JEE Chemistry Equilibrium MCQs Set 32

Answer: (c) I, II and III are HCl, NaOH and NaCl respectively.

Explanation:
1. Adding II or III to I raises the pH of I, so I is the acid HCl.
2. Adding III to II lowers the pH of II, which happens only if II is NaOH (pH 13) and III is NaCl (pH 7, dilution of NaOH).
3. So I, II, III = HCl, NaOH, NaCl.
4. HCl and NaCl mixed stay acidic, so they cannot give the pH of NaOH, and pH order is I (about 1) < III (7) < II (13).

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (c) \( 2\times10^{-4} \) M

Explanation:
1. Let the \( \ce{RNH3+} \) formed be y, so \( [\ce{OH-}] = 10^{-4} + y \).
2. \( K_b = y(10^{-4}+y)/0.01 = 2\times10^{-6} \) gives \( y^2 + 10^{-4}y - 2\times10^{-8} = 0 \).
3. Solving, \( y = 0.99\times10^{-4} \) M, so \( [\ce{OH-}] = 1.99\times10^{-4} \approx 2\times10^{-4} \) M.

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (c) 1.04 per cent decrease

Explanation:
1. \( K_b = 10^{-14}/(5\times10^{-10}) = 2\times10^{-5} \), pKb = 4.70. Initial pOH = 4.70, so pH = 9.30 (equal salt and base).
2. Moles: base 9 mmol, salt 9 mmol. 1 mmol HCl turns 1 mmol base into salt, giving base 8 mmol and salt 10 mmol.
3. New pOH = 4.70 + log(10/8) = 4.797, so pH = 9.203.
4. Change = (9.203 - 9.301)/9.301 = about 1.04 per cent decrease.

Teacher's Note:
1. A buffer resists pH change because the weak acid or base absorbs added H+ or OH-.
2. Use pH = pKa + log([salt]/[acid]) (or the pOH form for a base) with the moles present after the reaction.

Answer: (a) A, B and C only

Explanation:
1. In an alkaline solution \( [\ce{OH-}] > [\ce{H+}] \).
2. So pOH < pH (B true), pH is above the neutral value \( pK_w/2 \) (A true) and pOH is below it (C true).
3. (D is the opposite of (B), so it is false.)

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (c) C and D only

Explanation:
1. A salt of a strong acid and strong base (KClO4) gives pH 7 at all concentrations.
2. For a weak acid with weak base salt (CH3COONH4), \( pH = 7 + \tfrac12(pK_a - pK_b) \), with no concentration term.
3. For CH3COONa and C6H5NH3Cl, \( [\ce{H+}] \) depends on \( \sqrt{K_hC} \), so the pH changes with C.

Teacher's Note:
1. For a salt of a weak acid or base use Kh = Kw/Ka or Kw/Kb and the relation h = sqrt(Kh/C).
2. Salts of a weak acid with a weak base have a pH that does not depend on concentration.

Answer: (d) B and D only

Explanation:
1. Half ionised at pH 7.2 means pOH = pKb(In) = 6.8.
2. Ratio ionised/unionised = 5 gives pOH = 6.8 + log 5 = 7.5, so pH1 = 14 - 7.5 = 6.5.
3. For the new indicator, ionised/unionised = 4 at pOH 7.5, so pKb(new) = 7.5 - log 4 = 6.9.
4. 50 per cent ionised at pOH = 6.9, so pH2 = 7.1.

Teacher's Note:
1. An indicator changes colour around pH = pKin plus or minus 1.
2. Use pH = pKin + log([ionised]/[unionised]) for an acid-type indicator and the pOH form for a base type.

Answer: (b) PbS

Explanation:
1. Reaction quotient \( Q = (0.01)(0.1)/1^2 = 10^{-3} \).
2. A sulphide precipitates when Q is greater than \( K_{eq} \) (the reverse reaction is favoured).
3. \( 10^{-3} > 3\times10^{-7} \) only for PbS, while it is smaller than the other three K values.

