JEE Chemistry Ionic Equilibrium MCQs Set 22

Answer: (a) A - R, B - P, C - Q, D - S

Explanation:
1. Acid = 100 mmol.
2. (A) 25 mmol NaOH gives salt : acid = 25 : 75, so pH = pKa + log(1/3) = pKa - log 3 (R).
3. (B) 50 mmol gives 50 : 50, so pH = pKa (P).
4. (C) 75 mmol gives 75 : 25, so pH = pKa + log 3 (Q).
5. (D) 100 mmol is the equivalence point, with salt 0.5 M, so pH = 7 + (pKa + log 0.5)/2 = (1/2)[pKw + pKa - log 2] (S).

Teacher's Note:
1. Work out the salt to acid ratio in each case.
2. At the equivalence point use the salt hydrolysis formula.

Answer: (b) 11.3

Explanation:
1. HCl: [H+] = \( 10^{-2} \) M, so 200 mL gives 2 mmol H+.
2. NaOH: [OH-] = \( 10^{-2} \) M, so 300 mL gives 3 mmol OH-.
3. Excess OH- = 1 mmol in 500 mL = \( 2 \times 10^{-3} \) M.
4. pOH = 3 - 0.3 = 2.7, so pH = 11.3.

Teacher's Note:
1. Neutralise first, then find the excess.
2. log 2 = 0.3 gives pOH 2.7.

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

Explanation:
1. HCl is a strong acid (pH 1).
2. NH4Cl is an acidic salt (pH about 5), NaCl is neutral (pH 7) and NaCN is basic (pH about 11).
3. The order of increasing pH is HCl < NH4Cl < NaCl < NaCN.

Teacher's Note:
1. Classify each as acidic, neutral or basic.
2. Strong acid has the lowest pH.

Answer: (a) \( 1.2 \times 10^{-3} \) M

Explanation:
1. In pure water [Pb2+] = s = \( 6.7 \times 10^{-6} \) and [OH-] = 2s, so Ksp = 4s³ = 4 x (\( 6.7 \times 10^{-6} \))³ = \( 1.2 \times 10^{-15} \).
2. At pH 8, [OH-] = \( 10^{-6} \) M, so the solubility s = Ksp/[OH-]² = \( 1.2 \times 10^{-15} \) / \( 10^{-12} \) = \( 1.2 \times 10^{-3} \) M.

Teacher's Note:
1. A fixed pH fixes [OH-] and the solubility rises as the pH falls.
2. Find Ksp from the solubility in water first.

Answer: (a) Ls = S^(p+q) . p^p . q^q

Explanation:
1. ApBq gives pA + qB, so [A] = pS and [B] = qS.
2. Ls = (pS)^p (qS)^q = S^(p+q) p^p q^q.

Teacher's Note:
1. Raise each ion concentration to its coefficient.
2. Check with A2B3: 4 x 27 \( S^{5} \).

Answer: (a) 0.0175 %

Explanation:
1. After mixing, [CH3COOH] = 0.1 M and [HCl] = 0.1 M, so [H+] is about 0.1 M.
2. Ka = [H+][CH3COO-]/[CH3COOH] = 0.1 x (0.1 alpha)/0.1, so alpha = Ka/0.1 = \( 1.75 \times 10^{-4} \).
3. As a percentage this is 0.0175 %.

Teacher's Note:
1. HCl suppresses the ionisation of acetic acid strongly.
2. The pH of the solution is about 1.

Answer: (c) HgS

Explanation:
1. The ionic product for each is [M2+][S2-] = 10^-\( 3 \times 10^{-16} \) = \( 10^{-19} \).
2. Precipitation occurs when this exceeds Ksp, which happens for FeS (\( 10^{-23} \)), ZnS (\( 10^{-20} \)) and HgS (\( 10^{-54} \)) but not MnS (\( 10^{-15} \)).
3. HgS has the smallest Ksp and needs the lowest [S2-], so it precipitates first.

