JEE Chemistry Equilibrium MCQs Set 24

Answer: (b) levelling effect

Explanation:
1. A solvent that makes all strong acids or bases appear equally strong by complete ionisation shows a levelling effect.

Teacher's Note:
1. Water levels strong acids and strong bases but differentiates weak ones.

Answer: (a) 1 : 4

Explanation:
1. \([H^+] = \sqrt{K_aC}\), so equal pH needs \(K_{HA}C_{HA} = K_{HB}C_{HB}\).
2. \(C_{HA}/C_{HB} = K_{HB}/K_{HA} = 1/4\).

Teacher's Note:
1. The stronger acid must be the more dilute.

Answer: (c) \([OH^-] = 5 \times 10^{-5}\) M and \(\alpha = 0.5\)

Explanation:
1. The base is not weak enough for the approximation, so solve \(x^2/(10^{-4} - x) = 5 \times 10^{-5}\).
2. \(x = 5 \times 10^{-5}\) M, so \(\alpha = 0.5\) and pH = 9.70.

Teacher's Note:
1. When \(\alpha\) is large, use the quadratic.

Answer: (c) \([OH^-] = 4 \times 10^{-3}\) M and \([BH_2^{2+}] = 2.7 \times 10^{-8}\) M

Explanation:
1. \([OH^-] = \sqrt{8 \times 10^{-5} \times 0.2} = 4 \times 10^{-3}\) M and \([BH^+] \approx [OH^-]\).
2. \([BH_2^{2+}] = K_{b2}[BH^+]/[OH^-] = K_{b2} = 2.7 \times 10^{-8}\) M.

Teacher's Note:
1. The second ion's concentration is numerically \(K_{b2}\).

Answer: (d) NaCN and NaOBr

Explanation:
1. The weaker the acid, the stronger its conjugate base and the higher the pH.
2. HCN and HOBr are the two weakest acids, so NaCN and NaOBr give the highest pH.

Teacher's Note:
1. Rank by weakness of the parent acid.

Answer: (b) \(\frac12(pK_w + pK_{a3} + \log c)\)

Explanation:
1. A\(^{3-}\) is the conjugate base of HA\(^{2-}\), whose acid constant is \(K_{a3}\).
2. So \(K_h = K_w/K_{a3}\) and \(pH = \frac12(pK_w + pK_{a3} + \log c)\).

Teacher's Note:
1. Use the K\(_a\) of the last step.

Answer: (d) [Al(H\(_2\)O)\(_6\)]\(^{3+}\), H\(_2\)O, CH\(_3\)OH

Explanation:
1. In (I) the aluminium aqua ion gives up H\(^+\) to the iron hydroxo ion.
2. In (II) water donates H\(^+\) to O\(^{2-}\).
3. In (III) methanol donates H\(^+\) to the hydride.

Teacher's Note:
1. Identify which species loses a proton in each step.

Answer: (b) 3.1

Explanation:
1. \(K_{b1} = 2 \times 10^{-4}\).
2. \(x^2/(0.004 - x) = 2 \times 10^{-4}\) gives \(x = 8 \times 10^{-4}\) M.
3. \(pOH = 3.1\).

Teacher's Note:
1. The degree of ionisation is 20%, too large to ignore.

Answer: (a) 0.36 M

Explanation:
1. \([OH^-] = K_b[NH_3]/[NH_4^+] = 1.8 \times 10^{-5} \times 0.4/(2 \times 10^{-5})\).
2. \(= 0.36\) M.

Teacher's Note:
1. Common-ion calculation.

Answer: (b) \([OH^-] = 9.6 \times 10^{-7}\) M and \([H_2C_2O_4] \approx 1.8 \times 10^{-13}\) M

Explanation:
1. \(K_h = K_w/K_{a2} = 1.85 \times 10^{-10}\), so \([OH^-] = \sqrt{1.85 \times 10^{-10} \times 0.005} = 9.6 \times 10^{-7}\) M.
2. \([HC_2O_4^-] \approx 9.6 \times 10^{-7}\) M and \([H^+] = 1.04 \times 10^{-8}\) M.
3. \([H_2C_2O_4] = [H^+][HC_2O_4^-]/K_{a1} = 1.8 \times 10^{-13}\) M.

