JEE Chemistry Ionic Equilibrium MCQs Set 23

Answer: (d) 173.2 : 1

Explanation:
1. alpha = square root of (Ka/C), and C = 1 M for both.
2. alpha(acetic)/alpha(HCN) = square root of (\( 1.8 \times 10^{-5} \) / \( 6 \times 10^{-10} \)) = square root of (\( 3 \times 10^{4} \)) = 173.2.

Teacher's Note:
1. Same concentration means the ratio of alpha depends only on the Ka values.
2. Take the square root of the ratio of the Ka values.

Answer: (b) 4.7

Explanation:
1. [H+] = square root of (Ka C) = square root of (\( 8 \times 10^{-10} \) x 0.5) = \( 2 \times 10^{-5} \) M.
2. pH = 5 - log 2 = 4.7.

Teacher's Note:
1. Use [H+] = square root of (Ka C).
2. Boric acid is a Lewis acid that behaves as a weak monobasic acid in water.

Answer: (a) 0.522

Explanation:
1. H+ from H2SO4 = 2 x 0.1 x 50 = 10 mmol and from HCl = 0.4 x 50 = 20 mmol.
2. Total H+ = 30 mmol in 100 ml = 0.3 M, so pH = -log 0.3 = 0.522.

Teacher's Note:
1. Count two H+ from each H2SO4.
2. Add the millimoles and divide by the total volume.

Answer: (b) \( 1.65 \times 10^{-2} \) M

Explanation:
1. Let x be the amount of CHCl2COOH that ionises: x(0.01 + x)/(0.01 - x) = \( 3 \times 10^{-2} \).
2. \( x^{2} \) + 0.04x - \( 3 \times 10^{-4} \) = 0 gives x = \( 6.5 \times 10^{-3} \) M.
3. [H+] = 0.01 + 0.0065 = \( 1.65 \times 10^{-2} \) M.

Teacher's Note:
1. The acid is moderately strong, so the quadratic must be solved.
2. The strong acid HCl adds 0.01 M of H+.

Answer: (d) \( 1 \times 10^{-3} \) M

Explanation:
1. [H+] = square root of (Ka1 C1 + Ka2 C2) = square root of (\( 3.6 \times 10^{-7} \) + \( 6.4 \times 10^{-7} \)).
2. = square root of (\( 10^{-6} \)) = \( 10^{-3} \) M.

Teacher's Note:
1. Add the contributions of both weak acids.
2. Check that the ionisation is small.

Answer: (b) \( 2.7 \times 10^{-3} \) M

Explanation:
1. Only the first step matters: \( x^{2} \)/(0.01 - x) = \( 10^{-3} \).
2. \( x^{2} \) + \( 10^{-3} \) x - \( 10^{-5} \) = 0 gives x = (-\( 10^{-3} \) + \( 6.4 \times 10^{-3} \))/2 = \( 2.7 \times 10^{-3} \) M.

Teacher's Note:
1. K2 and K3 are much smaller than K1 and can be ignored.
2. Solve the quadratic.

Answer: (c) 11.3

Explanation:
1. Only the first ionisation counts: [OH-] = square root of (Kb1 C) = square root of (\( 4 \times 10^{-6} \)) = \( 2 \times 10^{-3} \) M.
2. pOH = 3 - 0.3 = 2.7, so pH = 11.3.

Teacher's Note:
1. Ignore the second step as Kb2 is very small.
2. pH = 14 - pOH.

Answer: (d) \( 1 \times 10^{-5} \) M

Explanation:
1. Kh = Kw/Ka = \( 10^{-14} \) / \( 1.8 \times 10^{-5} \) = \( 5.55 \times 10^{-10} \).
2. [OH-] = square root of (Kh C) = square root of (\( 5.55 \times 10^{-10} \) x 0.18) = \( 10^{-5} \) M.

Teacher's Note:
1. The anion of the weak acid hydrolyses to give OH-.
2. [OH-] = square root of (Kh C).

Answer: (b) 4.5

Explanation:
1. Kh = Kw/Kb = \( 5 \times 10^{-10} \).
2. [H+] = square root of (Kh C) = square root of (\( 5 \times 10^{-10} \) x 2) = \( 3.16 \times 10^{-5} \) M, so pH = 4.5.

Teacher's Note:
1. Ammonium ion is a weak acid.
2. pH = 7 - (pKb + log C)/2 = 7 - (4.7 + 0.3)/2 = 4.5.

