JEE Chemistry Ionic Equilibrium MCQs Set 13

Answer: (b) \(8.7\times10^{-7}\) M and \(4.0\times10^{-7}\) M

Explanation:
1. \([Ag^+]=[SCN^-]+[Br^-]\), with \([SCN^-]=K_1/[Ag^+]\) and \([Br^-]=K_2/[Ag^+]\).
2. \([Ag^+]^2=K_1+K_2=1.6\times10^{-12}\), so \([Ag^+]=1.265\times10^{-6}\ \text{M}\).
3. \([SCN^-]=1.1\times10^{-12}/1.265\times10^{-6}=8.7\times10^{-7}\) and \([Br^-]=5\times10^{-13}/1.265\times10^{-6}=4.0\times10^{-7}\ \text{M}\).

Teacher's Note:
1. Silver comes from both salts.
2. The larger Ksp gives the larger solubility.

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

Explanation:
1. Charge balance: \([F^-]=2[Mg^{2+}]+2[Sr^{2+}]\).
2. \([F^-]=\dfrac{2(9.5\times10^{-9}+4\times10^{-9})}{[F^-]^2}\), so \([F^-]^3=2.7\times10^{-8}\).
3. \([F^-]=3\times10^{-3}\ \text{M}\).

Teacher's Note:
1. Each divalent cation balances two fluoride ions.
2. Cube root at the end.

Answer: (a) \(6.67\times10^{-5}\ \text{M}\)

Explanation:
1. After mixing, both are 0.01 M and \(\ce{Ag+ + HCN <=> AgCN + H+}\) goes nearly to completion, giving \([H^+]\approx0.01\ \text{M}\).
2. The reverse reaction has \(K=K_{sp}/K_a=4.44\times10^{-7}\), so \([Ag^+][HCN]/[H^+]=4.44\times10^{-7}\).
3. With \([Ag^+]=[HCN]=x\): \(x^2=4.44\times10^{-7}\times0.01\), so \(x=6.67\times10^{-5}\ \text{M}\).

Teacher's Note:
1. The silver and the HCN left over are equal in amount.
2. The strong acid formed holds back the precipitation reaction.

Answer: (c) 19.5 kg

Explanation:
1. \(\ce{AgCl + Cl- <=> AgCl2-}\) has \(K=K_{sp}K_f=3\times10^{-5}\).
2. Required \([\ce{AgCl2-}]=0.010/100=10^{-4}\ \text{M}\), so \([Cl^-]=10^{-4}/3\times10^{-5}=3.33\ \text{M}\).
3. Total chloride needed \(\approx3.33\) mol per litre, so \(333\) mol for 100 L, mass \(=333\times58.5\approx19.5\ \text{kg}\).

Teacher's Note:
1. Combine Ksp and Kf into one equilibrium constant.
2. Multiply the molarity by the volume and the molar mass.

Answer: (b) \(K_f=2080\) and \(K_d=4.8\times10^{-4}\)

Explanation:
1. Overall formation constant \(K_f=K_1K_2K_3=130\times16\times1.0=2080\).
2. The dissociation constant is the reciprocal: \(K_d=1/2080=4.8\times10^{-4}\).

Teacher's Note:
1. Stepwise constants multiply, they do not add.
2. Dissociation is the reverse of formation.

Answer: (b) \(2.8\times10^{-3}\ \text{M}\)

Explanation:
1. \(\ce{AgBr + 2NH3 <=> Ag(NH3)2+ + Br-}\) has \(K=K_{sp}K_f=5\times10^{-5}\).
2. \(\dfrac{S^2}{(0.4-2S)^2}=5\times10^{-5}\), so \(\dfrac{S}{0.4-2S}=7.07\times10^{-3}\).
3. \(S=2.8\times10^{-3}\ \text{mol L}^{-1}\).

Teacher's Note:
1. Take the square root of both sides to avoid a quadratic.
2. Two ammonia molecules are used per silver ion.

Answer: (d) \(K=1.8\times10^{-16},\ K_w=10^{-14}\)

Explanation:
1. \([H^+]=[OH^-]=55.5\times1.8\times10^{-9}=10^{-7}\ \text{M}\), so \(K_w=10^{-14}\).
2. Ionisation constant \(K=\dfrac{[H^+][OH^-]}{[H_2O]}=\dfrac{10^{-14}}{55.5}=1.8\times10^{-16}\).

Teacher's Note:
1. Kw includes the concentration of water in the constant.
2. The ionisation constant divides by 55.5 M.

Answer: (d) \(1.25\times10^{-2}\)

Explanation:
1. Total \([H^+]=0.1\ \text{M}\). HCl gives 0.09 M, so the weak acids give only 0.01 M.
2. Acetic acid is suppressed (\([CH_3COO^-]=10^{-5}\times0.1/0.1=10^{-5}\)), so dichloroacetic acid gives \([A^-]=0.01\ \text{M}\) and \([HA]=0.08\ \text{M}\).
3. \(K_a=\dfrac{0.1\times0.01}{0.08}=1.25\times10^{-2}\).

