JEE Chemistry Ionic Equilibrium MCQs Set 21

Answer: (d) 12.70

Explanation:
1. [OH-] = \( 5 \times 10^{-2} \) M.
2. pOH = -log(\( 5 \times 10^{-2} \)) = 2 - log 5 = 2 - 0.7 = 1.3.
3. pH = 14 - 1.3 = 12.7.

Teacher's Note:
1. log 5 = 1 - log 2 = 0.7.
2. pH = 14 - pOH.

Answer: (b) S = (Ksp/6912)^(1/7)

Explanation:
1. Zr3(PO4)4 gives 3Zr4+ + 4PO4 3-, so [Zr4+] = 3S and [PO4 3-] = 4S.
2. Ksp = (3S)³ (4S)⁴ = 27 x 256 \( S^{7} \) = 6912 \( S^{7} \).
3. So S = (Ksp/6912)^(1/7).

Teacher's Note:
1. The total number of ions, 3 + 4 = 7, gives the power of S.
2. 27 x 256 = 6912.

Answer: (b) 6.9

Explanation:
1. For the salt of a weak acid and a weak base, pH = 7 + (pKa - pKb)/2.
2. pH = 7 + (3.2 - 3.4)/2 = 6.9.

Teacher's Note:
1. The base is slightly weaker (larger pKb) than the acid is strong, so pH is slightly below 7.
2. Remember the formula.

Answer: (c) 10 : 1

Explanation:
1. pKa = 5 and pH = 6.
2. pH = pKa + log(salt/acid), so log(salt/acid) = 1.
3. The ratio of salt to acid = 10 : 1.

Teacher's Note:
1. One unit above pKa means ten times more salt than acid.
2. Use Henderson's equation.

Answer: (c) 9.25

Explanation:
1. NH3 = 10 mmol and HCl = 5 mmol, so 5 mmol NH3 remains and 5 mmol NH4+ forms.
2. The ratio is 1:1, so pOH = pKb = 4.75.
3. pH = 14 - 4.75 = 9.25.

Teacher's Note:
1. Half neutralisation of a weak base gives pOH = pKb.
2. Then pH = 14 - pKb.

Answer: (a) CH3COOK

Explanation:
1. CH3COOK is the salt of a strong base and a weak acid, so it is basic.
2. FeCl3 is acidic. Pb(CH3COO)2 and Al(CN)3 are salts of weak bases, with Al(CN)3 being the salt of a weak base and a weak acid, so they are much less basic.

Teacher's Note:
1. Look for a strong base cation with a weak acid anion.
2. K+ does not hydrolyse.

Answer: (a) \( 3 \times 10^{-20} \)

Explanation:
1. For H2S going to 2H+ + S2-, K = K1 K2 = \( 1.2 \times 10^{-20} \).
2. [S2-] = K [H2S]/[H+]² = \( 1.2 \times 10^{-20} \) x 0.10 / (0.20)² = \( 3 \times 10^{-20} \) M.

Teacher's Note:
1. The H+ from HCl suppresses ionisation of H2S.
2. Use the overall constant.

Answer: (b) \( 1.1 \times 10^{-9} \) M

Explanation:
1. In the final 500 mL, [SO4 2-] = 50/500 = 0.1 M.
2. [Ba2+] at the onset = \( 10^{-10} \) / 0.1 = \( 10^{-9} \) M, giving \( 5 \times 10^{-10} \) mol in 500 mL.
3. These moles were in the original 450 mL, so the original concentration = \( 5 \times 10^{-10} \) / 0.45 = \( 1.1 \times 10^{-9} \) M.

Teacher's Note:
1. Account for the dilution of both ions.
2. The original volume is 500 - 50 = 450 mL.

Answer: (b) 0.18 L

Explanation:
1. For PbCl2, Ksp = 4s³ = \( 3.2 \times 10^{-8} \), so \( s^{3} \) = \( 8 \times 10^{-9} \) and s = \( 2 \times 10^{-3} \) mol/L.
2. Molar mass = 207 + 71 = 278 g/mol, so 0.1 g = \( 3.6 \times 10^{-4} \) mol.
3. Volume = \( 3.6 \times 10^{-4} \) / \( 2 \times 10^{-3} \) = 0.18 L.

