JEE Chemistry Ionic Equilibrium MCQs Set 15

Answer: (a) 11.3

Explanation:
1. \(\ce{H+}=4\times0.01=0.04\) mmol and \(\ce{OH-}=6\times0.01=0.06\) mmol.
2. Excess \(\ce{OH-}=0.02\) mmol in 10 mL, so \([OH^-]=2\times10^{-3}\ \text{M}\).
3. \(\text{pOH}=3-0.3=2.7\), so \(\text{pH}=11.3\).

Teacher's Note:
1. Base is in excess, so find pOH first.
2. Use total volume 10 mL.

Answer: (b) \(\ce{HCl}\) < \(\ce{NH4Cl}\) < \(\ce{NaCl}\) < \(\ce{NaCN}\)

Explanation:
1. \(\ce{HCl}\) is a strong acid with pH 1, \(\ce{NH4Cl}\) is weakly acidic (pH about 5), \(\ce{NaCl}\) is neutral (pH 7) and \(\ce{NaCN}\) is basic (pH about 11).
2. So the order of increasing pH is HCl, NH4Cl, NaCl, NaCN.

Teacher's Note:
1. Acid salts are below 7 and basic salts above.
2. Strong acid is the lowest of all.

Answer: (b) \(2.5\times10^{8}\)

Explanation:
1. \(\ce{BOH + H+ -> B+ + H2O}\) is the sum of \(\ce{BOH <=> B+ + OH-}\) and \(\ce{H+ + OH- -> H2O}\).
2. \(K=\dfrac{K_b}{K_w}=\dfrac{2.5\times10^{-6}}{10^{-14}}=2.5\times10^{8}\).

Teacher's Note:
1. Combine the two reactions and multiply their constants.
2. The reverse of water ionisation has constant 1/Kw.

Answer: (b) 9.35

Explanation:
1. \([CN^-]=2\times0.5=1.0\ \text{M}\).
2. \([OH^-]=\sqrt{K_bC}=\sqrt{10^{-9.3}\times1}=10^{-4.65}\ \text{M}\), so \(\text{pOH}=4.65\).
3. \(\text{pH}=14-4.65=9.35\).

Teacher's Note:
1. Barium cyanide has two cyanide ions per unit.
2. pOH = half of (pKb - log C).

Answer: (c) cationic hydrolysis of the salt formed

Explanation:
1. At equivalence exactly equal equivalents of acid and base have reacted, so there is no excess of either.
2. The salt contains the conjugate acid of the weak base, which hydrolyses to give \(\ce{H3O+}\).
3. The anion of a strong acid does not hydrolyse, so the solution is acidic.

Teacher's Note:
1. Equivalence point does not mean pH 7.
2. The cation of a weak base is a weak acid.

Answer: (d) 548 : 1

Explanation:
1. Relative strength of two acids at the same concentration is \(\sqrt{K_1/K_2}\).
2. \(\sqrt{1.2\times10^{-4}/4.0\times10^{-10}}=\sqrt{3\times10^{5}}=548\).

Teacher's Note:
1. Strength means hydrogen ion producing power, which goes as the square root of Ka.
2. The plain Ka ratio is 3e5.

Answer: (b) HA > HB > HC

Explanation:
1. A higher pH of the salt solution means a stronger conjugate base.
2. The stronger the conjugate base, the weaker its parent acid.
3. NaC has the highest pH, so HC is the weakest, and NaA the lowest, so HA is the strongest.

Teacher's Note:
1. Salt of a weak acid: higher pH means weaker acid.
2. Order from the lowest pH to the highest.

Answer: (d) BA

Explanation:
1. For a salt of a weak acid and a weak base, \(h=\sqrt{K_w/(K_aK_b)}\) approximately, and this has no concentration term.
2. So BA, which contains both a weak cation and a weak anion, has a concentration independent degree of hydrolysis.
3. For salts with only one hydrolysing ion (NaA or BCl), \(h=\sqrt{K_h/C}\), which depends on concentration.

Teacher's Note:
1. Both cation and anion hydrolyse in a weak acid and weak base salt.
2. Concentration cancels from the hydrolysis constant expression.

Answer: (c) increases by \(5\times10^{-3}\) unit

Explanation:
1. Buffer capacity \(=\dfrac{\text{moles of base added per litre}}{\Delta\text{pH}}\).
2. \(\Delta\text{pH}=\dfrac{0.5}{100}=5\times10^{-3}\).

Teacher's Note:
1. A large capacity means a very small change in pH.
2. Alkali raises the pH.