Teacher's Note:
1. The salt with the smallest Ksp precipitates first for the same ion type.
2. Compare the ionic product with each Ksp.

Answer: (c) 9

Explanation:
1. At the end point the salt NaA is present at 0.05 M.
2. Kh = \( 10^{-14} \) / (\( 5 \times 10^{-6} \)) = \( 2 \times 10^{-9} \).
3. [OH-] = square root of (\( 2 \times 10^{-9} \) x 0.05) = \( 10^{-5} \) M, so pOH = 5 and pH = 9.

Teacher's Note:
1. Equal volumes of acid and base double the volume, so the concentration is halved.
2. Use the salt hydrolysis formula.

Answer: (a) 0.01 %

Explanation:
1. Kh = Kw/Ka = \( 10^{-14} \)/\( 10^{-5} \) = \( 10^{-9} \).
2. h = square root of (Kh/C) = square root of (\( 10^{-9} \)/0.1) = \( 10^{-4} \).
3. As a percentage, h = 0.01 %.

Teacher's Note:
1. h = square root of (Kh / C) with C = 0.1 M.
2. Convert the fraction to a percentage.

Answer: (b) \( 8 \times 10^{-11} \)

Explanation:
1. HCl converts 0.08 mol of CH3NH2 into CH3NH3+, leaving 0.02 mol CH3NH2.
2. [OH-] = Kb x [base]/[salt] = \( 5 \times 10^{-4} \) x 0.02/0.08 = \( 1.25 \times 10^{-4} \) M.
3. [H+] = \( 10^{-14} \) / \( 1.25 \times 10^{-4} \) = \( 8 \times 10^{-11} \) M.

Teacher's Note:
1. This is a basic buffer, with unreacted base and its salt.
2. Use Kw to get [H+] from [OH-].

Answer: (a) CO2, H2CO3, HCO3-, CO3 2-

Explanation:
1. CO2 dissolves partly as CO2 (aq), and a small fraction forms H2CO3.
2. H2CO3 ionises in two steps to give HCO3- and CO3 2-.
3. So all four species are present.

Teacher's Note:
1. H2CO3 is a weak diprotic acid.
2. The CO3 2- concentration is very small.

Answer: (d) \( 2.7 \times 10^{-2} \) M

Explanation:
1. Base = 2.5 x 2/5 = 1 mmol, which needs 1 mmol HCl = 1/(2/15) = 7.5 mL.
2. Total volume = 10 mL, so the salt BH+ is 0.1 M.
3. Kh = Kw/Kb = \( 10^{-14} \)/\( 10^{-12} \) = \( 10^{-2} \), so \( x^{2} \)/(0.1 - x) = \( 10^{-2} \) gives \( x^{2} \) + 0.01x - \( 10^{-3} \) = 0, x = \( 2.7 \times 10^{-2} \) M.

Teacher's Note:
1. Kb is so small that the salt hydrolyses strongly.
2. The quadratic cannot be avoided here.

Answer: (d) MX > M3X > MX2

Explanation:
1. MX: s = square root of (\( 4.0 \times 10^{-8} \)) = \( 2 \times 10^{-4} \).
2. MX2: s = (\( 3.2 \times 10^{-14} \)/4)^(1/3) = \( 2 \times 10^{-5} \).
3. M3X: Ksp = 27 \( s^{4} \), so s = (\( 2.7 \times 10^{-15} \)/27)^(1/4) = \( 10^{-4} \).
4. Order: MX (\( 2 \times 10^{-4} \)) > M3X (\( 10^{-4} \)) > MX2 (\( 2 \times 10^{-5} \)).

Teacher's Note:
1. Compare molar solubilities, not Ksp values.
2. Use the correct formula for each salt type.

