JEE Chemistry Ionic Equilibrium MCQs Set 16

Answer: (c) 100 mL

Explanation:
1. Initial \([OH^-]=0.02\ \text{M}\), i.e. \(2\) mmol \(\ce{OH-}\) in 100 mL.
2. At pH 12, \([OH^-]=0.01\ \text{M}\), so the final volume \(=2/0.01=200\) mL.
3. Water added \(=200-100=100\) mL.

Teacher's Note:
1. The number of millimoles stays fixed.
2. Subtract the original volume from the final volume.

Answer: (c) 0.139

Explanation:
1. Before: \(\text{pH}=\text{p}K_a+\log\dfrac{0.05}{0.10}=\text{p}K_a-0.30\).
2. After: acid \(=0.11\) and salt \(=0.04\), so \(\text{pH}=\text{p}K_a+\log\dfrac{4}{11}=\text{p}K_a-0.4393\).
3. Change \(=0.4393-0.30=0.139\) (pH falls).

Teacher's Note:
1. The pKa cancels in the pH change.
2. Strong acid converts salt to acid.

Answer: (c) 3.92

Explanation:
1. \(\ce{HCOOH}=5\) mmol and \(\ce{NaOH}=3\) mmol, leaving 2 mmol acid and 3 mmol salt.
2. \(\text{pH}=\text{p}K_a+\log\dfrac32=3.745+0.176=3.92\).

Teacher's Note:
1. A buffer with salt exceeding acid has pH above pKa.
2. Use mmol for both.

Answer: (c) \(\sqrt{K_1/K_2}\)

Explanation:
1. \([H^+]=\sqrt{KC}\) for a weak acid.
2. Relative strength is the ratio of hydrogen ion concentrations at equal C: \(\sqrt{K_1C}/\sqrt{K_2C}=\sqrt{K_1/K_2}\).

Teacher's Note:
1. Strength relates to the hydrogen ion concentration produced.
2. At equal C the concentration cancels.

Answer: (c) 1.0 M

Explanation:
1. \(5.5=4.5+\log\dfrac{[\text{salt}]}{0.1}\), so \(\dfrac{[\text{salt}]}{0.1}=10\).
2. \([\text{salt}]=1.0\ \text{M}\).

Teacher's Note:
1. One pH unit above pKa means a tenfold salt to acid ratio.
2. Use the Henderson equation.

Answer: (c) 1.4

Explanation:
1. \(K_h=\dfrac{K_w}{K_aK_b}\).
2. \(K_h=\dfrac{10^{-14}}{4.0\times10^{-10}\times1.8\times10^{-5}}=\dfrac{10^{-14}}{7.2\times10^{-15}}=1.4\).

Teacher's Note:
1. Both ions hydrolyse, so Kw is divided by the product of Ka and Kb.
2. Kh does not depend on concentration here.

Answer: (d) \(8\times10^{-3}\ \text{M}\)

Explanation:
1. \(\ce{HgSO4}\) is a 1:1 salt, so \(S=\sqrt{K_{sp}}\).
2. \(S=\sqrt{6.4\times10^{-5}}=8\times10^{-3}\ \text{M}\).

Teacher's Note:
1. Square root for a 1:1 salt.
2. Check the power of ten.

Answer: (c) \(\ce{NH4CN}\)

Explanation:
1. The formula \(h=\sqrt{K_w/(K_aK_b)}\) applies where both cation and anion hydrolyse.
2. \(\ce{NH4CN}\) is a salt of the weak base ammonia and the weak acid HCN.
3. The other three salts have anions of strong acids.

Teacher's Note:
1. Weak acid plus weak base salts need both Ka and Kb.
2. Kw = 1e-14 gives the 1e-7 factor.

Answer: (d) 0.05 M \(\ce{NH3}\)

Explanation:
1. Sodium carbonate and silver nitrate both supply a common ion and reduce the solubility.
2. Ammonia forms the stable complex \(\ce{Ag(NH3)2+}\) with silver ions, pulling more salt into solution.
3. So the solubility is highest in ammonia.

Teacher's Note:
1. Complex formation increases solubility.
2. Common ions decrease it.