Teacher's Note:
1. Common-ion effect: the solubility is found by putting the common ion concentration into Ksp.
2. Precipitation starts only when the ionic product exceeds Ksp.

Answer: (c) 5

Explanation:
1. 2 g of \( \ce{H2} \) = 1 mol, gas volume = 3.24 - 1 = 2.24 L at 273 K.
2. \( P = nRT/V = 1\times0.0821\times273/2.24 = 10 \) atm.
3. \( K = [\ce{OH-}]^2 P/[\ce{I-}]^2 \), so \( [\ce{OH-}]^2 = 4\times10^{-9}\times0.25/10 = 10^{-10} \).
4. \( [\ce{OH-}] = 10^{-5} \) M, so pOH = 5.

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (d) 4

Explanation:
1. Since \( K_{a2} \) is tiny, treat the first step: \( x^2/(0.1 - x) = 0.05 \) gives \( x^2 + 0.05x - 0.005 = 0 \), so \( [\ce{H+}] = [\ce{HA-}] = 0.05 \) M.
2. \( [\ce{A^{2-}}] = K_{a2}[\ce{HA-}]/[\ce{H+}] = 2\times10^{-6} \) M.
3. \( [\ce{A^{2-}}]/[\ce{H+}] = 2\times10^{-6}/0.05 = 4\times10^{-5} \), which is 4 after multiplying by \( 10^5 \).

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (b) 1

Explanation:
1. Moles AgBr = \( 3.29\times10^{-6}/187.8 = 1.75\times10^{-8} \), concentration = \( 7.0\times10^{-8} \) M.
2. Solubility = \( \sqrt{4.9\times10^{-13}} = 7.0\times10^{-7} \) M.
3. Saturation = 10 per cent, and 10/10 = 1.

Teacher's Note:
1. Common-ion effect: the solubility is found by putting the common ion concentration into Ksp.
2. Precipitation starts only when the ionic product exceeds Ksp.

Answer: (a) 6

Explanation:
1. MnS starts to precipitate when \( [\ce{S^{2-}}] = 2.4\times10^{-13}/0.015 = 1.6\times10^{-11} \).
2. \( [\ce{S^{2-}}] = K_a[\ce{H2S}]/[\ce{H+}]^2 \), so \( [\ce{H+}]^2 = 9\times10^{-22}/1.6\times10^{-11} = 5.6\times10^{-11} \) and \( [\ce{H+}] = 7.5\times10^{-6} \).
3. Present \( [\ce{H+}] = 1.8\times10^{-5}\times0.25/0.15 = 3\times10^{-5} \), too high, so acetate must increase: [salt] = \( 1.8\times10^{-5}\times0.25/7.5\times10^{-6} = 0.6 \) M.
4. 0.6 x 10 = 6.

Teacher's Note:
1. Common-ion effect: the solubility is found by putting the common ion concentration into Ksp.
2. Precipitation starts only when the ionic product exceeds Ksp.

Answer: (b) I-R, II-Q, III-S, IV-P

Explanation:
1. \( \ce{AgIO3} \) needs far less \( \ce{Ag+} \), so all 0.005 mol of iodate precipitates first, using 0.005 mol \( \ce{Ag+} \).
2. The remaining 0.005 mol \( \ce{Ag+} \) forms about 0.0025 mol \( \ce{Ag2CrO4} \), leaving chromate near 0.0975 M.
3. \( [\ce{Ag+}] = \sqrt{10^{-8}/0.0975} = 3.2\times10^{-4} \) M (Q).
4. \( [\ce{IO3-}] = 10^{-13}/3.2\times10^{-4} = 3.1\times10^{-10} \) M (S).
5. Precipitate = 0.005 + 0.0023 = about 0.0075 mol (R) (values are rounded).
6. So I-R, II-Q, III-S, IV-P.

Teacher's Note:
1. Common-ion effect: the solubility is found by putting the common ion concentration into Ksp.
2. Precipitation starts only when the ionic product exceeds Ksp.

