.55.
For NH₃, pKb = 5. Using pH = 7 − ½ (pKb + log C), the pH of 0.1 M NH₄Cl is
- 9
- 6
- 5
- 4
Answer
C. 5
pH = 7 − ½ (5 − 1) = 7 − 2 = 5.
The molar solubility of Ag₂CrO₄ is 1.0 × 10⁻⁴ mol L⁻¹. Its Ksp is
- 1 × 10⁻¹²
- 4 × 10⁻¹²
- 4 × 10⁻⁸
- 1 × 10⁻⁸
Answer
B. 4 × 10⁻¹²
Ag₂CrO₄ ⇌ 2Ag⁺ + CrO₄²⁻, so Ksp = (2s)² s = 4s³ = 4 × 10⁻¹².
A sparingly soluble salt of the type AB has Ksp = 1.6 × 10⁻⁹ at 298 K. Its molar solubility is
- 4 × 10⁻⁹ mol L⁻¹
- 4 × 10⁻⁵ mol L⁻¹
- 1.2 × 10⁻³ mol L⁻¹
- 1.6 × 10⁻⁹ mol L⁻¹
Answer
B. 4 × 10⁻⁵ mol L⁻¹
s = √Ksp = √(16 × 10⁻¹⁰) = 4 × 10⁻⁵ mol L⁻¹.
A solution has [OH⁻] = 1 × 10⁻⁴ mol L⁻¹ at 298 K. Its pH is
- 4
- 7
- 14
- 10
Answer
D. 10
pOH = 4, so pH = 14 − 4 = 10.
An aqueous solution of Na₂CO₃ is
- basic
- acidic
- neutral
- amphoteric
Answer
A. basic
CO₃²⁻ is the anion of a weak acid and hydrolyses to give OH⁻.
The Ksp of AgCl is 1.0 × 10⁻¹⁰. Its solubility in 0.1 M NaCl is
- 1.0 × 10⁻⁵ mol L⁻¹
- 1.0 × 10⁻¹¹ mol L⁻¹
- 1.0 × 10⁻⁹ mol L⁻¹
- 1.0 × 10⁻¹⁰ mol L⁻¹
Answer
C. 1.0 × 10⁻⁹ mol L⁻¹
[Ag⁺] = Ksp/[Cl⁻] = 10⁻¹⁰/10⁻¹ = 10⁻⁹ mol L⁻¹.
Given below are two statements: Statement I: Water is amphoteric in the Brønsted-Lowry sense. Statement II: The conjugate acid of H₂O is OH⁻.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
H₂O can accept or donate H⁺, but its conjugate acid is H₃O⁺ and its conjugate base is OH⁻.
Given below are two statements: Statement I: The pH of pure water at 373 K is exactly 7. Statement II: Pure water is neutral at every temperature because [H⁺] = [OH⁻].
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
Kw rises with temperature, so the pH of pure water falls below 7, but [H⁺] = [OH⁻] still holds.
Given below are two statements: Statement I: The degree of ionisation of a weak acid increases on dilution. Statement II: The Ka of a weak acid increases on dilution.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
α = √(Ka/C) increases as C falls; Ka depends only on temperature.
Given below are two statements: Statement I: Addition of sodium acetate to acetic acid decreases the ionisation of acetic acid. Statement II: The pH of the acetic acid solution increases on adding sodium acetate.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
Common ion CH₃COO⁻ shifts the equilibrium back, so [H⁺] falls and pH rises.
Which of the following statements are correct? 1. BF₃ is a Lewis acid. 2. NH₃ is a Lewis base. 3. Cl⁻ is a Brønsted acid.
- 1, 2 and 3
- 2 and 3 only
- 1 and 2 only
- 1 and 3 only
Answer
C. 1 and 2 only
Cl⁻ cannot donate a proton; it is a Brønsted base.
Given below are two statements: Statement I: An aqueous solution of CH₃COONH₄ is nearly neutral. Statement II: The Ka of acetic acid and the Kb of ammonia are nearly equal.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
With pKa ≈ pKb the pH is about 7.
Given below are two statements: Statement I: The first ionisation constant of H₃PO₄ is smaller than its second. Statement II: Removal of H⁺ from a negatively charged ion is easier than from a neutral molecule.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are incorrect
Ka₁ > Ka₂ > Ka₃ because removal of H⁺ from a negative ion is harder.
Given below are two statements: Statement I: A mixture of CH₃COOH and CH₃COONa is an acidic buffer. Statement II: A mixture of HCl and NaCl is an acidic buffer.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
C. Statement I is correct but Statement II is incorrect
HCl is a strong acid, so HCl and NaCl do not make a buffer.