Answer: (b) \(10^{-2}\ \text{M}\ \ce{Ca^{2+}}\) and \(10^{-3}\ \text{M}\ \ce{F-}\)

Explanation:
1. Mixing equal volumes halves each concentration.
2. For \(10^{-2}\) and \(10^{-3}\): \(Q=(5\times10^{-3})(5\times10^{-4})^2=1.25\times10^{-9}>K_{sp}\), so a precipitate forms.
3. The other three give \(Q=1.25\times10^{-13}\), \(1.25\times10^{-12}\) and \(1.25\times10^{-14}\), all below \(K_{sp}\).

Teacher's Note:
1. Dilute each ion by half before computing Q.
2. Fluoride is squared.

Answer: (d) \(\text{p}K_a(\text{HCN})>\text{p}K_a(\text{CH}_3\text{COOH})\)

Explanation:
1. HCN is a much weaker acid than acetic acid (\(K_a=4\times10^{-10}\) against \(1.8\times10^{-5}\)).
2. A weaker acid has a larger pKa, so \(\text{p}K_a(\text{HCN})>\text{p}K_a(\text{CH}_3\text{COOH})\).
3. Its higher heat of ionisation is consistent with the less favourable ionisation.

Teacher's Note:
1. Weaker acid means a larger pKa.
2. Ionisation of HCN needs more energy.

Answer: (b) \(2\log3\)

Explanation:
1. At 1/4 neutralisation \(\text{pH}=\text{p}K_a+\log\tfrac13\).
2. At 3/4 neutralisation \(\text{pH}=\text{p}K_a+\log3\).
3. Difference \(=\log3-\log\tfrac13=2\log3\).

Teacher's Note:
1. The pKa cancels in the difference.
2. The ratios are salt to acid.

Answer: (a) less than 7

Explanation:
1. \(K_w\) increases with temperature, so neutral water has \([H^+]=\sqrt{K_w}>10^{-7}\ \text{M}\).
2. NaCl does not hydrolyse, so the solution is neutral but with pH below 7.

Teacher's Note:
1. Neutral pH is 7 only at 25 degrees C.
2. Heating water increases its ionisation.

Answer: (b) 3

Explanation:
1. \([H^+]=\dfrac{0.1\times1}{100}=10^{-3}\ \text{M}\).
2. \(\text{pH}=3\).

Teacher's Note:
1. Dilution factor is 100.
2. NaCl is a neutral salt.

Answer: (c) 2

Explanation:
1. \([H^+]=C\alpha=1.0\times0.01=0.01\ \text{M}\).
2. \(\text{pH}=2\).

Teacher's Note:
1. Hydrogen ion concentration equals C times alpha.
2. Do not confuse 1 percent with 0.1 percent.

Answer: (a) \([H^+]=3[PO_4^{3-}]\)

Explanation:
1. Since \(K_1\gg K_2\), almost all \(\ce{H+}\) comes from the first step, so \([H^+]=[H_2PO_4^-]\) and \([H^+]=\sqrt{K_1[H_3PO_4]}\) are correct.
2. In that case \([HPO_4^{2-}]=K_2\), also correct.
3. \([PO_4^{3-}]\) is vastly smaller than \([H^+]\), and there is no such factor of 3, so \([H^+]=3[PO_4^{3-}]\) is incorrect.

Teacher's Note:
1. The first ionisation dominates when K1 is much greater than K2.
2. Phosphate is present at negligible concentration.

Answer: (c) 50%

Explanation:
1. For a salt of a weak acid and a weak base \(h=\sqrt{K_w/(K_aK_b)}\) (approximately), which does not contain the concentration.
2. So the degree of hydrolysis is unchanged at 50%.

Teacher's Note:
1. No concentration term appears in h for this type of salt.
2. Dilution does not change h here.

Answer: (d) sodium acetate

Explanation:
1. Sodium acetate is the salt of a strong base and a weak acid, so its anion hydrolyses to give \(\ce{OH-}\).
2. Ammonium acetate is nearly neutral and the other two ammonium salts are acidic.

Teacher's Note:
1. Strong base plus weak acid gives a basic salt.
2. Ammonium salts of strong acids are acidic.

Answer: (c) 75 mL of M/5 HCl + 25 mL of M/5 NaOH

Explanation:
1. 75 mL of 0.2 M HCl \(=15\) mmol and 25 mL of 0.2 M NaOH \(=5\) mmol, leaving 10 mmol \(\ce{H+}\) in 100 mL.
2. \([H^+]=0.1\ \text{M}\), so \(\text{pH}=1\).
3. The first and third mixtures are basic, and the second has pH about 2 (excess 1 mmol in 100 mL).