Teacher's Note:
1. The stronger acid suppresses ionisation of the weaker acid.
2. Take care to include the total hydrogen ion concentration.

Answer: (a) 11.74

Explanation:
1. \(C=\dfrac{10\times0.935\times10}{17}=5.5\ \text{M}\).
2. \([OH^-]=\sqrt{K_bC}=\sqrt{5.5\times10^{-6}\times5.5}=5.5\times10^{-3}\ \text{M}\).
3. \(\text{pOH}=2.26\), so \(\text{pH}=11.74\).

Teacher's Note:
1. Molarity = 10 x density x mass percent / molar mass.
2. pH of a base = 14 - pOH.

Answer: (c) \(2.8\times10^{4}\) litre

Explanation:
1. Twice the pH is \(4.74\), so \([H^+]=1.8\times10^{-5}=K_a\).
2. When \([H^+]=K_a\), \([A^-]=[HA]\), so the acid is half ionised and \(C=2[H^+]=3.6\times10^{-5}\ \text{M}\).
3. \(V=1/3.6\times10^{-5}=2.8\times10^{4}\) litre.

Teacher's Note:
1. At this dilution the simple square-root formula does not work.
2. Half ionisation means the concentration equals twice the hydrogen ion concentration.

Answer: (c) pH \(=13.1\) and \(0.13\ \text{M}\)

Explanation:
1. \(n=\dfrac{PV}{RT}=\dfrac{959}{0.0821\times298}=39.2\ \text{mol per litre}\).
2. \([OH^-]=\sqrt{K_bC}=\sqrt{4.07\times10^{-4}\times39.2}=0.126\ \text{M}\), so \(\text{pOH}=0.90\) and \(\text{pH}=13.1\).
3. NaOH must be about 0.13 M to give the same hydroxide concentration.

Teacher's Note:
1. Convert gas volume to moles with the ideal gas equation.
2. NaOH is fully ionised, so its molarity equals [OH-].

Answer: (c) 1.65

Explanation:
1. \(K=\dfrac{[OH^-]^2\,p_{H_2}}{[I^-]^2}\), so \([OH^-]^2=\dfrac{1.2\times10^{-23}\times0.01}{0.6}=2\times10^{-25}\).
2. \([OH^-]=4.5\times10^{-13}\ \text{M}\), so \(\text{pOH}=12.35\).
3. \(\text{pH}=14-12.35=1.65\).

Teacher's Note:
1. Solids do not appear in the equilibrium expression.
2. A tiny [OH-] means a strongly acidic solution.

Answer: (d) 4.1

Explanation:
1. At equilibrium \([A]=[B]=0.33\ \text{M}\) and \([C]=[D]=0.67\ \text{M}\).
2. \(K=\dfrac{0.67\times0.67}{0.33\times0.33}=4.1\).

Teacher's Note:
1. Write the equilibrium amounts of each species.
2. Reactants and products have equal concentrations pair by pair.

Answer: (b) \(3.6\times10^{2}\) and \(9.8\times10^{3}\)

Explanation:
1. 95%: \(K=(0.95/0.05)^2=361\).
2. 99%: \(K=(0.99/0.01)^2=9801\).

Teacher's Note:
1. K is the square of the ratio of product to reactant.
2. Higher conversion means a much larger K.

Answer: (b) 2.24

Explanation:
1. \(\ce{H3PO4}=6\) mmol and \(\ce{NaOH}=3\) mmol, so 3 mmol \(\ce{H3PO4}\) and 3 mmol \(\ce{H2PO4-}\) in 70 mL (0.0429 M each).
2. \(\dfrac{x(0.0429+x)}{0.0429-x}=7.5\times10^{-3}\) gives \(x=[H^+]=5.7\times10^{-3}\ \text{M}\).
3. \(\text{pH}=2.24\). The simple Henderson value 2.12 is only approximate because \(K_1\) is large.

Teacher's Note:
1. A buffer with a fairly strong acid needs the quadratic.
2. Equal acid and salt gives pH close to pK1 only for weaker acids.

Answer: (c) 4.67

Explanation:
1. \(\ce{H3PO4}=6\) mmol and \(\ce{NaOH}=6\) mmol, so all is converted to \(\ce{NaH2PO4}\).
2. \(\text{p}K_1=2.12\) and \(\text{p}K_2=7.21\).
3. \(\text{pH}=\tfrac12(2.12+7.21)=4.67\).

Teacher's Note:
1. Amphiprotic ion: average of the two relevant pK values.
2. One equivalent of base gives the dihydrogen phosphate.

Answer: (a) 12.0

Explanation:
1. \(\ce{H3PO4}=4\) mmol and \(\ce{NaOH}=10\) mmol, which is 2.5 equivalents.
2. The mixture contains 2 mmol \(\ce{HPO4^{2-}}\) and 2 mmol \(\ce{PO4^{3-}}\).
3. \(\text{pH}=\text{p}K_3=12.0\).

