JEE Chemistry Ionic Equilibrium MCQs Set 02

Answer: (c) 11.5

Explanation:
1. \( pK_a \) of HCN \( =14-4.7=9.3 \).
2. For a salt of a weak acid and strong base, \( pH=7+\tfrac12(pK_a+\log C)=7+\tfrac12(9.3-0.30)\approx11.5 \).

Teacher's Note:
1. Use the hydrolysis formula for the salt of a weak acid and strong base.

Answer: (b) \( 10^{10} \)

Explanation:
1. \( K=\frac{[\ce{A-}]}{[\ce{HA}][\ce{OH-}]}=\frac{K_a}{K_w} \).
2. \( K=\frac{10^{-4}}{10^{-14}}=10^{10} \).

Teacher's Note:
1. The reaction is the reverse of the hydrolysis of A-, so K = 1/Kh = Ka/Kw.

Answer: (c) a and c

Explanation:
1. \( \ce{FeCl3} \) (weak base, strong acid) is acidic, so (a) is wrong.
2. Ammonium acetate has \( K_a\approx K_b \), so it is almost neutral, not acidic, so (c) is wrong.
3. (b) and (d) are correct.

Teacher's Note:
1. Check the parent acid and base of every salt.

Answer: (d) \( 7.559\times10^{-6} \)

Explanation:
1. \( h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{K_w}{K_aC}} \).
2. \( h=\sqrt{\frac{10^{-14}}{3.5\times10^{-3}\times0.05}}=\sqrt{5.71\times10^{-11}}=7.559\times10^{-6} \).

Teacher's Note:
1. Kh = Kw/Ka for the salt of a weak acid.

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

Explanation:
1. \( K_h=\frac{10^{-14}}{1.4\times10^{-9}}=7.14\times10^{-6} \).
2. \( h=\sqrt{K_h/C}=\sqrt{7.14\times10^{-6}/0.01}=2.7\times10^{-2} \).

Teacher's Note:
1. KCN is a salt of a weak acid, so use Kh = Kw/Ka.

Answer: (c) \( 1.68\times10^{-12} \)

Explanation:
1. \( [\ce{CrO4^2-}]=\frac12(1.5\times10^{-4})=0.75\times10^{-4} \) M.
2. \( K_{sp}=[\ce{Ag+}]^2[\ce{CrO4^2-}]=(1.5\times10^{-4})^2(0.75\times10^{-4})\approx1.69\times10^{-12} \).

Teacher's Note:
1. Remember that the silver ion concentration is twice the solubility.

Answer: (a) 0.0064 g

Explanation:
1. \( s=\sqrt{K_{sp}}=\sqrt{2.5\times10^{-9}}=5\times10^{-5} \) mol/L.
2. Mass \( =5\times10^{-5}\times128=6.4\times10^{-3} \) g.

Teacher's Note:
1. For a 1:1 salt, solubility is the square root of Ksp.

Answer: (c) \( 1\times10^{-10} \)

Explanation:
1. Solubility \( =1.43\times10^{-3} \) g/L \( =\frac{1.43\times10^{-3}}{143}=10^{-5} \) mol/L.
2. \( K_{sp}=s^2=10^{-10} \).

Teacher's Note:
1. Convert g/100 mL to mol/L before squaring.

Answer: (d) \( 2916S^8 \)

Explanation:
1. One formula unit gives 3 \( \ce{Li+} \), 3 \( \ce{Na+} \) and 2 \( \ce{AlF6^3-} \).
2. \( K_{sp}=(3S)^3(3S)^3(2S)^2=27\times27\times4\,S^8=2916S^8 \).

Teacher's Note:
1. Raise each ion concentration to the power of its coefficient.

Answer: (a) \( 4.9\times10^{-3} \)

Explanation:
1. Molar mass of \( \ce{CaCO3} \) is 100, so \( s=\frac{7.0}{100}=0.07 \) M.
2. \( K_{sp}=s^2=4.9\times10^{-3} \).

Teacher's Note:
1. The residue mass per litre gives the solubility directly.

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

Explanation:
1. Carbonate or silver ions already present reduce solubility (common ion effect).
2. Ammonia forms the soluble complex \( \ce{[Ag(NH3)2]+} \), removing \( \ce{Ag+} \) and dissolving more salt.

Teacher's Note:
1. Complex formation increases solubility.

Answer: (d) \( \ce{Ag2S} \) > CuS > HgS

Explanation:
1. CuS: \( s=\sqrt{10^{-37}}\approx3\times10^{-19} \) M.
2. \( \ce{Ag2S} \): \( s=(10^{-44}/4)^{1/3}\approx1.4\times10^{-15} \) M.
3. HgS: \( s=\sqrt{10^{-54}}=10^{-27} \) M.
4. So \( \ce{Ag2S} \)> CuS > HgS.