Teacher's Note:
1. Volume = moles / molar solubility.
2. Add the mass of both Cl atoms to the molar mass.

Answer: (a) 3.3

Explanation:
1. [H+] of the pH 5 solution = \( 10^{-5} \) M and of the pH 3 solution = \( 10^{-3} \) M.
2. After mixing equal volumes, [H+] = (\( 10^{-5} \) + \( 10^{-3} \))/2 = \( 5.05 \times 10^{-4} \) M.
3. pH = 4 - log 5.05 = 3.3.

Teacher's Note:
1. Average the concentrations, not the pH values.
2. The stronger acid dominates the mixture.

Answer: (d) Ag3PO4 (Ksp = \( 1.8 \times 10^{-18} \))

Explanation:
1. Solubilities cannot be compared by Ksp alone for salts of different types, so find s for each.
2. MnS: s = \( 9 \times 10^{-19} \); ZnS: s = \( 2.6 \times 10^{-8} \); Bi2S3: s = (Ksp/108)^(1/5) = \( 1.6 \times 10^{-15} \); Ag3PO4: s = (Ksp/27)^(1/4) = \( 1.6 \times 10^{-5} \).
3. Ag3PO4 has the largest s.

Teacher's Note:
1. Compare molar solubilities, not Ksp values, for salts with different formulas.
2. Use the correct power for each type.

Answer: (c) 11 + log 3

Explanation:
1. Up to the first equivalence point CO3 2- is converted into HCO3-.
2. At 1/4th of this, 1/4 of the carbonate has changed, so [HCO3-] : [CO3 2-] = 1 : 3.
3. pH = pKa2 + log([CO3 2-]/[HCO3-]) = 11 + log 3.

Teacher's Note:
1. The relevant acid-base pair is HCO3- and CO3 2-, with pKa2.
2. Use fractions of the first equivalence point.

Answer: (a) 1.20

Explanation:
1. At 80% red, [In-]/[HIn] = 20/80 = 1/4, so pH1 = 5 + log(1/4) = 5 - 0.60 = 4.40.
2. At 80% blue, the ratio is 4, so pH2 = 5 + 0.60 = 5.60.
3. Change = 5.60 - 4.40 = 1.20.

Teacher's Note:
1. The change is 2 x log 4 = 1.2.
2. Use pH = pKa + log (base / acid).

Answer: (d) Statement-1 is false, statement-2 is true

Explanation:
1. On dilution of a saturated solution of a sparingly soluble salt, more solid dissolves to restore the solubility product, so the moles of Sr2+ in solution increase, not decrease. Statement-1 is false.
2. Ksp depends only on temperature, so statement-2 is true.

Teacher's Note:
1. The concentration stays fixed by Ksp but the total moles dissolved increase on dilution.
2. Ksp is a constant at constant temperature.

Answer: (a) Statement-1 is true, statement-2 is true and statement-2 is a correct explanation for statement-1

Explanation:
1. [H+] = C alpha, and alpha rises on dilution as alpha is proportional to 1/square root of C.
2. The concentration falls by a factor n while alpha rises only by a factor square root of n, so the product C alpha decreases.
3. So [H+] decreases because the volume increase outweighs the rise in alpha.

Teacher's Note:
1. Dilution raises alpha but lowers [H+].
2. [H+] = square root of (Ka x C).

Answer: (d) A, B, C and D

Explanation:
1. (A) Kb of A- = Kw/Ka = \( 10^{-14} \) / \( 2 \times 10^{-6} \) = \( 5 \times 10^{-9} \), correct.
2. (B) HA + OH- going to A- + H2O has K = Ka/Kw = \( 2 \times 10^{-6} \)/\( 10^{-14} \) = \( 2 \times 10^{8} \), correct.
3. (C) [H+] = square root of (\( 2 \times 10^{-6} \) x 0.1) = \( 4.47 \times 10^{-4} \) M, so pH = 3.35, correct.
4. (D) A- is the conjugate base of a weak acid, so it is basic, correct.

