JEE Chemistry Ionic Equilibrium MCQs Set 17

Answer: (a) PbS only

Explanation:
1. \(Q=\dfrac{[M^{2+}][H_2S]}{[H^+]^2}=\dfrac{0.01\times0.1}{1}=10^{-3}\).
2. A sulphide precipitates if \(Q>K_{eq}\), that is, if \(K_{eq}<10^{-3}\).
3. Only PbS (\(3\times10^{-7}\)) satisfies this.

Teacher's Note:
1. A smaller Keq means the sulphide is less soluble in acid.
2. Compare Q with Keq.

Answer: (d) \(\ce{HPO4^{2-}}\)

Explanation:
1. The conjugate base is formed by removing one proton.
2. \(\ce{H2PO4- -> HPO4^{2-} + H+}\).

Teacher's Note:
1. Remove just one proton.
2. Do not skip to the phosphate ion.

Answer: (a) \(3.98\times10^{-6}\ \text{M}\)

Explanation:
1. \([H^+]=10^{-5.4}=10^{-6}\times10^{0.6}\).
2. \(10^{0.6}=3.98\), so \([H^+]=3.98\times10^{-6}\ \text{M}\).

Teacher's Note:
1. Write 10^-5.4 as 10^-6 times 10^0.6.
2. The result lies between 1e-6 and 1e-5.

Answer: (b) \(5\times10^{-15}\)

Explanation:
1. For \(\ce{H2A <=> 2H+ + A^{2-}}\) the overall constant is the product of the two steps.
2. \(K=K_1K_2=1.0\times10^{-5}\times5.0\times10^{-10}=5.0\times10^{-15}\).

Teacher's Note:
1. Stepwise constants multiply.
2. Overall constant is much smaller than either step.

Answer: (c) 9.5

Explanation:
1. 50% ionised means \([HA]=[A^-]\), so \(\text{pH}=\text{p}K_a=4.5\).
2. \(\text{pOH}=14-4.5=9.5\).

Teacher's Note:
1. At half ionisation pH = pKa.
2. The question asks for pOH.

Answer: (c) 7.01

Explanation:
1. For a salt of a weak acid and a weak base, \(\text{pH}=7+\tfrac12(\text{p}K_a-\text{p}K_b)\).
2. \(\text{pH}=7+\tfrac12(4.80-4.78)=7.01\).

Teacher's Note:
1. The pH is independent of concentration for such a salt.
2. Almost equal Ka and Kb give a nearly neutral solution.

Answer: (b) \(5.1\times10^{-5}\ \text{M}\)

Explanation:
1. Precipitation starts when \([Ba^{2+}][CO_3^{2-}]=K_{sp}\).
2. \([Ba^{2+}]=5.1\times10^{-9}/10^{-4}=5.1\times10^{-5}\ \text{M}\).

Teacher's Note:
1. Use the carbonate concentration directly.
2. Ignore hydrolysis of carbonate here.

Answer: (c) (ii) only

Explanation:
1. An acid donates a proton.
2. In (ii) \(\ce{H2PO4-}\) gives a proton to water, so it acts as an acid.
3. In (i) it is formed as a conjugate base, and in (iii) it accepts a proton to form \(\ce{H3PO4}\), acting as a base.

Teacher's Note:
1. Look at which side of the equation the species is on and whether it loses a proton.
2. Amphiprotic ions can do either.

Answer: (b) the concentrations of \(\ce{H+}\) and \(\ce{HCO3-}\) are approximately equal

Explanation:
1. The first ionisation gives equal \([H^+]\) and \([HCO_3^-]\), each about \(\sqrt{4.2\times10^{-7}\times0.034}=1.2\times10^{-4}\ \text{M}\).
2. The second ionisation is tiny: \([CO_3^{2-}]\approx K_2=4.8\times10^{-11}\ \text{M}\), far below \([HCO_3^-]\).
3. So only the statement about \(\ce{H+}\) and \(\ce{HCO3-}\) is correct.

Teacher's Note:
1. For a diprotic acid with K1 much greater than K2, only the first step matters for [H+].
2. The second ion concentration equals K2.

Answer: (b) 9.0 L

Explanation:
1. Initially \([H^+]=0.1\ \text{M}\) in 1 L, i.e. 0.1 mol.
2. At pH 2, \([H^+]=0.01\ \text{M}\), so the final volume \(=0.1/0.01=10\) L.
3. Water added \(=10-1=9\) L.

Teacher's Note:
1. A tenfold change in concentration means a tenfold final volume.
2. Subtract the original volume.

