JEE Chemistry Ionic Equilibrium MCQs Set 01

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Chapter-wise Objective Questions: Ionic Equilibrium

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Multiple Choice Questions

Question 1: The normality of 2 M sulphuric acid is
(a) 2 N
(b) 4 N
(c) N/2
(d) N/4
Show Answer & Explanation

Answer: (b) 4 N

Explanation:
1. Sulphuric acid is dibasic, so n-factor = 2.
2. \( N=M\times n=2\times2=4 \) N.

Teacher's Note:
1. Normality = molarity times basicity.

Question 2: If pH = 3.31, the approximate value of \( [\ce{H+}] \) is
(a) \( 3.39\times10^{-4} \)
(b) \( 5\times10^{-4} \)
(c) \( 3.0\times10^{-3} \)
(d) none
Show Answer & Explanation

Answer: (b) \( 5\times10^{-4} \)

Explanation:
1. \( [\ce{H+}]=10^{-3.31}=10^{-4}\times10^{0.69}\approx4.9\times10^{-4}\approx5\times10^{-4} \) M.

Teacher's Note:
1. Write the pH as 4 - 0.69.

Question 3: The basicities of \( \ce{H3PO3} \) and \( \ce{H3PO2} \) are respectively
(a) 1 and 2
(b) 2 and 3
(c) 3 and 2
(d) 2 and 1
Show Answer & Explanation

Answer: (d) 2 and 1

Explanation:
1. In phosphorous acid only the two P-OH hydrogens are ionisable, so basicity = 2.
2. In hypophosphorous acid only one P-OH hydrogen is ionisable, so basicity = 1.

Teacher's Note:
1. Hydrogens bonded to phosphorus are not acidic.

Question 4: The pH of a solution containing \( 2\times10^{-3} \) moles of \( \ce{OH-} \) ions in 2 litres of solution is
(a) \( 3 \)
(b) \( 3+\log2 \)
(c) \( 3-\log2 \)
(d) \( 11 \)
Show Answer & Explanation

Answer: (d) \( 11 \)

Explanation:
1. \( [\ce{OH-}]=\frac{2\times10^{-3}}2=10^{-3} \) M, so \( pOH=3 \).
2. \( pH=14-3=11 \).

Teacher's Note:
1. Find the concentration first.

Question 5: The pH of tomato juice is 4.4. The concentration of \( \ce{H3O+} \) will be
(a) \( 39\times10^{-4} \)
(b) \( 3.9\times10^{-5} \)
(c) \( 3.9\times10^{-4} \)
(d) \( 3.9\times10^5 \)
Show Answer & Explanation

Answer: (b) \( 3.9\times10^{-5} \)

Explanation:
1. \( [\ce{H3O+}]=10^{-4.4}=10^{-5}\times10^{0.6}=3.98\times10^{-5}\approx3.9\times10^{-5} \) M.

Teacher's Note:
1. Write 4.4 as 5 - 0.6.

Question 6: The amount of acetic acid present in 100 mL of 0.1 M solution is
(a) 0.30 g
(b) 3.0 g
(c) 0.60 g
(d) none
Show Answer & Explanation

Answer: (c) 0.60 g

Explanation:
1. Moles \( =0.1\times0.1=0.01 \); molar mass 60 g/mol.
2. Mass \( =0.6 \) g.

Teacher's Note:
1. Moles times molar mass.

Question 7: The number of milli equivalents of acid in 100 mL of 0.5 N HCl solution is
(a) 50
(b) 100
(c) 25
(d) 200
Show Answer & Explanation

Answer: (a) 50

Explanation:
1. Milli equivalents \( =N\times V(\text{mL})=0.5\times100=50 \).

Teacher's Note:
1. meq = normality times volume in mL.

Question 8: If the molar concentration of \( \ce{PbI2} \) is \( 1.5\times10^{-3} \) mol/L, the concentration of iodide ions in g ion/L is
(a) \( 3.0\times10^{-3} \)
(b) \( 6.0\times10^{-3} \)
(c) \( 0.3\times10^{-3} \)
(d) \( 0.6\times10^{-6} \)
Show Answer & Explanation

Answer: (a) \( 3.0\times10^{-3} \)

Explanation:
1. Each formula unit gives two iodide ions.
2. \( [\ce{I-}]=2\times1.5\times10^{-3}=3.0\times10^{-3} \) g ion/L.

Teacher's Note:
1. Count the ions per formula unit.

