JEE Chemistry Ionic Equilibrium MCQs Set 14

Answer: (d) 9.77

Explanation:
1. The ion is amphiprotic between \(\ce{H2PO4-}\) and \(\ce{PO4^{3-}}\).
2. \(\text{pH}=\tfrac12(\text{p}K_2+\text{p}K_3)=\tfrac12(7.20+12.35)=9.77\).

Teacher's Note:
1. Use the pK values on either side of the ion.
2. The concentration drops out of the formula.

Answer: (c) \(5\times10^{-3}\ \text{M}\)

Explanation:
1. The mixture is 0.5 M \(\ce{NaHCO3}\) (amphiprotic).
2. \([H^+]=\sqrt{K_1K_2}=\sqrt{2.5\times10^{-17}}=5\times10^{-9}\ \text{M}\).
3. \([CO_3^{2-}]=\dfrac{K_2[HCO_3^-]}{[H^+]}=\dfrac{5\times10^{-11}\times0.5}{5\times10^{-9}}=5\times10^{-3}\ \text{M}\).

Teacher's Note:
1. Carbonate equals carbonic acid concentration in an amphiprotic solution.
2. Use K2 and the hydrogen ion concentration.

Answer: (c) \(3\times10^{-7}\ \text{M}\)

Explanation:
1. \(K_b=K_w/K_2=2\times10^{-10}\).
2. The hydrolysis is so weak that water supplies hydroxide too: \([OH^-]^2=K_w+K_bC=10^{-14}+2\times10^{-10}\times4\times10^{-4}=9\times10^{-14}\).
3. \([OH^-]=3\times10^{-7}\ \text{M}\).

Teacher's Note:
1. Do not ignore the ionisation of water when the salt is dilute.
2. Use the proton balance.

Answer: (d) 0.91

Explanation:
1. \(K=\dfrac{[Fe(OH)^{2+}][H^+]}{[Fe^{3+}]}\) with the ratio \([Fe(OH)^{2+}]/[Fe^{3+}]\le5/95\).
2. \([H^+]\ge K\times19=6.5\times10^{-3}\times19=0.1235\ \text{M}\).
3. \(\text{pH}\le0.91\).

Teacher's Note:
1. A low pH is needed to suppress hydrolysis.
2. Convert 95 percent to a ratio of 19 to 1.

Answer: (d) \(9.2\times10^{-5}\ \text{M}\), \(9.2\times10^{-5}\ \text{M}\), \(8.9\times10^{-16}\ \text{M}\); pH \(=9.96\)

Explanation:
1. \([OH^-]=[N_2H_5^+]=\sqrt{K_{b1}C}=\sqrt{8.5\times10^{-7}\times0.01}=9.2\times10^{-5}\ \text{M}\).
2. \([N_2H_6^{2+}]=K_{b2}=8.9\times10^{-16}\ \text{M}\) because \([OH^-]=[N_2H_5^+]\).
3. \(\text{pOH}=4.04\), so \(\text{pH}=9.96\).

Teacher's Note:
1. The second step contributes negligibly.
2. For a dibasic base the second conjugate acid concentration equals Kb2.

Answer: (c) 7.3

Explanation:
1. For the first indicator \(\text{p}K_{in}=7.2\), and \(\text{pH}=7.2+\log5=7.9\).
2. For the new indicator at pH 7.9: \(7.9=\text{p}K_{in}'+\log4\), so \(\text{p}K_{in}'=7.9-0.60=7.3\).
3. At 50% ionisation pH equals pK, so the pH is 7.3.

Teacher's Note:
1. Half ionised means pH equals pKin.
2. Work step by step through the two indicators.

Answer: (a) 7.81

Explanation:
1. Buffer contents: \(\ce{HPO4^{2-}}=8\) mmol and \(\ce{PO4^{3-}}=2\) mmol. Oxidation releases 4 mmol \(\ce{H+}\).
2. First 2 mmol \(\ce{H+}\) convert all \(\ce{PO4^{3-}}\) to \(\ce{HPO4^{2-}}\) (10 mmol). The other 2 mmol convert 2 mmol of \(\ce{HPO4^{2-}}\) to \(\ce{H2PO4-}\), leaving 8 mmol \(\ce{HPO4^{2-}}\) and 2 mmol \(\ce{H2PO4-}\).
3. \(\text{pH}=7.21+\log(8/2)=7.81\).