Answer: (a) pH starts high, changes slowly, falls steeply near the equivalence point and then stays low

Explanation:
1. The flask initially contains a strong base, so the pH is high (about 13).
2. On adding acid the pH falls slowly until near the equivalence point, where one drop causes a steep fall.
3. After that the pH stays low and levels off as excess acid accumulates.

Teacher's Note:
1. Titrating a base with an acid gives a falling curve.
2. Strong acid with strong base has a very steep equivalence jump.

Answer: (a) (a), (b) and (c)

Explanation:
1. (a) \(\ce{H+}=80\) mmol and \(\ce{OH-}=40\) mmol, leaving 40 mmol in 800 mL, so \([H^+]=0.05\ \text{M}\) and \(\text{pH}=1.3\): true.
2. (b) \(K_w\) rises with temperature: true. (c) \(\text{pH}=\text{p}K_a\) means \([A^-]=[HA]\), so 50% dissociated: true.
3. (d) The common ion effect is an application of Le Chatelier principle, so (d) is false.

Teacher's Note:
1. Check each statement separately.
2. pH equal to pKa means half dissociation.

Answer: (b) \(3\times10^{-22}\ \text{M}\)

Explanation:
1. \([OH^-]\approx0.2\ \text{M}\) from NaOH.
2. \(S=\dfrac{K_{sp}}{[OH^-]^3}=\dfrac{2.4\times10^{-24}}{8\times10^{-3}}=3\times10^{-22}\ \text{M}\).

Teacher's Note:
1. Cube the hydroxide concentration for a trihydroxide.
2. Aluminium hydroxide also dissolves in excess base but this question uses simple Ksp.

Answer: (a) \(S=\left(\dfrac{K_{sp}}{6912}\right)^{1/7}\)

Explanation:
1. \(\ce{Zr3(PO4)4 <=> 3Zr^{4+} + 4PO4^{3-}}\), so \([Zr^{4+}]=3S\) and \([PO_4^{3-}]=4S\).
2. \(K_{sp}=(3S)^3(4S)^4=27\times256\,S^7=6912S^7\).

Teacher's Note:
1. Seven ions in all, so the power is seven.
2. Multiply 3 cubed by 4 to the fourth.

Answer: (c) 5.35

Explanation:
1. \(K_h=K_w/K_b=10^{-9}\).
2. \([H^+]=\sqrt{10^{-9}\times0.02}=\sqrt{2\times10^{-11}}=4.47\times10^{-6}\ \text{M}\).
3. \(\text{pH}=5.35\).

Teacher's Note:
1. Ammonium chloride is acidic.
2. Use the hydrolysis formula.

Answer: (c) 25 mL

Explanation:
1. \(\ce{Na2CO3}\) \(=3\) mmol needs \(6\) mmol \(\ce{HCl}\) (two protons per carbonate), so \(M_{HCl}=6/25=0.24\ \text{M}\).
2. NaOH \(=30\times0.2=6\) mmol needs 6 mmol HCl.
3. Volume \(=6/0.24=25\) mL.

Teacher's Note:
1. Carbonate takes two protons to go all the way to carbonic acid.
2. The HCl amount is the same as in the first titration.

Answer: (b) 0.01%

Explanation:
1. \(K_h=K_w/K_a=10^{-9}\).
2. \(h=\sqrt{K_h/C}=\sqrt{10^{-8}}=10^{-4}\).
3. Percentage hydrolysis \(=0.01\%\).

Teacher's Note:
1. Convert the fraction to percent.
2. Hydrolysis of this salt is tiny.

Answer: (a) 8.98

Explanation:
1. Equal volumes are mixed, so the salt concentration is \(0.05\ \text{M}\).
2. \([OH^-]=\sqrt{\dfrac{K_w}{K_a}C}=\sqrt{1.79\times10^{-9}\times0.05}=9.45\times10^{-6}\ \text{M}\).
3. \(\text{pOH}=5.02\), so \(\text{pH}=8.98\).

Teacher's Note:
1. A weak acid and strong base give a basic end point.
2. Do not forget the dilution.

Answer: (b) \(2.7\times10^{-2}\ \text{M}\)

Explanation:
1. Base \(=2.5\times0.4=1\) mmol. HCl needed \(=1/(2/15)=7.5\) mL, so the final volume is 10 mL and the salt concentration is 0.1 M.
2. \(K_h=K_w/K_b=10^{-2}\), which is not small, so solve \(\dfrac{x^2}{0.1-x}=10^{-2}\): \(x^2+0.01x-0.001=0\).
3. \(x=[H^+]=\dfrac{-0.01+\sqrt{10^{-4}+4\times10^{-3}}}{2}=2.7\times10^{-2}\ \text{M}\).

