JEE Chemistry Ionic Equilibrium MCQs Set 32

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Multiple Choice Questions

Question 1: Which of the following mixtures of acids (equal volumes mixed) has the highest pH? Concentrations: (1) M/10 H2SO4, M/20 HNO3, M/10 HClO4; (2) M/20 H2SO4, M/10 HNO3, M/20 HClO4; (3) M/20 H2SO4, M/10 HNO3, M/40 HClO4; (4) M/20 H2SO4, M/5 HNO3, M/5 HClO4.
(a) (1)
(b) (2)
(c) (3)
(d) (4)
Show Answer & Explanation

Answer: (c) (3)

Explanation:
1. The highest pH has the smallest total [H+] (H2SO4 gives two H+).
2. Totals (before mixing): (1) 0.2 + 0.05 + 0.1 = 0.35; (2) 0.1 + 0.1 + 0.05 = 0.25; (3) 0.1 + 0.1 + 0.025 = 0.225; (4) 0.1 + 0.2 + 0.2 = 0.5.
3. The smallest is (3).
4. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 2: If 100 mL of a solution of pH = 3 and 400 mL of a solution of pH = 3 are mixed, the pH of the mixture is:
(a) 3.2
(b) 3.0
(c) 3.5
(d) 2.8
Show Answer & Explanation

Answer: (b) 3.0

Explanation:
1. Both solutions have [H+] = 10−3 M, so the mixture also has 10−3 M.
2. pH = 3.0.
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 3: 10−6 M HCl is diluted to 100 times. Its pH is closest to:
(a) 6.0
(b) 8.0
(c) 6.95
(d) 9.5
Show Answer & Explanation

Answer: (c) 6.95

Explanation:
1. After dilution the acid contributes 10−8 M, which is less than the 10−7 M from water, so water cannot be ignored.
2. Solving x(x + 10−8) = 10−14 gives [H+] ≈ 1.05 × 10−7 M and pH ≈ 6.98, closest to 6.95.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 4: The pH of a solution is increased from 3 to 6. Its H+ ion concentration will be:
(a) Reduced to half
(b) Doubled
(c) Reduced by 1000 times
(d) Increased by 1000 times
Show Answer & Explanation

Answer: (c) Reduced by 1000 times

Explanation:
1. [H+] changes from 10−3 to 10−6 M, i.e. it falls by a factor of 1000.
2. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 5: An aqueous solution whose pH = 0 is:
(a) Basic
(b) Acidic
(c) Neutral
(d) Amphoteric
Show Answer & Explanation

Answer: (b) Acidic

Explanation:
1. pH = 0 means [H+] = 1 M, a strongly acidic solution.
2. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 6: Five solutions of KOH were prepared as: first 0.1 mol in 1 L, second 0.2 mol in 2 L, third 0.3 mol in 3 L, fourth 0.4 mol in 4 L, fifth 0.5 mol in 5 L. The pH of the resultant solution (all mixed) is:
(a) 2
(b) 1
(c) 13
(d) 7
Show Answer & Explanation

Answer: (c) 13

Explanation:
1. Each solution is 0.1 M KOH.
2. Mixing equal-concentration solutions leaves 0.1 M, so pOH = 1 and pH = 13.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 7: For N/10 H2SO4, the pH value is:
(a) 1
(b) 0.586
(c) 0.856
(d) None of these
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. N/10 H2SO4 gives 0.1 g-equivalent of H+ per litre, i.e. [H+] = 0.1 M.
2. pH = 1.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 8: H2X is a dibasic acid that dissociates completely in water. The molarity of an aqueous solution of this acid which has a pH of 1 is:
(a) 0.1
(b) 0.05
(c) 0.2
(d) 0.5
Show Answer & Explanation

Answer: (b) 0.05

Explanation:
1. pH 1 means [H+] = 0.1 M.
2. Each H2X gives 2H+, so M = 0.05.
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 9: One litre of solution contains 1 mol of HOCl (Ka = 10−8) and 1 mol of NaOH. The pH of the solution is:
(a) 8
(b) 11
(c) 5
(d) 2
Show Answer & Explanation

Answer: (b) 11

Explanation:
1. They react completely to give 1 M NaOCl.
2. Hydrolysis of the anion: pH = 7 + ½(pKa + log C) = 7 + ½(8 + 0) = 11.
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 10: 0.001 mol of the strong electrolyte M(OH)2 has been dissolved to make 20 mL of its saturated solution. Its pH will be (Kw = 1 × 10−14):
(a) 13
(b) 3.3
(c) 11
(d) 9.8
Show Answer & Explanation

Answer: (a) 13

Explanation:
1. [M(OH)2] = 0.001/0.020 = 0.05 M, so [OH−] = 0.1 M.
2. pOH = 1 and pH = 13.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 11: The pH of 0.1 M solutions of the following salts increases in the order:
(a) NaCl < NH4Cl < NaCN < HCl
(b) NaCN < NH4Cl < NaCl < HCl
(c) HCl < NaCl < NaCN < NH4Cl
(d) HCl < NH4Cl < NaCl < NaCN
Show Answer & Explanation