Answer: (a) 0.50 mol

Explanation:
1. \(\text{pOH}=14-8.26=5.74=4.74+\log\dfrac{[NH_4^+]}{[NH_3]}\), so \(\dfrac{[NH_4^+]}{[NH_3]}=10\).
2. \([NH_4^+]=1.0\ \text{M}\) for 0.1 M ammonia.
3. Each mole of \(\ce{(NH4)2SO4}\) gives 2 mol \(\ce{NH4+}\), so 0.50 mol is needed.

Teacher's Note:
1. Do not forget that the sulphate salt contains two ammonium ions.
2. A one unit difference from pKb means a tenfold ratio.

Answer: (a) increase

Explanation:
1. \(\ce{CO2 + H2O <=> H2CO3 <=> H+ + HCO3-}\).
2. Loss of \(\ce{CO2}\) shifts the equilibrium to the left and lowers \([H^+]\).
3. Lower \([H^+]\) means higher pH.

Teacher's Note:
1. Carbon dioxide dissolved in water is an acid.
2. Breathing out acid raises the blood pH.

Answer: (c) 150 mL

Explanation:
1. \(\text{pH}-\text{p}K_a=0.3\), so \(\dfrac{[\text{salt}]}{[\text{acid}]}=10^{0.3}\approx2\).
2. With \(v\) mL of acid: salt \(=0.2(300-v)\) mmol and acid \(=0.1v\) mmol, so \(\dfrac{0.2(300-v)}{0.1v}=2\).
3. \(300-v=v\), so \(v=150\) mL.

Teacher's Note:
1. Both volumes add up to 300 mL.
2. Ratio 2 means salt in millimoles is twice the acid in millimoles.

Answer: (a) \(2.8-\log0.54\)

Explanation:
1. Concentration \(=0.3\ \text{M}\); with 90% dissociation \([LaC^-]=2\times0.9\times0.3=0.54\ \text{M}\).
2. \([OH^-]=10^{-5.6}\), so \(K_h=\dfrac{[OH^-]^2}{0.54}=\dfrac{10^{-11.2}}{0.54}\) and \(K_a=K_w/K_h=0.54\times10^{-2.8}\).
3. \(\text{p}K_a=2.8-\log0.54\) (which is about 3.07).

Teacher's Note:
1. Each formula unit gives two lactate ions.
2. Kh = Kw/Ka for the lactate ion.

Answer: (a) phenolphthalein (8.3 to 10.0)

Explanation:
1. The equivalence point of a weak acid with a strong base lies in the basic region (pH above 7).
2. Phenolphthalein changes colour between pH 8.3 and 10, which covers this region.

Teacher's Note:
1. Match the indicator range to the pH at equivalence.
2. Methyl orange is for strong acid and weak base.

Answer: (d) methyl orange (3.1 to 4.4)

Explanation:
1. The equivalence point of a weak base with a strong acid is acidic.
2. Methyl orange changes colour at pH 3.1 to 4.4, in the acidic region.

Teacher's Note:
1. Use an indicator whose range is on the same side of 7 as the equivalence point.
2. Phenolphthalein would change too late.

Answer: (a) \(1\times10^{-5}\ \text{M}\)

Explanation:
1. \(K_{sp}=108S^5\).
2. \(S=(1.08\times10^{-23}/108)^{1/5}=(10^{-25})^{1/5}=10^{-5}\ \text{M}\).

Teacher's Note:
1. Remember the factors 2 squared times 3 cubed equal 108.
2. Fifth root of 1e-25 is 1e-5.

Answer: (d) \(\ce{Cl-}\) ions are precipitated first

Explanation:
1. AgCl needs \([Ag^+]=1.7\times10^{-10}/0.05=3.4\times10^{-9}\ \text{M}\).
2. \(\ce{Ag2CrO4}\) needs \([Ag^+]=\sqrt{1.9\times10^{-12}/0.05}=6.2\times10^{-6}\ \text{M}\).
3. AgCl needs a much lower silver concentration, so chloride precipitates first.

Teacher's Note:
1. Compare the silver concentration needed for each.
2. You cannot compare Ksp values of different salt types directly.

Answer: (d) \(S_0>S_2>S_1>S_3\)

Explanation:
1. \(S_0=\sqrt{K_{sp}}\approx1.3\times10^{-5}\ \text{M}\) is the largest.
2. \(S_1=K_{sp}/0.02=8.5\times10^{-9}\), \(S_2=K_{sp}/0.01=1.7\times10^{-8}\), \(S_3=K_{sp}/0.05=3.4\times10^{-9}\) (taking \(K_{sp}=1.7\times10^{-10}\)).
3. So \(S_0>S_2>S_1>S_3\).

Teacher's Note:
1. A higher common ion concentration gives a lower solubility.
2. Calcium chloride gives twice the chloride of sodium chloride.