Answer: (c) HNO3 and CH3COONa, and CH3COOH and CH3COONa

Explanation:
1. HNO3 with CH3COONa (in suitable volume ratio, with less HNO3) converts some acetate to acetic acid, giving CH3COOH + CH3COONa, a buffer.
2. CH3COOH with CH3COONa is a buffer directly.
3. HNO3 with CH3COOH is two acids, and KOH with CH3COONa is a base with a salt, so neither is a buffer.

Teacher's Note:
1. A buffer needs a weak acid with its conjugate base.
2. A strong acid can generate the weak acid in situ.

Answer: (b) \( 1.1 \times 10^{-10} \)

Explanation:
1. In 0.1 M AgNO3, [Ag+] = 0.1 M (silver from the salt is negligible).
2. [CrO4 2-] = s = Ksp/[Ag+]² = \( 1.1 \times 10^{-12} \) / (0.1)² = \( 1.1 \times 10^{-10} \) M.

Teacher's Note:
1. The common ion Ag+ is squared in Ksp.
2. The solubility falls by a factor of about 1000.

Answer: (c) 4.47

Explanation:
1. AB gives A+ + B-, and B- reacts with H+ to give HB.
2. S = [A+] and total B = [B-] + [HB] = [B-](1 + [H+]/Ka) = [B-](1 + \( 10^{-3} \)/\( 10^{-8} \)) = [B-] x (1 + \( 10^{5} \)).
3. Ksp = S x [B-] = \( S^{2} \)/(1 + \( 10^{5} \)), so \( S^{2} \) = \( 2 \times 10^{-10} \) x 100001 = \( 2 \times 10^{-5} \).
4. S = \( 4.47 \times 10^{-3} \) M, so Y = 4.47.

Teacher's Note:
1. At low pH the anion of a weak acid is protonated and more salt dissolves.
2. Use Ksp = \( S^{2} \) / (1 + [H+]/Ka).

Answer: (a) \( 6.022 \times 10^{7} \)

Explanation:
1. pH = 13 gives [H+] = \( 10^{-13} \) mol/L = \( 10^{-16} \) mol/mL.
2. Number of ions = \( 10^{-16} \) x \( 6.022 \times 10^{23} \) = \( 6.022 \times 10^{7} \).

Teacher's Note:
1. Convert litres to millilitres before using Avogadro's number.
2. Number of ions = moles x Avogadro's number.

Answer: (a) 2.87

Explanation:
1. [H+] = square root of (Ka C) = square root of (\( 1.8 \times 10^{-5} \) x 0.1) = \( 1.34 \times 10^{-3} \) M.
2. pH = 3 - log 1.34 = 2.87.

Teacher's Note:
1. Use [H+] = square root of (Ka C) for a weak acid.
2. The acid ionises about 1.3% only.

Answer: (c) 11.30

Explanation:
1. 0.1 M Ba(OH)2 diluted 100 times gives \( 10^{-3} \) M.
2. [OH-] = \( 2 \times 10^{-3} \) M, so pOH = 3 - 0.3 = 2.7 and pH = 11.3.

Teacher's Note:
1. Ba(OH)2 gives two OH- per formula unit.
2. Dilution of 100 times divides the concentration by 100.

Answer: (d) \( 1 \times 10^{-8} \)

Explanation:
1. [H+] = \( 10^{-4.5} \) = \( 3.16 \times 10^{-5} \) M.
2. Ka = [H+]²/C = (\( 3.16 \times 10^{-5} \))² / 0.10 = \( 10^{-8} \).

Teacher's Note:
1. Use Ka = [H+]² / C for a weak acid.
2. The ionisation is negligible compared with 0.10 M.

Answer: (b) 10

Explanation:
1. alpha = square root of (Ka/C), so alpha is proportional to 1/square root of C.
2. On 100 times dilution C becomes C/100, so alpha2/alpha1 = square root of 100 = 10.

Teacher's Note:
1. Dilution increases the degree of dissociation.
2. alpha varies as the inverse square root of concentration.