Question 9: If \( \alpha \) is the degree of dissociation of a weak dibasic organic acid and y is the hydrogen ion concentration, the initial concentration of the acid is
(a) \( \frac{\alpha(y)^{-1}}2 \) M
(b) \( y(\alpha)^{-1} \) M
(c) \( \frac{y(\alpha)^{-1}}2 \) M
(d) none of them
Show Answer & Explanation

Answer: (c) \( \frac{y(\alpha)^{-1}}2 \) M

Explanation:
1. For a dibasic acid, \( [\ce{H+}]=2C\alpha \) (two hydrogen ions per molecule).
2. \( C=\frac y{2\alpha} \).

Teacher's Note:
1. Count two H+ per molecule.

Question 10: The degree of dissociation of acetic acid is given by \( \alpha=0.1\times C^{-1} \) (C = concentration of the acid). The pH of the solution is
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (a) 1

Explanation:
1. \( [\ce{H+}]=C\alpha=0.1 \) M, whatever the value of C.
2. \( pH=1 \).

Teacher's Note:
1. H+ concentration = C times alpha.

Question 11: Ostwald's dilution law is not applicable to strong electrolytes because
(a) strong electrolytes are completely ionised
(b) strong electrolytes are volatile
(c) strong electrolytes are unstable
(d) strong electrolytes often contain metal ions
Show Answer & Explanation

Answer: (a) strong electrolytes are completely ionised

Explanation:
1. The law assumes an equilibrium between ionised and unionised molecules.
2. A strong electrolyte is completely ionised, so there is no such equilibrium.

Teacher's Note:
1. The law is based on an equilibrium constant.

Question 12: The value of \( K_a \) for \( 10^{-2} \) M HCN having pOH = 10 is
(a) \( 10^{-4} \)
(b) \( 10^{-2} \)
(c) \( 10^{-5} \)
(d) none of them
Show Answer & Explanation

Answer: (d) none of them

Explanation:
1. \( pH=14-10=4 \), so \( [\ce{H+}]=10^{-4} \).
2. \( K_a=\frac{[\ce{H+}]^2}C=\frac{10^{-8}}{10^{-2}}=10^{-6} \), which is not among the first three options.

Teacher's Note:
1. Find [H+] from pOH.

Question 13: Which of the following will occur if a 1.0 M solution of a weak acid is diluted to 0.01 M at constant temperature?
(a) Percentage ionisation will increase
(b) [H+] will decrease to 0.01 M
(c) \( K_a \) will increase
(d) pH will decrease by 2 units
Show Answer & Explanation

Answer: (a) Percentage ionisation will increase

Explanation:
1. \( \alpha=\sqrt{K_a/C} \) increases on dilution.
2. \( K_a \) is constant at fixed temperature. \( [\ce{H+}]=\sqrt{K_aC} \) falls only tenfold, so the pH rises by 1 unit, not 2.

Teacher's Note:
1. Apply Ostwald's dilution law.

Question 14: The pH of a 0.15 M solution of HOCl (\( K_a=9.6\times10^{-6} \)) is
(a) 4.42
(b) 2.92
(c) 3.42
(d) none
Show Answer & Explanation

Answer: (b) 2.92

Explanation:
1. \( [\ce{H+}]=\sqrt{K_aC}=\sqrt{9.6\times10^{-6}\times0.15}=1.2\times10^{-3} \) M.
2. \( pH=-\log(1.2\times10^{-3})=2.92 \).

Teacher's Note:
1. Use [H+] = square root of Ka times C.

Question 15: If \( K_a \) of HCN is \( 4\times10^{-10} \), then the pH of a \( 2.5\times10^{-1} \) M HCN solution is
(a) 4.2
(b) 4.7
(c) 0.47
(d) 5.0
Show Answer & Explanation

Answer: (d) 5.0

Explanation:
1. \( [\ce{H+}]=\sqrt{4\times10^{-10}\times0.25}=10^{-5} \) M.
2. \( pH=5.0 \).

Teacher's Note:
1. Use [H+] = square root of Ka times C.

Question 16: The molarity of nitrous acid (\( K_a=4.5\times10^{-4} \)) at which its pH becomes 2 is (using \( [\ce{H+}]^2=K_aC \))
(a) 0.3333
(b) 0.4444
(c) 0.6666
(d) 0.2222
Show Answer & Explanation

Answer: (d) 0.2222

Explanation:
1. \( [\ce{H+}]=10^{-2} \), so \( C=\frac{[\ce{H+}]^2}{K_a}=\frac{10^{-4}}{4.5\times10^{-4}}=0.2222 \) M.