Teacher's Note:
1. Follow the protons through the successive buffer pairs.
2. After the third step is used up, the second pair controls the pH.

Answer: (d) 11.22

Explanation:
1. At one quarter of equivalence \([BH^+]/[B]=1/3\), so \(\text{pOH}=14-9.24=4.76=\text{p}K_b+\log(1/3)\), giving \(\text{p}K_b=5.24\).
2. The salt present is 6 meq, which is one quarter of the total base, so the total BOH is 24 meq. In 50 mL its concentration is 0.48 M.
3. \(\text{pOH}=\tfrac12(5.24-\log0.48)=2.78\), so \(\text{pH}=11.22\).

Teacher's Note:
1. At the 1/4 point the salt is a quarter of the total amount of base.
2. Converting all salt back gives the whole base in solution.

Answer: (d) 0.954

Explanation:
1. At 1/4 neutralisation \(\text{pH}=\text{p}K_a+\log(1/3)\).
2. At 3/4 neutralisation \(\text{pH}=\text{p}K_a+\log3\).
3. Difference \(=2\log3=0.954\).

Teacher's Note:
1. The pKa cancels.
2. The salt to acid ratios are 1/3 and 3.

Answer: (a) \(K_a=1.77\times10^{-5}\); \(1/11\) and \(10/11\)

Explanation:
1. At 1/3 neutralisation salt:acid \(=1:2\), so \(4.45=\text{p}K_a-0.301\), giving \(\text{p}K_a=4.751\) and \(K_a=1.77\times10^{-5}\).
2. A 2 unit change needs the salt to acid ratio to change by a factor of 100, from 0.1 to 10.
3. Ratio 0.1 means \(1/11\) neutralised and ratio 10 means \(10/11\) neutralised.

Teacher's Note:
1. Use fractional neutralisation to find the salt to acid ratio.
2. A ratio of 1:10 corresponds to 1/11 neutralisation.

Answer: (c) 5.85

Explanation:
1. First equivalence: HA is neutralised (12.5 mL of NaOH), giving \(\ce{A-}=0.04\ \text{M}\) and \(\ce{HB}=0.08\ \text{M}\) in 62.5 mL.
2. Proton balance: \([H^+]^2(1+[A^-]/K_a^{HA})=K_a^{HB}[HB]\), so \([H^+]^2=\dfrac{6.31\times10^{-9}\times0.08}{1+0.04/1.58\times10^{-4}}\approx2.0\times10^{-12}\).
3. \([H^+]=1.4\times10^{-6}\ \text{M}\), so \(\text{pH}=5.85\).

Teacher's Note:
1. The simple average of pKa values is only approximate here.
2. Consider that the A- formed takes protons back from water.

Answer: (d) 10.48

Explanation:
1. Second equivalence: 7.5 mmol of NaOH used (37.5 mL), volume \(87.5\) mL, \(\ce{A-}=0.0286\ \text{M}\) and \(\ce{B-}=0.0571\ \text{M}\).
2. \([OH^-]^2=K_w\left(\dfrac{[A^-]}{K_a^{HA}}+\dfrac{[B^-]}{K_a^{HB}}\right)=10^{-14}(180+9.06\times10^{6})=9.06\times10^{-8}\).
3. \([OH^-]=3.0\times10^{-4}\ \text{M}\), so \(\text{pOH}=3.52\) and \(\text{pH}=10.48\).

Teacher's Note:
1. Both anions are basic and add their effects.
2. The weaker acid gives the much stronger conjugate base.

Answer: (d) \(1.12\times10^{-6}\ \text{M}\)

Explanation:
1. At pH 9, \([OH^-]=10^{-5}\ \text{M}\). \([Zn(OH)_2(aq)]=K_1=10^{-6}\).
2. \([Zn(OH)^+]=K_2K_1/[OH^-]=10^{-8}\), \([Zn^{2+}]=K_3[Zn(OH)^+]/[OH^-]=10^{-7}\), \([Zn(OH)_3^-]=K_4K_1[OH^-]=10^{-8}\), \([Zn(OH)_4^{2-}]\approx10^{-12}\).
3. Sum \(=1.12\times10^{-6}\ \text{M}\).