Answer: (d) \(\ce{Al(OH)3}\)

Explanation:
1. For \(\ce{Al(OH)3}\): \([OH^-]=(2.7\times10^{-23}/0.1)^{1/3}=6.5\times10^{-8}\ \text{M}\).
2. For \(\ce{Zn(OH)2}\): \([OH^-]=(3.2\times10^{-14}/0.1)^{1/2}=5.7\times10^{-7}\ \text{M}\).
3. Aluminium hydroxide needs less hydroxide, so it precipitates first.

Teacher's Note:
1. Compute the hydroxide needed for each hydroxide.
2. Different formula types require the power to match.

Answer: (b) \(2\times10^{-8}\ \text{M}\)

Explanation:
1. \(\ce{Ag+}\) forms 0.02 M \(\ce{Ag(NH3)2+}\) using 0.04 mol \(\ce{NH3}\), leaving \([NH_3]=0.10\ \text{M}\).
2. \([Ag^+]=\dfrac{[Ag(NH_3)_2^+]}{K_f[NH_3]^2}=\dfrac{0.02}{10^{8}\times0.01}\).
3. \([Ag^+]=2\times10^{-8}\ \text{M}\).

Teacher's Note:
1. Subtract the ammonia used in the complex.
2. Formation constant sits in the denominator for free silver.

Answer: (c) 1.2 mol

Explanation:
1. \(\ce{AgCl + 2NH3 <=> Ag(NH3)2+ + Cl-}\) has \(K=K_{sp}K_f=10^{-2}\).
2. \(\dfrac{0.1\times0.1}{[NH_3]^2}=10^{-2}\) gives \([NH_3]=1.0\ \text{M}\) free.
3. Total ammonia \(=1.0+2(0.1)=1.2\) mol.

Teacher's Note:
1. Add the ammonia bound in the complex to the free ammonia.
2. Two ammonia molecules per silver ion.

Answer: (b) \(3.33\times10^{-6}\)

Explanation:
1. Complexed \(\ce{SCN-}=1.51\times10^{-3}-10^{-5}=1.5\times10^{-3}\ \text{M}\) \(=[M(SCN)^{2+}]\).
2. Free \([M^{3+}]=2\times10^{-3}-1.5\times10^{-3}=5\times10^{-4}\ \text{M}\).
3. \(K_d=\dfrac{[M^{3+}][SCN^-]}{[M(SCN)^{2+}]}=\dfrac{5\times10^{-4}\times10^{-5}}{1.5\times10^{-3}}=3.33\times10^{-6}\).

Teacher's Note:
1. Only one thiocyanate per metal in this complex.
2. Dissociation constant is the reciprocal of formation.

Answer: (b) IV < II < III < I

Explanation:
1. I: \([Ag^+]=\sqrt{10^{-10}}=10^{-5}\ \text{M}\). II: \([Ag^+]=\sqrt{10^{-16}}=10^{-8}\ \text{M}\).
2. III: \([Ag^+]=K_{inst}\dfrac{1}{(0.1)^2}=10^{-6}\ \text{M}\). IV: \([Ag^+]=10^{-21}\times\dfrac1{(0.1)^2}=10^{-19}\ \text{M}\).
3. Increasing order: IV (\(10^{-19}\)) < II (\(10^{-8}\)) < III (\(10^{-6}\)) < I (\(10^{-5}\)).

Teacher's Note:
1. Dissociation of a complex is governed by the free ligand concentration.
2. Compare the powers of ten carefully.

Answer: (b) 2.47

Explanation:
1. Initial pH \(=\tfrac12(4.74+1)=2.87\).
2. After adding: acid \(=0.5\) mmol and salt \(=2.0\) mmol, so \(\text{pH}=4.74+\log4=5.34\).
3. Change \(=5.34-2.87=2.47\).

Teacher's Note:
1. Find both the initial and final pH.
2. Use the weak acid formula first, then the buffer formula.

Answer: (a) 0.01625 M

Explanation:
1. \(\ce{HCl}=9\) mmol and \(\ce{NaOH}=8.9\) mmol, so \(\ce{H+}\) in excess \(=0.1\) mmol.
2. Final volume \(=140\) mL and \([H^+]=0.01\ \text{M}\) needs \(1.4\) mmol \(\ce{H+}\) in total.
3. Sulphuric acid supplies \(1.3\) mmol \(\ce{H+}\), so \(2\times40M=1.3\) and \(M=0.01625\ \text{M}\).

Teacher's Note:
1. Sulphuric acid provides two hydrogen ions per molecule.
2. Use the final volume of 140 mL.

Answer: (a) AOH

Explanation:
1. \(K_h=K_w/K_b\), so the weakest base has the largest \(K_h\).
2. AOH has the smallest \(K_b\), so its salt hydrolyses the most.

Teacher's Note:
1. Weaker the base, more the hydrolysis of its cation.
2. Kh and Kb are inversely related.