Answer: (c) 0.56 %

Explanation:
1. For a salt of a weak acid and a weak base h = square root of (Kw/(Ka Kb)).
2. h = square root of (\( 10^{-14} \) / \( 3.24 \times 10^{-10} \)) = \( 5.56 \times 10^{-3} \) = 0.56 %.

Teacher's Note:
1. The extent of hydrolysis does not depend on concentration.
2. Here the pH is 7 because Ka = Kb.

Answer: (c) 8.3

Explanation:
1. For an amphiprotic anion pH = (pK1 + pK2)/2.
2. pK1 = 6.3 and pK2 = 10.3, so pH = (6.3 + 10.3)/2 = 8.3.

Teacher's Note:
1. The pH of an ampholyte does not depend on its concentration.
2. Use the mean of the two pK values.

Answer: (b) 4.19

Explanation:
1. The hydrogen phthalate ion is amphiprotic.
2. pH = (pK1 + pK2)/2 = (2.94 + 5.44)/2 = 4.19.

Teacher's Note:
1. Use the average of pK1 and pK2.
2. The concentration does not matter.

Answer: (b) 9

Explanation:
1. Equal acid and salt gives a buffer with pH = pKa.
2. pKa = 9, so pH = 9.

Teacher's Note:
1. pH = pKa + log (salt / acid) with ratio 1.
2. This is a buffer at its maximum capacity.

Answer: (c) 9.56

Explanation:
1. pOH = pKb + log(salt/base) = 4.74 + log(0.1/0.2) = 4.74 - 0.30 = 4.44.
2. pH = 14 - 4.44 = 9.56.

Teacher's Note:
1. Use pOH for a basic buffer.
2. The base is double the salt, so pH is above pKw - pKb.

Answer: (b) 5.04

Explanation:
1. Acid = 0.4 x 0.5 = 0.2 mol and salt = 0.4 mol.
2. pH = 4.74 + log(0.4/0.2) = 4.74 + 0.30 = 5.04.

Teacher's Note:
1. pKa = 5 - 0.26 = 4.74.
2. Use moles in the Henderson equation.

Answer: (d) 0.05

Explanation:
1. pOH = 14 - 9.26 = 4.74 = pKb, so [NH4+] = [NH3].
2. NH4+ needed = 0.1 mol. // Each (NH4)2SO4 gives 2 NH4+, so x = 0.05 mol.

Teacher's Note:
1. Ammonium sulphate supplies two NH4+ ions per formula unit.
2. Equal base and salt gives pOH = pKb.

Answer: (c) \( 9 \times 10^{-6} \) M

Explanation:
1. [OH-] = Kb [NH3]/[NH4+] = \( 1.8 \times 10^{-5} \) x 0.050/0.100.
2. = \( 9 \times 10^{-6} \) M.

Teacher's Note:
1. The common ion effect suppresses ionisation of NH3.
2. Use the Kb expression directly.

Answer: (d) 8.78

Explanation:
1. HCl = 15 mmol and NH3 = 20 mmol, so NH3 left = 5 mmol and NH4+ = 15 mmol.
2. pOH = 4.74 + log(15/5) = 4.74 + 0.48 = 5.22, so pH = 14 - 5.22 = 8.78.

Teacher's Note:
1. The strong acid is the limiting reagent.
2. Use the buffer equation for a base.

Answer: (c) 3.33

Explanation:
1. Each component = 10 mmol. A pH rise of 0.3 means log(salt/acid) = 0.3, so the ratio is 2.
2. (10 + x)/(10 - x) = 2 gives x = 3.33 mmol.

Teacher's Note:
1. NaOH converts acid into salt one-for-one.
2. A change of 0.3 pH needs the ratio to double.

Answer: (c) \( 1 \times 10^{-5} \) M

Explanation:
1. At the equivalence point the salt concentration is 0.1 M (volume doubled).
2. Kh = \( 10^{-9} \), so [OH-] = square root of (\( 10^{-9} \) x 0.1) = \( 10^{-5} \) M.

Teacher's Note:
1. Halve the concentration when equal volumes are mixed.
2. The end point solution is basic.

Answer: (b) \( 2 \times 10^{-9} \) M

Explanation:
1. The salt concentration is 0.05 M. Kh = \( 10^{-14} \) / \( 2 \times 10^{-5} \) = \( 5 \times 10^{-10} \).
2. [OH-] = square root of (\( 5 \times 10^{-10} \) x 0.05) = \( 5 \times 10^{-6} \) M.
3. [H+] = \( 10^{-14} \) / \( 5 \times 10^{-6} \) = \( 2 \times 10^{-9} \) M.

Teacher's Note:
1. The solution is basic at the equivalence point.
2. Get [H+] from Kw at the end.