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Ionic Equilibrium question

2025 · 23 Jan · Shift 1 · Q13
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  5. /2025 · 23 Jan · Shift 1 · Q13

Ionic Equilibrium question

2025 · 23 Jan · Shift 1 · Q13

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
Which of the following happens when NH4OH\mathrm{NH}_4 \mathrm{OH}NH4​OH is added gradually to the solution containing 1 M A2+\mathrm{A}^{2+}A2+ and 1MB3+1 \mathrm{M} \mathrm{B}^{3+}1MB3+ ions? Given : Ksp [A(OH)2]=9×10−10\mathrm{K}_{\text {sp }}\left[\mathrm{A}(\mathrm{OH})_2\right]=9 \times 10^{-10}Ksp ​[A(OH)2​]=9×10−10 and Ksp[B(OH)3]=27×10−18\mathrm{K}_{\mathrm{sp}}\left[\mathrm{B}(\mathrm{OH})_3\right]=27 \times 10^{-18}Ksp​[B(OH)3​]=27×10−18 at 298 K.
  1. A
    A(OH)2\mathrm{A}(\mathrm{OH})_2A(OH)2​ will precipitate before B(OH)3\mathrm{B}(\mathrm{OH})_3B(OH)3​
  2. B
    A(OH)2\mathrm{A}(\mathrm{OH})_2A(OH)2​ and B(OH)3\mathrm{B}(\mathrm{OH})_3B(OH)3​ will precipitate together
  3. C
    Both A(OH)2\mathrm{A}(\mathrm{OH})_2A(OH)2​ and B(OH)3\mathrm{B}(\mathrm{OH})_3B(OH)3​ do not show precipitation with NH4OH\mathrm{NH}_4 \mathrm{OH}NH4​OH
  4. D
    B(OH)3\mathrm{B}(\mathrm{OH})_3B(OH)3​ will precipitate before A(OH)2\mathrm{A}(\mathrm{OH})_2A(OH)2​
View written solutionFree

Correct answer: D

  1. Condition for precipitation

A salt starts precipitating when its ionic product becomes equal to or exceeds its solubility product.

  1. For A(OH)2\mathrm{A}(\mathrm{OH})_2A(OH)2​

Dissolution equilibrium:

A(OH)2(s)⇌A2++2OH−\mathrm{A}(\mathrm{OH})_2(s) \rightleftharpoons \mathrm{A}^{2+} + 2\mathrm{OH}^-A(OH)2​(s)⇌A2++2OH−

So,

Ksp=[A2+][OH−]2K_{sp}=[\mathrm{A}^{2+}][\mathrm{OH}^-]^2Ksp​=[A2+][OH−]2

Given:

Ksp=9×10−10,[A2+]=1 MK_{sp}=9\times 10^{-10}, \quad [\mathrm{A}^{2+}] = 1\,\mathrm{M}Ksp​=9×10−10,[A2+]=1M

At the point of first precipitation,

[OH−]2=9×10−101=9×10−10[\mathrm{OH}^-]^2 = \frac{9\times 10^{-10}}{1}=9\times 10^{-10}[OH−]2=19×10−10​=9×10−10

Thus,

[OH−]=3×10−5 M[\mathrm{OH}^-] = 3\times 10^{-5}\,\mathrm{M}[OH−]=3×10−5M
  1. For B(OH)3\mathrm{B}(\mathrm{OH})_3B(OH)3​

Dissolution equilibrium:

B(OH)3(s)⇌B3++3OH−\mathrm{B}(\mathrm{OH})_3(s) \rightleftharpoons \mathrm{B}^{3+} + 3\mathrm{OH}^-B(OH)3​(s)⇌B3++3OH−

So,

Ksp=[B3+][OH−]3K_{sp}=[\mathrm{B}^{3+}][\mathrm{OH}^-]^3Ksp​=[B3+][OH−]3

Given:

Ksp=27×10−18,[B3+]=1 MK_{sp}=27\times 10^{-18}, \quad [\mathrm{B}^{3+}] = 1\,\mathrm{M}Ksp​=27×10−18,[B3+]=1M

At the point of first precipitation,

[OH−]3=27×10−181=27×10−18[\mathrm{OH}^-]^3 = \frac{27\times 10^{-18}}{1}=27\times 10^{-18}[OH−]3=127×10−18​=27×10−18

Thus,

[OH−]=3×10−6 M[\mathrm{OH}^-] = 3\times 10^{-6}\,\mathrm{M}[OH−]=3×10−6M
  1. Compare the OH−^-− concentration required
  • For A(OH)2\mathrm{A}(\mathrm{OH})_2A(OH)2​: [OH−]=3×10−5[\mathrm{OH}^-] = 3\times 10^{-5}[OH−]=3×10−5
  • For B(OH)3\mathrm{B}(\mathrm{OH})_3B(OH)3​: [OH−]=3×10−6[\mathrm{OH}^-] = 3\times 10^{-6}[OH−]=3×10−6

Since a smaller hydroxide ion concentration is needed for B(OH)3\mathrm{B}(\mathrm{OH})_3B(OH)3​, it will precipitate first on gradual addition of NH4OH\mathrm{NH_4OH}NH4​OH.

  1. Check options
  • A: False
  • B: False
  • C: False
  • D: True

Therefore, the correct option is:

D: B(OH)3 will precipitate before A(OH)2\boxed{\text{D: } \mathrm{B}(\mathrm{OH})_3 \text{ will precipitate before } \mathrm{A}(\mathrm{OH})_2}D: B(OH)3​ will precipitate before A(OH)2​​
  1. Comparison with stored correct answer

Stored correct answer: D

My derived answer: D

They agree.

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