Answer: (a) A-s, B-r, C-q, D-p

Explanation:
1. (A) OH- 0.3 mol, H+ 0.4 mol, excess H+ 0.1 mol in 500 L: [H+] = \( 2\times10^{-4} \), pH = 3.7 (s).
2. (B) 15 mmol acid + 15 mmol NaOH gives 0.1 M acetate: \( pH = \tfrac12(14 + 4.74 + \log 0.1) = 8.87 \) (r).
3. (C) Gives 0.25 M \( \ce{RNH2} \) with \( K_b = 10^{-5} \): [OH-] = \( \sqrt{10^{-5}\times0.25} \), pOH = 2.8, pH = 11.2 (q).
4. (D) Gives \( \ce{H2A-} \) (amphiprotic): \( pH = \tfrac12(4 + 7) = 5.5 \) (p).

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (b) 19

Explanation:
1. Total phosphate = 0.98/98 = 0.01 mol in 0.1 L = 0.1 M.
2. At pH 5 the species ratio is \( \ce{H3PO4} : \ce{H2PO4-} : \ce{HPO4^{2-}} : \ce{PO4^{3-}} = [\ce{H+}]^3 : K_1[\ce{H+}]^2 : K_1K_2[\ce{H+}] : K_1K_2K_3 = 10^{-15} : 10^{-13} : 10^{-16} : 10^{-23} \).
3. The sum is \( 1.0101\times10^{-13} \), giving \( \ce{H2PO4-} \) = 0.0990 M, \( \ce{HPO4^{2-}} \) = \( 9.9\times10^{-5} \) M, \( \ce{H3PO4} \) = \( 9.9\times10^{-4} \) M, \( \ce{PO4^{3-}} \) = \( 9.9\times10^{-12} \) M.
4. Sum of pC = 1.005 + 4.005 + 3.005 + 11.005 = 19.02, so 19.

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (c) 64

Explanation:
1. \( \ce{H2CO3} \) = 10 mmol. Buffer capacity is maximum when pH = pKa. For pH above 7 this is \( pK_{a2} = 10.25 \), so x = 10.25.
2. The half-way point of the second stage needs 10 mmol (first stage) + 5 mmol = 15 mmol KOH, so V = 150 mL.
3. (150 + 10.25) x 0.4 = 64.1, so 64.

Teacher's Note:
1. A buffer resists pH change because the weak acid or base absorbs added H+ or OH-.
2. Use pH = pKa + log([salt]/[acid]) (or the pOH form for a base) with the moles present after the reaction.

Answer: (b) A, B and C only

Explanation:
1. In basic \( \ce{NH3} \) even acetic acid is completely ionised (A true).
2. HF is a stronger acid than \( \ce{HNO3} \), so \( \ce{HNO3} \) accepts a proton from it (B true).
3. Acetic acid is a poor proton acceptor, so \( \ce{H2SO4} \) ionises only slightly in it (C true).
4. Water levels HI and HCl to the same strength (D false).

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (b) 0.54

Explanation:
1. Before adding base: \( x^2/(1.6\times10^{-3} - x) = 8\times10^{-5} \) gives \( [\ce{OH-}] = 3.2\times10^{-4} \) M, pOH 3.49, pH 10.51.
2. After adding 1.0 x 10^-3 M of \( \ce{OH-} \) (and 5 x 10^-4 M \( \ce{Ba^{2+}} \)), the charge balance \( [\ce{OH-}] = [\ce{BH+}] + 2[\ce{BH2^{2+}}] + 2[\ce{Ba^{2+}}] \) gives \( [\ce{OH-}] = 1.11\times10^{-3} \) M, pH 11.04.
3. Change = 11.04 - 10.51 = 0.54.

Teacher's Note:
1. Keep the stoichiometry exact first, then apply the equilibrium expression.
2. Check approximations by comparing the answer with the quantities you neglected.