Teacher's Note:
1. Salts of different type cannot be compared by Ksp alone; calculate s.

Answer: (d) \( 4\times10^{-4} \) M

Explanation:
1. \( K_{sp}=4s^3=4(0.01)^3=4\times10^{-6} \).
2. In 0.1 M NaCl: \( s=\frac{K_{sp}}{(0.1)^2}=4\times10^{-4} \) M.

Teacher's Note:
1. The common chloride ion governs the solubility in NaCl.

Answer: (b) \( 2.89\times10^{-2} \) M

Explanation:
1. \( K_{sp}=4s^3 \Rightarrow s=(3\times10^{-6})^{1/3}=1.44\times10^{-2} \) M.
2. \( [\ce{M+}]=2s=2.89\times10^{-2} \) M.

Teacher's Note:
1. Do not forget the factor 2 for the cation.

Answer: (d) \( \ce{PQ2} \), \( \ce{PQ3} \), PQ

Explanation:
1. PQ: \( s=\sqrt{4\times10^{-20}}=2\times10^{-10} \).
2. \( \ce{PQ2} \): \( s=(3.2\times10^{-14}/4)^{1/3}=2\times10^{-5} \).
3. \( \ce{PQ3} \): \( s=(2.7\times10^{-35}/27)^{1/4}=10^{-9} \).
4. Order: \( \ce{PQ2} \) > \( \ce{PQ3} \) > PQ.

Teacher's Note:
1. Compute each solubility with the correct Ksp formula.

Answer: (c) 8.47

Explanation:
1. \( K_{sp}=27s^4 \Rightarrow s=(1.04\times10^{-24})^{1/4}\approx1.0\times10^{-6} \) M.
2. \( [\ce{OH-}]=3s\approx3.0\times10^{-6} \), so pOH = 5.52.
3. pH = 14 - 5.52 = 8.48, close to 8.47.

Teacher's Note:
1. Hydroxide concentration is three times the solubility.

Answer: (c) s(\( \ce{AgBrO3} \)) < s(\( \ce{Ag2SO4} \))

Explanation:
1. \( \ce{AgBrO3} \): \( s=\sqrt{5.5\times10^{-5}}=7.4\times10^{-3} \) M.
2. \( \ce{Ag2SO4} \): \( s=(2\times10^{-5}/4)^{1/3}=1.7\times10^{-2} \) M.

Teacher's Note:
1. Different salt types need explicit solubility calculation.

Answer: (a) 9

Explanation:
1. \( [\ce{OH-}]=\sqrt{\frac{10^{-11}}{0.1}}=10^{-5} \) M.
2. pOH = 5, so pH = 9.

Teacher's Note:
1. Precipitation starts when the ionic product equals Ksp.

Answer: (b) CuS precipitates

Explanation:
1. Ionic product \( =0.01\times8.1\times10^{-21}=8.1\times10^{-23} \).
2. This is less than \( K_{sp} \) of ZnS (\( 3\times10^{-22} \)), so ZnS does not precipitate.
3. It is far greater than \( K_{sp} \) of CuS, so CuS precipitates.

Teacher's Note:
1. Compare the ionic product with Ksp for each salt.

Answer: (b) No precipitate forms

Explanation:
1. pH 9 gives \( [\ce{OH-}]=10^{-5} \) M.
2. Ionic product \( =10^{-3}\times(10^{-5})^2=10^{-13} \), which is below \( 8.9\times10^{-12} \).
3. So no precipitate forms.

Teacher's Note:
1. Q less than Ksp means the solution is unsaturated.

Answer: (d) \( 3.0\times10^{-5} \) M

Explanation:
1. \( [\ce{OH-}]^2=\frac{9\times10^{-12}}{0.010}=9\times10^{-10} \).
2. \( [\ce{OH-}]=3.0\times10^{-5} \) M.

Teacher's Note:
1. Use Ksp = [Mg2+][OH-]^2.

Answer: (b) \( 2.4\times10^{-7} \) M

Explanation:
1. AgCl starts when \( [\ce{Ag+}]=\frac{10^{-10}}{0.06}=1.67\times10^{-9} \) M.
2. At this \( [\ce{Ag+}] \), \( [\ce{I-}]=\frac{4\times10^{-16}}{1.67\times10^{-9}}=2.4\times10^{-7} \) M.

Teacher's Note:
1. AgI is already precipitating, so iodide is limited by the common silver ion.