Teacher's Note:
1. Second hydrolysis step is tiny.

Answer: (d) 25 mL

Explanation:
1. pOH = 7.69 and \(pOH = pK_{b2} + \log([BH_2^{2+}]/[BH^+])\), so the ratio is \(10^{0.3} = 2\).
2. BH\(_2^{2+}\) = 5 mmol, so BH\(^+\) = 2.5 mmol.
3. Volume = 2.5/0.1 = 25 mL.

Teacher's Note:
1. Use the second dissociation constant for this pair.

Answer: (b) orange

Explanation:
1. \(pH = \frac12(pK_w - pK_b - \log C) = \frac12(14 - 4.74 + 1) = 5.13\).
2. That lies between 3.1 and 6.3, so the colour is orange.

Teacher's Note:
1. Find the salt's pH and compare it with the indicator range.

Answer: (b) 10/11

Explanation:
1. \([In^-]/[HIn] = 10^{6-5} = 10\).
2. Basic fraction = 10/(10 + 1) = 10/11.

Teacher's Note:
1. Convert the ratio into a fraction.

Answer: (d) 5 : 1

Explanation:
1. Half-way colour means \(pH = pK_{In} = 5.45\).
2. \(5.45 = 4.75 + \log r\) gives \(\log r = 0.7\), so \(r = 5\).

Teacher's Note:
1. Equal amounts of the two coloured forms means pH equals \(pK_{In}\).

Answer: (b) 6.173

Explanation:
1. Na\(_2\)CO\(_3\) = 2.5 mmol; HCl = 4 mmol.
2. 2.5 mmol converts to HCO\(_3^-\), then 1.5 mmol more gives 1.5 mmol H\(_2\)CO\(_3\) and 1.0 mmol HCO\(_3^-\).
3. \(pH = 6.35 + \log(1.0/1.5) = 6.17\).

Teacher's Note:
1. Second-stage buffer: H\(_2\)CO\(_3\) and HCO\(_3^-\).

Answer: (a) 0.8 M

Explanation:
1. After mixing, [Mg\(^{2+}\)] = 0.05 M and [NH\(_4^+\)] = 0.4 M.
2. \([OH^-] = \sqrt{1.62 \times 10^{-11}/0.05} = 1.8 \times 10^{-5}\) M.
3. \([NH_3] = K_b[NH_4^+]/[OH^-] = 0.4\) M after mixing, so 0.8 M before.

Teacher's Note:
1. Remember the dilution on mixing equal volumes.

Answer: (b) 5.21

Explanation:
1. MnS is the harder to precipitate: \([S^{2-}] \ge 2.5 \times 10^{-13}/0.01 = 2.5 \times 10^{-11}\) M.
2. \([H^+]^2 \le 9.6 \times 10^{-21} \times 0.1/(2.5 \times 10^{-11}) = 3.84 \times 10^{-11}\), so \([H^+] \le 6.2 \times 10^{-6}\) M.
3. \(pH \ge 5.21\).

Teacher's Note:
1. Precipitating both means satisfying the harder one.

Answer: (b) (iii) < (ii) < (iv) < (i)

Explanation:
1. In NaBr, \(s = K_{sp}/0.2\); in AgNO\(_3\), \(s = K_{sp}/0.1\); pure water gives \(\sqrt{K_{sp}}\).
2. NH\(_3\) forms the complex [Ag(NH\(_3\))\(_2\)]\(^+\) and increases solubility the most.

Teacher's Note:
1. Common ions reduce solubility; complexing agents raise it.

Answer: (c) phenolphthalein (8 to 9.6)

Explanation:
1. The equivalence point lies in the basic region (pH above 7), where phenolphthalein changes colour.

Teacher's Note:
1. Match the indicator's range with the equivalence-point pH.

Answer: (b) HSO\(_4^-\)

Explanation:
1. HSO\(_4^-\) can donate a proton (giving SO\(_4^{2-}\)) and can accept one (giving H\(_2\)SO\(_4\)).
2. The other three can only accept a proton, or practically cannot act as acids.

Teacher's Note:
1. Amphiprotic species have both a transferable H and a lone pair.