Teacher's Note:
1. The base is so weak that the salt hydrolyses strongly.
2. Use the quadratic since Kh is large.

Answer: (b) \(\ce{MX}>\ce{M3X}>\ce{MX2}\)

Explanation:
1. MX: \(S=\sqrt{4\times10^{-8}}=2\times10^{-4}\ \text{M}\).
2. \(\ce{MX2}\): \(S=(3.2\times10^{-14}/4)^{1/3}=2\times10^{-5}\ \text{M}\).
3. \(\ce{M3X}\): \(K_{sp}=27S^4\), so \(S=(10^{-16})^{1/4}=10^{-4}\ \text{M}\). Order: MX > M3X > MX2.

Teacher's Note:
1. Compare molar solubilities, not Ksp.
2. Different formula types use different powers.

Answer: (c) 8

Explanation:
1. \(K_h=K_w/K_a=10^{-10}\).
2. \([OH^-]=\sqrt{K_hC}=\sqrt{10^{-10}\times10^{-2}}=10^{-6}\ \text{M}\).
3. \(\text{pOH}=6\), so \(\text{pH}=8\).

Teacher's Note:
1. The salt of a weak acid is basic.
2. Use the hydrolysis formula for the anion.

Answer: (c) \(\ce{HNO3}\) with \(\ce{CH3COONa}\), and \(\ce{CH3COOH}\) with \(\ce{CH3COONa}\)

Explanation:
1. \(\ce{CH3COOH}\) with \(\ce{CH3COONa}\) is the usual acid buffer pair.
2. \(\ce{HNO3}\) is a strong acid that converts part of \(\ce{CH3COONa}\) to \(\ce{CH3COOH}\); with the right volume ratio (less acid than acetate) the mixture contains both and is a buffer.
3. \(\ce{HNO3}\) with \(\ce{CH3COOH}\) and \(\ce{KOH}\) with \(\ce{CH3COONa}\) do not give a weak acid and its conjugate base.

Teacher's Note:
1. A buffer needs a weak acid and its conjugate base.
2. A strong acid can create the weak acid in situ from its salt.

Answer: (d) 7

Explanation:
1. CuCl dissolves to give \([Cl^-]=\sqrt{10^{-6}}=10^{-3}\ \text{M}\) (the 0.1 mol is only partly dissolved).
2. \([Ag^+]=K_{sp}(\text{AgCl})/[Cl^-]=1.6\times10^{-10}/10^{-3}=1.6\times10^{-7}\ \text{M}\).
3. So \(x=7\).

Teacher's Note:
1. The chloride from the more soluble CuCl dominates.
2. Silver chloride is a common ion case.

Answer: (b) \(1\times10^{-4}\)

Explanation:
1. The rate is proportional to \([H^+]\). For the strong acid \([H^+]=1\ \text{M}\).
2. The weak acid gives \([H^+]=0.01\ \text{M}\).
3. \(K_a=\dfrac{[H^+]^2}{C}=\dfrac{10^{-4}}{1}=1\times10^{-4}\).

Teacher's Note:
1. Acid catalysis rate depends on hydrogen ion concentration.
2. Use Ka = [H+]^2 / C for a weak acid.

Answer: (b) (i), (iii) and (iv) only

Explanation:
1. (i) Both have four electron domains on chlorine: \(sp^3\). True.
2. (ii) Chlorine and water give \(\ce{HCl}\) and \(\ce{HOCl}\), not \(\ce{HClO4}\). False.
3. (iii) \(\ce{HClO4}\) is a very strong acid, so \(\ce{ClO4-}\) is a very weak base. True. (iv) The perchlorate ion is resonance stabilised, which explains the higher acidity. True.

Teacher's Note:
1. Chlorine reacts with water to give HCl and HOCl.
2. The stronger the acid, the weaker its conjugate base.

Answer: (a) P - 1; Q - 5; R - 4; S - 1

Explanation:
1. P is a buffer (equal acid and salt): its \([H^+]=K_a[\text{acid}]/[\text{salt}]\) does not change on dilution, so effect 1.
2. Q is sodium acetate: \([OH^-]=\sqrt{K_hC}\) falls to \(1/\sqrt2\), so \([H^+]=K_w/[OH^-]\) rises \(\sqrt2\) times (effect 5). R is ammonium chloride: \([H^+]=\sqrt{K_hC}\) falls to \(1/\sqrt2\) times (effect 4).
3. S: excess solid keeps the solution saturated, so concentrations (and \([H^+]\)) do not change (effect 1).

Teacher's Note:
1. A buffer ratio is unchanged on dilution.
2. For a hydrolysed salt the ion concentration scales with the square root of C.