Teacher's Note:
1. Conjugate pairs obey Ka x Kb = Kw.
2. All four statements are true.

Answer: (b) B and C only

Explanation:
1. (A) is wrong: [H+] is approximately [HA-] and [A2-] is about Ka2 = \( 10^{-8} \), so they are not related by a factor 2.
2. (B) [H+] = square root of (\( 10^{-5} \) x 0.1) = \( 10^{-3} \) M, so pH = 3, correct.
3. (C) H2A = 5 mmol, so adding 5 mmol of NaHA gives equal acid and conjugate base, so pH = pKa1 = 5, an increase of 2 units. Correct.
4. (D) Complete neutralisation of a dibasic acid needs 10 mmol NaOH = 100 mL, so (D) is wrong.

Teacher's Note:
1. A dibasic acid needs two equivalents of base.
2. Equal acid and conjugate base gives pH = pKa.

Answer: (d) A, B and the 99.9% precipitation of Pb2+ when Zn2+ just starts precipitating are all correct

Explanation:
1. Minimum [S2-] needed: PbS: \( 10^{-11} \)/\( 10^{-2} \) = \( 10^{-9} \); Ag2S: \( 10^{-12} \)/(\( 10^{-2} \))² = \( 10^{-8} \); ZnS: \( 10^{-8} \)/\( 10^{-2} \) = \( 10^{-6} \); Cr2S3: [S2-]³ = \( 10^{-10} \)/(\( 10^{-2} \))² = \( 10^{-6} \), so [S2-] = \( 10^{-2} \).
2. The order of precipitation is Pb2+, Ag+, Zn2+, Cr3+, so Pb2+ is first and Cr3+ is last, while Zn2+ comes after Ag+.
3. When Zn2+ just starts, [S2-] = \( 10^{-6} \) and [Pb2+] left = \( 10^{-11} \)/\( 10^{-6} \) = \( 10^{-5} \) M, which is 99.9% precipitated.
4. So (a), (b) and the 99.9% statement are correct, and (c) is wrong.

Teacher's Note:
1. Compute the [S2-] needed for each ion.
2. The smallest [S2-] means the first to precipitate.

Answer: (a) adding 1 ml of water

Explanation:
1. The solution is a buffer whose pH depends on the ratio [NH4Cl]/[NH4OH].
2. Adding water dilutes both equally, so the ratio is unchanged, so the pH stays (nearly) constant.
3. Adding NH4Cl or NH4OH changes the ratio.

Teacher's Note:
1. Dilution of a buffer does not change the ratio of salt to base.
2. pH of a buffer stays almost constant on dilution.

Answer: (a) \( 10^{-4} \) M (Ag+) and \( 10^{-4} \) M (Cl-)

Explanation:
1. On mixing equal volumes, each concentration is halved.
2. (a) gives \( 5 \times 10^{-5} \) M each, so the ionic product = \( 2.5 \times 10^{-9} \), greater than Ksp, so AgCl precipitates.
3. (b) gives \( 2.5 \times 10^{-11} \) and (c) and (d) are even smaller, so no precipitate.

Teacher's Note:
1. Halve each concentration after mixing equal volumes.
2. Compare the ionic product with Ksp.

Answer: (b) 8.65

Explanation:
1. HX = 0.1 mol needs 0.1 mol NaOH, which is 400 mL of 0.25 M NaOH.
2. The total volume is 500 mL, so [X-] = 0.1/0.5 = 0.2 M.
3. Kh = \( 10^{-14} \)/\( 10^{-4} \) = \( 10^{-10} \), [OH-] = square root of (\( 10^{-10} \) x 0.2) = \( 4.47 \times 10^{-6} \), pOH = 5.35 and pH = 8.65.

Teacher's Note:
1. Find the concentration of the salt at the equivalence point.
2. The solution is basic at the equivalence point.