Teacher's Note:
1. Add all zinc containing species.
2. Each species follows from one equilibrium constant.

Answer: (b) pH \(=9.99\) and \(S=2.5\times10^{-9}\ \text{M}\)

Explanation:
1. \(S=\dfrac{K_{sp}}{[OH^-]^2}+K_c[OH^-]^2\), which is minimum when \(\dfrac{dS}{d[OH^-]}=0\), i.e. \([OH^-]^4=K_{sp}/K_c\).
2. \([OH^-]=(1.2\times10^{-17}/0.13)^{1/4}=9.8\times10^{-5}\ \text{M}\), so \(\text{pOH}=4.01\) and \(\text{pH}=9.99\).
3. \(S_{min}=2\sqrt{K_{sp}K_c}=2\sqrt{1.56\times10^{-18}}=2.5\times10^{-9}\ \text{M}\).

Teacher's Note:
1. Solubility is the sum of the zinc ion and zincate contributions.
2. Set the derivative to zero, or note both terms are equal at the minimum.

Answer: (b) \(9.6\times10^{-3}\ \text{M}\)

Explanation:
1. \(S=[Ag^+]+[Ag(NH_3)^+]+[Ag(NH_3)_2^+]=[Cl^-]\), and \([Ag^+]=K_{sp}/S\).
2. \(S^2=K_{sp}(1+K_1[NH_3]+K_1K_2[NH_3]^2)\) with \([NH_3]=0.20-2S\approx0.181\ \text{M}\).
3. \(S^2=1.7\times10^{-10}\times5.5\times10^{5}\), so \(S=9.6\times10^{-3}\ \text{M}\).

Teacher's Note:
1. Include the ammonia used up by the complex.
2. The diammine complex is the main contributor.

Answer: (b) CdS

Explanation:
1. \([Ag^+]=\dfrac{K_{inst}[Ag(CN)_2^-]}{[CN^-]^2}=\dfrac{10^{-20}\times0.02}{0.04}=5\times10^{-21}\ \text{M}\), so \(\ce{Ag2S}\) needs \([S^{2-}]>10^{-50}/(2.5\times10^{-41})=4\times10^{-10}\ \text{M}\).
2. \([Cd^{2+}]=\dfrac{7.8\times10^{-18}\times0.02}{(0.2)^4}=9.75\times10^{-14}\ \text{M}\), so CdS needs \([S^{2-}]>7.1\times10^{-28}/9.75\times10^{-14}=7.3\times10^{-15}\ \text{M}\).
3. The sulphide ion concentration required for CdS is much smaller, so CdS precipitates first.

Teacher's Note:
1. Compare the sulphide concentration needed to start precipitation.
2. Use free metal ion concentration from the instability constant.

Answer: (d) \(S=\sqrt{K_{sp}\left(1+\dfrac{[H^+]}{K_2}+\dfrac{[H^+]^2}{K_1K_2}\right)}\)

Explanation:
1. \(S=[M^{2+}]=[A^{2-}]+[HA^-]+[H_2A]\).
2. \([HA^-]=[H^+][A^{2-}]/K_2\) and \([H_2A]=[H^+]^2[A^{2-}]/(K_1K_2)\), so \(S=[A^{2-}]\left(1+\dfrac{[H^+]}{K_2}+\dfrac{[H^+]^2}{K_1K_2}\right)\).
3. With \(K_{sp}=[M^{2+}][A^{2-}]=S[A^{2-}]\) we get \(S^2=K_{sp}\left(1+\dfrac{[H^+]}{K_2}+\dfrac{[H^+]^2}{K_1K_2}\right)\).

Teacher's Note:
1. Total anion in all protonation states equals the cation concentration.
2. The solubility rises as [H+] increases.

Answer: (c) (\(\ce{HCl}\), \(\ce{Cl-}\)) and (\(\ce{CH3COOH2+}\), \(\ce{CH3COOH}\))

Explanation:
1. A conjugate pair differs by exactly one proton.
2. \(\ce{HCl}\) loses a proton to give \(\ce{Cl-}\), so (HCl, Cl-) is a pair.
3. \(\ce{CH3COOH}\) gains a proton to give \(\ce{CH3COOH2+}\), so (\(\ce{CH3COOH2+}\), \(\ce{CH3COOH}\)) is the other pair.