Which of the following statements are correct? 1. Precipitation occurs when the ionic product exceeds Ksp. 2. The solubility of AgCl decreases in NaCl solution. 3. The Ksp of AgCl decreases in NaCl solution.
- 2 and 3 only
- 1, 2 and 3
- 1 and 3 only
- 1 and 2 only
Answer
D. 1 and 2 only
Ksp depends only on temperature, so statement 3 is incorrect.
Given below are two statements: Statement I: The aqueous solution of CH₃COONa is acidic. Statement II: Hydrolysis of CH₃COO⁻ gives OH⁻.
- Both Statement I and Statement II are correct
- Both Statement I and Statement II are incorrect
- Statement I is correct but Statement II is incorrect
- Statement I is incorrect but Statement II is correct
Answer
D. Statement I is incorrect but Statement II is correct
The solution is basic, so Statement I is incorrect.
Assertion (A): Aqueous Na₂CO₃ solution is basic. Reason (R): CO₃²⁻ undergoes hydrolysis to give OH⁻ ions.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
The hydrolysis of CO₃²⁻ produces OH⁻.
Assertion (A): HI is a stronger acid than HF in water. Reason (R): Iodine is more electronegative than fluorine.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
C. A is true but R is false
A is true because the H–I bond is weaker; R is false since fluorine is more electronegative.
Assertion (A): A solution of NaCl has pH 7 at 298 K. Reason (R): The Kw of water is 1.0 × 10⁻¹⁴ at 298 K.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
B. Both A and R are true but R is not the correct explanation of A
Both are true, but neutral pH of NaCl is due to neither ion hydrolysing, not to the value of Kw.
Assertion (A): The Ka of a weak acid increases when the solution is diluted. Reason (R): The degree of ionisation of a weak acid increases on dilution.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
D. A is false but R is true
Ka is unchanged by dilution, so A is false; R is true.
Assertion (A): Ammonium chloride solution is acidic. Reason (R): The Cl⁻ ion undergoes hydrolysis to release H⁺.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
C. A is true but R is false
A is true because NH₄⁺ hydrolyses; Cl⁻ is the conjugate base of a strong acid and does not hydrolyse.
Assertion (A): Precipitation occurs when the ionic product exceeds the solubility product. Reason (R): The excess ions leave the solution as solid until the ionic product equals Ksp.
- Both A and R are true and R is the correct explanation of A
- Both A and R are true but R is not the correct explanation of A
- A is true but R is false
- A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
This is the reason precipitation stops at Q = Ksp.
Match List I with List II. List I: a. NaCl b. CH₃COONa c. NH₄Cl d. CH₃COONH₄ List II: i. Only the anion hydrolyses ii. Only the cation hydrolyses iii. Neither ion hydrolyses iv. Both ions hydrolyse
- a-i, b-iii, c-ii, d-iv
- a-iii, b-i, c-ii, d-iv
- a-iii, b-ii, c-i, d-iv
- a-iii, b-i, c-iv, d-ii
Answer
B. a-iii, b-i, c-ii, d-iv
NaCl is a strong acid-strong base salt; CH₃COONa has a weak acid anion; NH₄Cl has a weak base cation.
Match List I with List II. List I: a. H₂SO₄ b. NH₄⁺ c. H₂O d. H₃O⁺ List II: i. NH₃ ii. HSO₄⁻ iii. OH⁻ iv. H₂O
- a-i, b-ii, c-iii, d-iv
- a-ii, b-i, c-iii, d-iv
- a-iv, b-i, c-iii, d-ii
- a-ii, b-i, c-iv, d-iii
Answer
B. a-ii, b-i, c-iii, d-iv
Each acid loses H⁺ to give its conjugate base.
Match List I with List II. List I: a. 0.1 M HCl b. 0.01 M HCl c. 0.01 M NaOH d. 0.1 M NaOH List II: i. 12 ii. 13 iii. 1 iv. 2
- a-iv, b-iii, c-i, d-ii
- a-iii, b-iv, c-ii, d-i
- a-iii, b-i, c-iv, d-ii
- a-iii, b-iv, c-i, d-ii
Answer
D. a-iii, b-iv, c-i, d-ii
pH values at 298 K are 1, 2, 12 and 13.