Answer: (d) HCl < NH4Cl < NaCl < NaCN

Explanation:
1. HCl is strongly acidic (pH 1), NH4Cl is weakly acidic (about 5), NaCl is neutral (7) and NaCN is basic (about 11).
2. Order: HCl < NH4Cl < NaCl < NaCN.
3. Answer: (d)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 12: In a buffer solution, the ratio of the concentration of NH4Cl and NH4OH is 1 : 1 and it changes to 2 : 1. What will be the effect on the pH of the buffer?
(a) Increase
(b) Decrease
(c) No effect
(d) None of these
Show Answer & Explanation

Answer: (b) Decrease

Explanation:
1. pOH = pKb + log([NH4Cl]/[NH4OH]).
2. The ratio rises from 1 to 2, so pOH rises by log 2 = 0.3 and pH decreases.
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 13: What is the suitable indicator for the titration of NaOH and oxalic acid?
(a) Methyl orange
(b) Methyl red
(c) Phenolphthalein
(d) Starch solution
Show Answer & Explanation

Answer: (c) Phenolphthalein

Explanation:
1. Oxalic acid is a weak acid, so the end point (sodium oxalate solution) is basic; phenolphthalein (pH range 8 to 9.8) is suitable.
2. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 14: Which can act as a buffer?
(a) NH4OH + NaOH
(b) HCl + NaCl
(c) 40 mL of 0.1 M NaCN + 20 mL of 0.1 M HCl
(d) None of these
Show Answer & Explanation

Answer: (c) 40 mL of 0.1 M NaCN + 20 mL of 0.1 M HCl

Explanation:
1. 4 mmol CN− + 2 mmol H+ gives 2 mmol HCN and 2 mmol CN− left.
2. This is a weak acid with its conjugate base, a buffer.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 15: The buffering action of a mixture of CH3COOH and CH3COONa is maximum when the ratio of salt to acid is equal to:
(a) 1.0
(b) 100.0
(c) 10.0
(d) 0.1
Show Answer & Explanation

Answer: (a) 1.0

Explanation:
1. Buffer capacity is maximum when [salt] = [acid], i.e. pH = pKa.
2. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 16: Which indicator works in the pH range 8 to 9.8?
(a) Phenolphthalein
(b) Methyl orange
(c) Methyl red
(d) Litmus
Show Answer & Explanation

Answer: (a) Phenolphthalein

Explanation:
1. Phenolphthalein changes from colourless to pink between about pH 8.3 and 10 (8 to 9.8 in many texts); methyl orange (3.1 to 4.4), methyl red (4.2 to 6.3) and litmus (5 to 8) work elsewhere.
2. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 17: A basic buffer will obey the equation pOH − pKb = 1 only under the condition:
(a) [Conjugate acid] : [base] = 1 : 10
(b) [Conjugate acid] = [base]
(c) [Conjugate acid] : [base] = 10 : 1
(d) None of these
Show Answer & Explanation

Answer: (c) [Conjugate acid] : [base] = 10 : 1

Explanation:
1. pOH = pKb + log([conjugate acid]/[base]).
2. pOH − pKb = 1 means the ratio is 10 : 1.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 18: In which of the following titrations is methyl orange the best indicator?
(a) CH3COOH + NaOH
(b) H2C2O4 + NaOH
(c) HCl + NaOH
(d) CH3COOH + NH4OH
Show Answer & Explanation

Answer: (c) HCl + NaOH

Explanation:
1. Methyl orange works in the pH range 3.1 to 4.4 and is used when the end point is acidic or neutral and sharp, as in strong acid with strong base titration.
2. It is unsuitable when a weak acid is present, since the end point is above pH 7.
3. Answer: (c)

Teacher's Note:
1. Choose the indicator whose range covers the steep jump.
2. Strong acid versus strong base has a very sharp jump.

Question 19: The H+ ion concentration in 0.001 M acetic acid is 1.34 × 10−4 g ion L−1. The H+ ion concentration of a solution obtained when 0.164 g of CH3COONa is added to a litre of 0.001 M CH3COOH will be:
(a) 9 × 10−6
(b) 18 × 10−6
(c) 4.5 × 10−6
(d) 5 × 10−6
Show Answer & Explanation

Answer: (a) 9 × 10−6

Explanation:
1. Ka = (1.34 × 10−4)2/0.001 = 1.8 × 10−5.
2. CH3COONa = 0.164/82 = 0.002 mol L−1.
3. [H+] = Ka[acid]/[salt] = 1.8 × 10−5 × 0.001/0.002 = 9 × 10−6.
4. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 20: When 1.0 mL of dilute HCl is added to 100 mL of a buffer solution of pH 4.0, the pH of the solution:
(a) Becomes 7
(b) Does not change (appreciably)
(c) Becomes 2
(d) Becomes 10
Show Answer & Explanation

Answer: (b) Does not change (appreciably)

Explanation:
1. A buffer resists a change in pH on addition of a small amount of acid or base.
2. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 21: Which of the following solutions does NOT act as a buffer?
(a) H3PO4 + NaH2PO4
(b) NaHCO3 + H2CO3
(c) NH4Cl + HCl
(d) CH3COOH + CH3COONa
Show Answer & Explanation