Teacher's Note:
1. 2.5 equivalents is the half way point of the third step.
2. Hydrolysis of phosphate raises the pH slightly but is ignored here.

Answer: (b) 8.35

Explanation:
1. \(\ce{Na2CO3}=2\) mmol and \(\ce{HCl}=2\) mmol, giving 2 mmol \(\ce{NaHCO3}\).
2. \(\text{pH}=\tfrac12(\text{p}K_1+\text{p}K_2)=\tfrac12(6.38+10.32)=8.35\).

Teacher's Note:
1. Work out which species are present after the reaction.
2. Amphiprotic salts or equal acid-salt pairs give simple pH values.

Answer: (c) 9.60

Explanation:
1. \(\ce{Na3PO4}=0.8\) mmol and \(\ce{HCl}=1.6\) mmol, giving 0.8 mmol \(\ce{HPO4^{2-}}\).
2. \(\text{pH}=\tfrac12(\text{p}K_2+\text{p}K_3)=\tfrac12(7.21+12.00)=9.60\).

Teacher's Note:
1. Work out which species are present after the reaction.
2. Amphiprotic salts or equal acid-salt pairs give simple pH values.

Answer: (c) 9.60

Explanation:
1. \(\ce{PO4^{3-} + H2PO4- -> 2HPO4^{2-}}\): 5 mmol of each gives 10 mmol \(\ce{HPO4^{2-}}\).
2. \(\text{pH}=\tfrac12(\text{p}K_2+\text{p}K_3)=9.60\).

Teacher's Note:
1. Work out which species are present after the reaction.
2. Amphiprotic salts or equal acid-salt pairs give simple pH values.

Answer: (d) 7.21

Explanation:
1. \(\ce{H3PO4 + PO4^{3-} -> H2PO4- + HPO4^{2-}}\): 4 mmol of each product forms.
2. \(\text{pH}=\text{p}K_2=7.21\).

Teacher's Note:
1. Work out which species are present after the reaction.
2. Amphiprotic salts or equal acid-salt pairs give simple pH values.

Answer: (c) 5.16

Explanation:
1. Total acetate \(=150\) mmol. At pH 5: \([A^-]/[HA]=K_a/[H^+]=1.8\), so \(HA=53.6\) and \(A^-=96.4\) mmol.
2. Reduction uses \(4\times3=12\) mmol \(\ce{H+}\), supplied by acid: \(HA=41.6\) mmol and \(A^-=108.4\) mmol.
3. \(\text{pH}=4.745+\log(108.4/41.6)=5.16\).

Teacher's Note:
1. The buffer replaces the hydrogen ion consumed by converting acid into salt.
2. Find the starting ratio from the initial pH.

Answer: (d) acid \(0.070\ \text{M}\) and salt \(0.127\ \text{M}\)

Explanation:
1. Let the acid be \(A\) mmol and salt \(S=1.82A\) mmol. Adding 1 mmol of base needs \(\dfrac{S+1}{A-1}\le1.82\times10^{0.1}=2.29\), which gives \(A\ge6.99\) mmol.
2. Adding 1 mmol of acid needs \(\dfrac{S-1}{A+1}\ge1.82\times10^{-0.1}=1.446\), which gives only \(A\ge6.53\) mmol. The base condition is stricter.
3. So \(A=7.0\) mmol and \(S=12.7\) mmol in 100 mL: 0.070 M and 0.127 M.

Teacher's Note:
1. Check both directions of pH change and use the stricter one.
2. Concentration in mmol per mL equals molarity.

Answer: (d) 3.82

Explanation:
1. \(\text{p}K_a=3.745\).
2. \(\text{pH}=3.745+\log(0.06/0.05)=3.745+0.079=3.82\).

Teacher's Note:
1. Salt exceeds acid, so the pH exceeds pKa.
2. Use the Henderson equation.

Answer: (c) 3.85

Explanation:
1. New concentrations: acid \(=0.005\ \text{M}\), salt \(=0.006\ \text{M}\).
2. Dissociation is no longer negligible: \(x(0.006+x)/(0.005-x)=1.8\times10^{-4}\) gives \(x=[H^+]=1.4\times10^{-4}\ \text{M}\).
3. \(\text{pH}=3.85\).

Teacher's Note:
1. Dilution changes the pH of a buffer slightly once the acid starts to ionise appreciably.
2. Solve the quadratic rather than the Henderson equation.

Answer: (a) \(3.73\times10^{-2}\ \text{M}\)

Explanation:
1. \(\ce{PO4^{3-} + H2O <=> HPO4^{2-} + OH-}\) has \(K_b=K_w/K_3=2.22\times10^{-2}\).
2. \(\dfrac{x^2}{0.1-x}=2.22\times10^{-2}\) gives \(x^2+0.0222x-0.00222=0\).
3. \(x=[OH^-]=3.73\times10^{-2}\ \text{M}\).

Teacher's Note:
1. Phosphate is a fairly strong base because K3 is tiny.
2. The quadratic is needed since x is not small compared with 0.1.