Answer: (d) A and D only

Explanation:
1. \( \ce{O^{2-}} \) is a far stronger base than \( \ce{S^{2-}} \), so pH1 > pH2.
2. Acid strength of the hydrides rises down the group: \( \ce{H2Te} \) > \( \ce{H2Se} \) > \( \ce{H2S} \).
3. So the basicity of the anions falls: \( \ce{S^{2-}} \) > \( \ce{Se^{2-}} \) > \( \ce{Te^{2-}} \) and pH2 > pH3 > pH4.
4. Hence (A) and (D) are correct, (B) and (C) are wrong.

Teacher's Note:
1. For a salt of a weak acid or base use Kh = Kw/Ka or Kw/Kb and the relation h = sqrt(Kh/C).
2. Salts of a weak acid with a weak base have a pH that does not depend on concentration.

Answer: (b) A, C and D only

Explanation:
1. Buffer capacity is greatest when salt = base (half-equivalence), so it rises, peaks and falls (A true, B false).
2. More buffer components mean more capacity (C true).
3. The buffer depends on the \( \ce{H+} \) added, not on the anion, so \( \ce{H2SO4} \) of equal [H3O+] behaves the same (D true).

Teacher's Note:
1. A buffer resists pH change because the weak acid or base absorbs added H+ or OH-.
2. Use pH = pKa + log([salt]/[acid]) (or the pOH form for a base) with the moles present after the reaction.

Answer: (b) A and D only

Explanation:
1. \( \log([\ce{In-}]/[\ce{HIn}]) = pH - pK_{In} = 0.6 \), so the ratio is 3.98.
2. Per cent ionised = 3.98/4.98 = 80 per cent, more than 25 (D true, C false).
3. So the pink colour is visible (A true, B false).

Teacher's Note:
1. An indicator changes colour around pH = pKin plus or minus 1.
2. Use pH = pKin + log([ionised]/[unionised]) for an acid-type indicator and the pOH form for a base type.

Answer: (d) \( 2.34\times10^{-4} \) M

Explanation:
1. \( [\ce{CO3^{2-}}] = 4\times10^{-12}/[\ce{Ag+}]^2 \) and \( [\ce{CrO4^{2-}}] = 2.4\times10^{-12}/[\ce{Ag+}]^2 \).
2. Charge balance: \( [\ce{Ag+}] = 2[\ce{CO3^{2-}}] + 2[\ce{CrO4^{2-}}] = 2(6.4\times10^{-12})/[\ce{Ag+}]^2 \).
3. \( [\ce{Ag+}]^3 = 1.28\times10^{-11} \), so \( [\ce{Ag+}] = 2.34\times10^{-4} \) M.

Teacher's Note:
1. Common-ion effect: the solubility is found by putting the common ion concentration into Ksp.
2. Precipitation starts only when the ionic product exceeds Ksp.

Answer: (c) \( 2.34\times10^{-6} \) M

Explanation:
1. Carbonate used is negligible compared with 7.5 mmol, so \( [\ce{CO3^{2-}}] \approx 1.5 \) M.
2. \( [\ce{Ag+}]^2 = 8.2\times10^{-12}/1.5 = 5.47\times10^{-12} \).
3. \( [\ce{Ag+}] = 2.34\times10^{-6} \) M.

Teacher's Note:
1. Common-ion effect: the solubility is found by putting the common ion concentration into Ksp.
2. Precipitation starts only when the ionic product exceeds Ksp.

Answer: (b) \( 6.4\times10^{5} \)

Explanation:
1. Both salts share the same \( [\ce{Ag+}] \).
2. \( [\ce{Cl-}] = K_{sp}(\ce{AgCl})/[\ce{Ag+}] \) and \( [\ce{CN-}] = K_{sp}(\ce{AgCN})/[\ce{Ag+}] \).
3. Ratio = \( 1.6\times10^{-10}/2.5\times10^{-16} = 6.4\times10^{5} \).

Teacher's Note:
1. Common-ion effect: the solubility is found by putting the common ion concentration into Ksp.
2. Precipitation starts only when the ionic product exceeds Ksp.