Answer: (c) NH4Cl + HCl

Explanation:
1. NH4Cl + HCl has a strong acid, not a weak acid or base with its conjugate; it cannot resist pH change.
2. The other three are weak acid with conjugate base pairs.
3. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 22: On addition of NaOH to a CH3COOH solution, 60% of the acid is neutralised. If pKa of CH3COOH is 4.7, the pH of the resulting solution is:
(a) More than 4.7 but less than 5.0
(b) Less than 4.7 but more than 4.0
(c) More than 5.0
(d) Remains unchanged
Show Answer & Explanation

Answer: (a) More than 4.7 but less than 5.0

Explanation:
1. [salt]/[acid] = 60/40 = 1.5.
2. pH = 4.7 + log 1.5 = 4.7 + 0.18 = 4.88.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 23: Half of the formic acid solution is neutralised on addition of a KOH solution to it. If Ka(HCOOH) = 2 × 10−4, the pH of the solution is (log 2 = 0.3010):
(a) 3.6990
(b) 10.3010
(c) 3.85
(d) 4.3010
Show Answer & Explanation

Answer: (a) 3.6990

Explanation:
1. At half neutralisation [salt] = [acid], so pH = pKa.
2. pKa = 4 − log 2 = 3.699.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 24: The Henderson equation pH − pKa = 1 will be applicable to an acidic buffer when:
(a) [Acid] = [Conjugate base]
(b) [Acid] × 10 = [Conjugate base]
(c) [Acid] = [Conjugate base] × 10
(d) None of these
Show Answer & Explanation

Answer: (b) [Acid] × 10 = [Conjugate base]

Explanation:
1. pH = pKa + log([base]/[acid]); pH − pKa = 1 gives [base]/[acid] = 10.
2. So [Conjugate base] = 10 [Acid].
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 25: When 0.02 mol of NaOH is added to a litre of buffer solution, its pH changes from 5.75 to 5.80. What is its buffer capacity?
(a) 0.4
(b) 0.05
(c) −0.05
(d) 2.5
Show Answer & Explanation

Answer: (a) 0.4

Explanation:
1. Buffer capacity = moles of base added per litre/change in pH.
2. = 0.02/0.05 = 0.4.
3. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 26: Calculate the pH of the buffer obtained on mixing 300 cc of 0.3 M NH3 and 500 cc of 0.5 M NH4Cl. Kb for NH3 = 1.8 × 10−5.
(a) 8.1187
(b) 9.8117
(c) 8.8117
(d) None of these
Show Answer & Explanation

Answer: (c) 8.8117

Explanation:
1. NH3 = 90 mmol and NH4+ = 250 mmol.
2. pOH = pKb + log(250/90) = 4.745 + 0.444 = 5.189.
3. pH = 14 − 5.189 = 8.81.
4. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 27: What will be the pH of the solution if 0.01 mol of HCl is dissolved in a buffer solution containing 0.02 mol of propanoic acid (Ka = 1.34 × 10−5) and 0.0152 mol of its salt at 25°C? [log(0.173) = −0.76]
(a) 3.11
(b) 4.11
(c) 5.11
(d) 6.11
Show Answer & Explanation

Answer: (b) 4.11

Explanation:
1. HCl converts salt to acid: salt = 0.0152 − 0.01 = 0.0052 mol; acid = 0.02 + 0.01 = 0.03 mol.
2. pH = pKa + log(0.0052/0.03) = 4.873 + log 0.173 = 4.873 − 0.76 = 4.11.
3. Answer: (b)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 28: Calculate the ratio of the pH of a solution containing 1 mol of CH3COONa + 1 mol of HCl per litre to that of another solution containing 1 mol of CH3COONa + 1 mol of acetic acid per litre:
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) 2 : 3
Show Answer & Explanation

Answer: (c) 1 : 2

Explanation:
1. First solution: CH3COONa + HCl gives 1 M CH3COOH, so pH = ½(pKa − log 1) = ½pKa.
2. Second: a 1 : 1 buffer, pH = pKa.
3. Ratio = ½pKa : pKa = 1 : 2.
4. Answer: (c)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

Question 29: N/10 acetic acid was titrated with N/10 NaOH. When 25%, 50% and 75% of the titration is over, the pH values of the solution will be (Ka = 10−5):
(a) 5 + log 1/3, 5, 5 + log 3
(b) 5 + log 3, 4, 5 + log 1/3
(c) 5 − log 1/3, 5, 5 − log 3
(d) 5 − log 1/3, 4, 5 + log 1/3
Show Answer & Explanation

Answer: (a) 5 + log 1/3, 5, 5 + log 3

Explanation:
1. pH = pKa + log([salt]/[acid]) with pKa = 5.
2. 25%: log(25/75) = log 1/3; 50%: log 1 = 0; 75%: log 3.
3. pH values: 5 + log 1/3, 5, 5 + log 3.
4. Answer: (a)

Teacher's Note:
1. Write the relevant relation first.
2. Check the answer by substituting back.

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