Teacher's Note:
1. Find which reagent is in excess in each mixture.
2. Divide excess millimoles by total volume.

Answer: (c) 5.3

Explanation:
1. The salt \(\ce{NH4Cl}\) formed has concentration \(0.05\ \text{M}\).
2. \([H^+]=\sqrt{\dfrac{K_w}{K_b}C}=\sqrt{5.56\times10^{-10}\times0.05}=5.3\times10^{-6}\ \text{M}\).
3. \(\text{pH}=5.3\), which is acidic.

Teacher's Note:
1. Salt of weak base with strong acid is acidic.
2. Equivalence does not mean neutral.

Answer: (c) \(1.8\times10^{9}\)

Explanation:
1. The second reaction is the first plus \(\ce{OH- + H3O+ -> 2H2O}\), whose constant is \(1/K_w\).
2. \(K=K_a/K_w=1.8\times10^{-5}/10^{-14}=1.8\times10^{9}\).

Teacher's Note:
1. Add the reactions and multiply the constants.
2. The neutralisation of water has constant 1e14.

Answer: (d) 10.3

Explanation:
1. \(\ce{KOH}=8\) mmol and \(\ce{HCOOH}=16\) mmol, leaving 8 mmol acid and 8 mmol salt.
2. \(\text{pH}=\text{p}K_a=3.70\).
3. \(\text{pOH}=14-3.70=10.3\).

Teacher's Note:
1. Equal acid and salt gives pH = pKa.
2. The question asks for pOH.

Answer: (c) 3 to 5

Explanation:
1. \(K_a\) of HX \(=10^{-14}/10^{-6}=10^{-8}\).
2. At the equivalence point \([HX]=0.05\ \text{M}\), so \([H^+]=\sqrt{10^{-8}\times0.05}=2.2\times10^{-5}\ \text{M}\) and \(\text{pH}=4.65\).
3. The best indicator changes colour around 4.65, so the range 3 to 5 is the most suitable.

Teacher's Note:
1. The equivalence point is acidic because HX is a weak acid.
2. Choose the range that contains the equivalence point pH.

Answer: (a) \(4.8\times10^{-4}\ \text{M}\)

Explanation:
1. 0.05 mol of \(\ce{NO2}\) gives 0.025 M \(\ce{HNO2}\) and 0.025 M \(\ce{HNO3}\) (fully ionised).
2. Let \(x=[NO_2^-]\): \(\dfrac{x(0.025+x)}{0.025-x}=5\times10^{-4}\), so \(x^2+0.0255x-1.25\times10^{-5}=0\).
3. \(x=4.8\times10^{-4}\ \text{M}\).

Teacher's Note:
1. Nitric acid suppresses the ionisation of nitrous acid.
2. Solve the quadratic.

Answer: (a) \(\ce{Ag3PO4}\)

Explanation:
1. \(\ce{MnS}\): \(S=\sqrt{8\times10^{-37}}=9\times10^{-19}\ \text{M}\). \(\ce{ZnS}\): \(S=2.6\times10^{-8}\ \text{M}\).
2. \(\ce{Bi2S3}\): \(S=(10^{-72}/108)^{1/5}\approx1.0\times10^{-15}\ \text{M}\).
3. \(\ce{Ag3PO4}\): \(K_{sp}=27S^4\), so \(S=(6.7\times10^{-20})^{1/4}=1.6\times10^{-5}\ \text{M}\), the highest.

Teacher's Note:
1. Compare molar solubilities, not Ksp values.
2. The power depends on the formula type.

Answer: (c) \(S_1>S_2=S_3>S_4\)

Explanation:
1. \(S_1=\sqrt{K_{sp}}\) is the largest.
2. In 0.01 M \(\ce{CaCl2}\), \([Cl^-]=0.02\ \text{M}\), the same as in 0.02 M NaCl, so \(S_2=S_3=K_{sp}/0.02\).
3. In 0.05 M \(\ce{AgNO3}\), \(S_4=K_{sp}/0.05\), which is smaller still.

Teacher's Note:
1. Count chloride ions from calcium chloride twice.
2. Equal common ion concentration means equal solubility.