Teacher's Note:
1. Rearrange the weak acid formula.

Question 17: The correct statement for the weak acid HCN at \( 25^\circ \)C is
(a) \( \alpha=\frac{K_a}{[\ce{H+}]} \)
(b) \( \alpha=\frac{K_a\times[\ce{OH-}]}{K_w} \)
(c) both (a) and (b)
(d) \( K_b=C\alpha^2 \)
Show Answer & Explanation

Answer: (c) both (a) and (b)

Explanation:
1. For a weak acid \( K_a=[\ce{H+}]\alpha \) (approximately), so \( \alpha=\frac{K_a}{[\ce{H+}]} \).
2. Since \( [\ce{H+}]=\frac{K_w}{[\ce{OH-}]} \), also \( \alpha=\frac{K_a[\ce{OH-}]}{K_w} \). Both are correct.

Teacher's Note:
1. Express alpha using Ka and [H+].

Question 18: For water at \( 25^\circ \)C, \( 2\times10^{-7} \) moles per litre is the correct answer for which one of the following?
(a) \( [\ce{H+}]+[\ce{OH-}] \)
(b) \( [\ce{H+}]^2 \)
(c) \( [\ce{OH-}]^2 \)
(d) \( [\ce{H+}]-[\ce{OH-}] \)
Show Answer & Explanation

Answer: (a) \( [\ce{H+}]+[\ce{OH-}] \)

Explanation:
1. In pure water at 25 degrees C, \( [\ce{H+}]=[\ce{OH-}]=10^{-7} \) M.
2. The sum is \( 2\times10^{-7} \); the difference is zero and the squares are \( 10^{-14} \).

Teacher's Note:
1. Both concentrations are 10^-7 M.

Question 19: At \( 25^\circ \)C, the dissociation constant for pure water is given by
(a) \( (55.4\times10^{14})^{-1} \)
(b) \( 1\times10^{-14} \)
(c) \( \frac{1\times10^{-14}}{18} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( (55.4\times10^{14})^{-1} \)

Explanation:
1. \( K=\frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]}=\frac{10^{-14}}{55.4} \).
2. This equals \( (55.4\times10^{14})^{-1} \).

Teacher's Note:
1. Divide Kw by the molar concentration of water.

Question 20: The ionic product of water is equal to
(a) dissociation constant of water \( \times[\ce{H2O}] \)
(b) dissociation constant of water \( \times[\ce{H+}] \)
(c) product of \( [\ce{H2O}] \) and \( [\ce{H+}] \)
(d) product of \( [\ce{OH-}]^2 \) and \( [\ce{H+}] \)
Show Answer & Explanation

Answer: (a) dissociation constant of water \( \times[\ce{H2O}] \)

Explanation:
1. \( K=\frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]} \), and \( K_w=[\ce{H+}][\ce{OH-}]=K\times[\ce{H2O}] \).

Teacher's Note:
1. Water is in large excess, so its concentration is constant.

Question 21: At \( 90^\circ \)C pure water has \( [\ce{H3O+}]=10^{-6.7} \) mol/L. The value of \( K_w \) at \( 90^\circ \)C is
(a) \( 10^{-6} \)
(b) \( 10^{-12} \)
(c) \( 10^{-67} \)
(d) \( 10^{-13.4} \)
Show Answer & Explanation

Answer: (d) \( 10^{-13.4} \)

Explanation:
1. In pure water \( [\ce{OH-}]=[\ce{H3O+}] \), so \( K_w=(10^{-6.7})^2=10^{-13.4} \).

Teacher's Note:
1. Square the hydrogen ion concentration.

Question 22: At 373 K, the pH of pure water can be
(a) less than 7
(b) greater than 7
(c) equal to 7
(d) equal to 0
Show Answer & Explanation

Answer: (a) less than 7

Explanation:
1. At 373 K the ionic product of water is about \( 5.5\times10^{-13} \), so \( [\ce{H+}]\approx7.4\times10^{-7} \) and the pH is about 6.1.
2. The water is still neutral, but its pH is less than 7.

Teacher's Note:
1. Kw increases with temperature.