Teacher's Note:
1. Write acid first and its conjugate base second.
2. The acid of one pair is on the left and the other acid is on the right.

Answer: (a) \(\ce{PO4^{3-}}\) and \(\ce{C2O4^{2-}}\)

Explanation:
1. A Bronsted base accepts a proton.
2. \(\ce{PO4^{3-}}\) accepts a proton in the forward reaction to give \(\ce{HPO4^{2-}}\).
3. \(\ce{C2O4^{2-}}\) accepts a proton in the reverse reaction to give \(\ce{HC2O4-}\).

Teacher's Note:
1. Check both directions of the equilibrium.
2. The two bases are on opposite sides of the equation.

Answer: (b) \(\ce{4NH3 + [Cu(H2O)4]^{2+} <=> [Cu(NH3)4]^{2+} + 4H2O}\)

Explanation:
1. The other three involve transfer of a proton, so they fit both definitions.
2. In the copper reaction ammonia donates a lone pair to the metal ion (a Lewis acid) with no proton transfer.
3. So it is a Lewis acid-base reaction only.

Teacher's Note:
1. Complex formation is the classic Lewis acid-base reaction.
2. Look for proton transfer to decide Bronsted behaviour.

Answer: (d) \(\ce{CH3NH2 + H2O <=> CH3NH3+ + OH-}\)

Explanation:
1. A base accepts a proton from water and releases hydroxide.
2. \(\ce{CH3NH2}\) takes a proton from water to give \(\ce{CH3NH3+}\) and \(\ce{OH-}\).
3. The other equations do not describe a real proton transfer or contain a wrong species.

Teacher's Note:
1. Weak bases produce hydroxide by taking a proton from water.
2. The conjugate acid of methylamine is methylammonium.

Answer: (d) \(\ce{H2O}\), \(\ce{H2PO3-}\), \(\ce{HPO4^{2-}}\)

Explanation:
1. An amphiprotic species can both donate and accept a proton.
2. \(\ce{H2O}\), \(\ce{H2PO3-}\) (phosphorous acid is dibasic) and \(\ce{HPO4^{2-}}\) can all do both.
3. \(\ce{H3O+}\) can only donate, \(\ce{H2PO2-}\) can only accept (monobasic parent acid) and \(\ce{SO4^{2-}}\) can only accept, which rules out the other options.

Teacher's Note:
1. Look for a species with an ionisable hydrogen and also a basic site.
2. Check each set member by member.

Answer: (b) \(5+\log\dfrac53\)

Explanation:
1. Total HA \(=10+20=30\) mmol and total \(\ce{A-}=20+30=50\) mmol.
2. \(\text{pH}=\text{p}K_a+\log\dfrac{[A^-]}{[HA]}=5+\log\dfrac{50}{30}\).
3. \(\text{pH}=5+\log\dfrac53\).

Teacher's Note:
1. Add the millimoles of each component before taking the ratio.
2. Volume cancels in the ratio.

Answer: (c) 0.50

Explanation:
1. \(\text{pOH}=1\), so \([OH^-]=0.1\ \text{M}\).
2. \([Ba(OH)_2]=0.05\ \text{M}\) because each unit gives two hydroxides.
3. In 10 mL: \(0.05\times10=0.50\) mmol.

Teacher's Note:
1. Divide by two for a dihydroxide.
2. Millimoles = molarity times millilitres.

Answer: (d) pH \(=6.7\) and water is acidic

Explanation:
1. \([H^+]=[OH^-]=2\times10^{-7}\ \text{M}\), so \(\text{pH}=\text{pOH}=7-0.3=6.7\).
2. \(\text{pH}+\text{pOH}=\text{p}K_w=13.4\).
3. Pure water is always neutral because \([H^+]=[OH^-]\), so calling it acidic is incorrect.

Teacher's Note:
1. Neutral means equal H+ and OH-, not pH 7.
2. The statement that water is acidic is the false one.