Match List I with List II. List I: a. pH = pKa + log ([salt]/[acid]) b. α = √(Ka/C) c. [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ d. Ksp = 4s³ List II: i. Ostwald's dilution law ii. Henderson-Hasselbalch equation iii. Ionic product of water iv. Salt of the type AB₂
- a-ii, b-i, c-iii, d-iv
- a-i, b-ii, c-iii, d-iv
- a-iii, b-i, c-ii, d-iv
- a-ii, b-i, c-iv, d-iii
Answer
A. a-ii, b-i, c-iii, d-iv
Each expression is matched with its law or case.
Match List I with List II. List I: a. Arrhenius acid b. Brønsted-Lowry base c. Lewis acid d. Lewis base List II: i. Electron pair donor ii. Electron pair acceptor iii. Proton acceptor iv. Gives H⁺ in water
- a-iv, b-iii, c-i, d-ii
- a-iv, b-iii, c-ii, d-i
- a-iv, b-ii, c-iii, d-i
- a-iii, b-iv, c-ii, d-i
Answer
B. a-iv, b-iii, c-ii, d-i
Definitions of the four concepts.
Match List I with List II. List I: a. Acidic buffer b. Basic buffer c. Salt of weak acid and strong base d. Salt of strong acid and weak base List II: i. pOH = pKb + log ([salt]/[base]) ii. pH = pKa + log ([salt]/[acid]) iii. pH = 7 + ½ (pKa + log C) iv. pH = 7 − ½ (pKb + log C)
- a-ii, b-i, c-iii, d-iv
- a-ii, b-i, c-iv, d-iii
- a-iii, b-i, c-ii, d-iv
- a-i, b-ii, c-iii, d-iv
Answer
A. a-ii, b-i, c-iii, d-iv
Standard pH relations for the four systems.
Equal volumes of 4 × 10⁻⁵ M AgNO₃ and 4 × 10⁻⁵ M NaCl are mixed. Ksp of AgCl is 1.0 × 10⁻¹⁰. The ionic product after mixing and the result are
- Q = 4 × 10⁻¹⁰; no precipitate
- Q = 1.6 × 10⁻⁹; precipitate forms
- Q = 1 × 10⁻¹⁰; no precipitate
- Q = 4 × 10⁻¹⁰; precipitate forms
Answer
D. Q = 4 × 10⁻¹⁰; precipitate forms
After mixing, each concentration is 2 × 10⁻⁵ M, so Q = 4 × 10⁻¹⁰ > Ksp.
A 0.01 M solution of a weak base has Kb = 1.0 × 10⁻⁶. The pH at 298 K is
- 4
- 10
- 8
- 11
Answer
B. 10
[OH⁻] = √(Kb C) = √(10⁻⁸) = 10⁻⁴ M; pOH = 4 and pH = 10.
A weak acid is 1% ionised in 0.1 M solution. At the same temperature, in 0.001 M solution its degree of ionisation is nearly
- 100%
- 10%
- 1%
- 0.1%
Answer
B. 10%
Ka = Cα² = 0.1 × 10⁻⁴ = 10⁻⁵; at 0.001 M, α = √(10⁻⁵/10⁻³) = 0.1.
100 mL of 0.2 M HCl is mixed with 100 mL of 0.1 M NaOH. Taking log 5 = 0.70, the pH of the mixture is nearest to
- 1.30
- 2.30
- 1.70
- 1.00
Answer
A. 1.30
HCl left = 20 − 10 = 10 mmol in 200 mL = 0.05 M; pH = −log (5 × 10⁻²) = 2 − 0.70 = 1.30.
Which of the following mixtures acts as a buffer?
- 0.1 M NaOH and 0.1 M NaCl
- 0.1 M NaCl alone
- 0.1 M CH₃COOH and 0.1 M CH₃COONa
- 0.1 M HCl and 0.1 M NaCl
Answer
C. 0.1 M CH₃COOH and 0.1 M CH₃COONa
A weak acid together with its salt of a strong base is an acidic buffer.
A salt is made from a weak acid with pKa = 4 and a weak base with pKb = 6. Using pH = 7 + ½ (pKa − pKb), the pH of its aqueous solution is
- 8
- 5
- 7
- 6
Answer
D. 6
pH = 7 + ½ (4 − 6) = 7 − 1 = 6 (acidic since the acid is the stronger of the two).
Mg(OH)₂ has Ksp = 4 × 10⁻¹² at 298 K. Taking log 2 = 0.30, the pH of its saturated solution is nearest to
- 9.70
- 10.30
- 3.70
- 11.00
Answer
B. 10.30
4s³ = 4 × 10⁻¹² gives s = 10⁻⁴ M; [OH⁻] = 2 × 10⁻⁴ M; pOH = 4 − 0.30 = 3.70; pH = 10.30.