Question 23: The common ion effect is shown by which of the following sets of solutions?
(a) \( \ce{BaCl2}+\ce{Ba(NO3)2} \)
(b) \( \ce{NaCl}+\ce{HCl} \)
(c) \( \ce{NH4OH}+\ce{NH4Cl} \)
(d) none
Show Answer & Explanation

Answer: (c) \( \ce{NH4OH}+\ce{NH4Cl} \)

Explanation:
1. The common ion effect is the suppression of the ionisation of a weak electrolyte by a strong electrolyte that gives a common ion.
2. \( \ce{NH4OH} \) is a weak base and \( \ce{NH4Cl} \) supplies \( \ce{NH4+} \).

Teacher's Note:
1. The weak electrolyte's ionisation must be suppressed.

Question 24: Which of the following is a true statement?
(a) The ionisation constant and the ionic product of water are the same
(b) Water is a strong electrolyte
(c) The value of the ionic product of water is less than that of its ionisation constant
(d) At 298 K the number of \( \ce{H+} \) ions in a litre of water is \( 6.023\times10^{16} \)
Show Answer & Explanation

Answer: (d) At 298 K the number of \( \ce{H+} \) ions in a litre of water is \( 6.023\times10^{16} \)

Explanation:
1. At 298 K, \( [\ce{H+}]=10^{-7} \) mol/L, so the number of ions is \( 10^{-7}\times6.023\times10^{23}=6.023\times10^{16} \) per litre.
2. Water is a very weak electrolyte and its ionic product (\( 10^{-14} \)) is larger than its ionisation constant (\( 1.8\times10^{-16} \)).

Teacher's Note:
1. Multiply the concentration by Avogadro's number.

Question 25: \( \ce{H2S} \) is a weak acid that ionises into \( 2\ce{H+} \) and \( \ce{S^2-} \). If HCl is added to this solution so that the pH becomes less, what will happen?
(a) The \( \ce{S^2-} \) ion concentration decreases
(b) The concentration of \( \ce{S^2-} \) is not affected
(c) The \( \ce{S^2-} \) ion concentration increases
(d) It is not possible to add HCl to the solution
Show Answer & Explanation

Answer: (a) The \( \ce{S^2-} \) ion concentration decreases

Explanation:
1. Added \( \ce{H+} \) shifts the equilibrium of \( \ce{H2S} \) to the left (common ion effect).
2. So the \( \ce{S^2-} \) concentration decreases.

Teacher's Note:
1. Use Le Chatelier's principle.

Question 26: Which of the following is not an acidic salt?
(a) \( \ce{NaHSO4} \)
(b) \( \ce{HCOONa} \)
(c) \( \ce{NaH2PO3} \)
(d) none of them
Show Answer & Explanation

Answer: (b) \( \ce{HCOONa} \)

Explanation:
1. Acidic salts still contain ionisable hydrogen from an acid. \( \ce{HCOONa} \) is the salt of a monobasic acid; its hydrogen is not ionisable, so it is a normal salt.

Teacher's Note:
1. Look for replaceable hydrogen atoms.

Question 27: The process of neutralisation invariably results in the production of
(a) \( \ce{H+} \) ions
(b) \( \ce{OH-} \) ions
(c) both \( \ce{H+} \) and \( \ce{OH-} \) ions
(d) molecules of water
Show Answer & Explanation

Answer: (d) molecules of water

Explanation:
1. Neutralisation is \( \ce{H+ + OH- -> H2O} \).

Teacher's Note:
1. Acid plus base gives salt and water.

Question 28: Which of the following is an acid salt?
(a) \( \ce{Na2S} \)
(b) \( \ce{Na2SO3} \)
(c) \( \ce{NaHSO3} \)
(d) \( \ce{Na2SO4} \)
Show Answer & Explanation

Answer: (c) \( \ce{NaHSO3} \)

Explanation:
1. \( \ce{NaHSO3} \) still contains an ionisable hydrogen (from \( \ce{H2SO3} \)).

Teacher's Note:
1. An acid salt has replaceable hydrogen.

Question 29: The mixed salt among the following is
(a) \( \ce{CH(OH)COONa-CH(OH)COONa} \) (sodium tartrate)
(b) \( \ce{NaKSO4} \)
(c) \( \ce{CaCl2} \)
(d) all
Show Answer & Explanation

Answer: (b) \( \ce{NaKSO4} \)

Explanation:
1. A mixed salt contains more than one cation or anion from the same acid. \( \ce{NaKSO4} \) has two different cations (Na+ and K+).

Teacher's Note:
1. Mixed salts have two different cations or two different anions.

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