{ The correct order of solubility of the given salts in water at } 298 {~K} { is } Salt {K}_{ {sp }} at {2 9 8} K AgBr 5.0 × 10^{-13} {Zn}({OH})₂ 1.0 × 10^{-15} {Hg}₂ {Cl}₂ 1.3 × 10^{-18} [NEET 2026 Re-exam]

Correct Answer is : Zn(OH)2>AgBr>Hg2Cl2{Zn}({OH})₂>{AgBr}>{Hg}₂ {Cl}₂Zn(OH)2>AgBr>Hg2Cl2

{Zn}({OH})₂>{AgBr}>{Hg}₂ {Cl}₂

{Hg}₂ {Cl}₂>{Zn}({OH})₂>{AgBr}

{AgBr}>{Zn}({OH})₂>{Hg}₂ {Cl}₂

{Hg}₂ {Cl}₂>{AgBr}>{Zn}({OH})₂

Correct Answer is : Zn(OH)2>AgBr>Hg2Cl2{Zn}({OH})₂>{AgBr}>{Hg}₂ {Cl}₂Zn(OH)2>AgBr>Hg2Cl2

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Class 11 NEET Chemistry Equilibrium Questions Part 2

Section: Chemistry

Question : 1 of 33

Marks: +1, -0

\text { The correct order of solubility of the given salts in water at } 298 \mathrm{~K} \text { is }

Solution:

Concept:

Solubility (S) is calculated from

K_{sp}

using the dissociation stoichiometry. Compare S values directly.

Explanation:

For

\mathrm{AgBr}

(1:1 salt):

K_{sp} = S^2 = 5.0 \times 10^{-13}

S = \sqrt{5.0 \times 10^{-13}} = 7.07 \times 10^{-7} \, \text{mol/L}

For

\mathrm{Zn(OH)_2}

(1:2 salt):

K_{sp} = 4S^3 = 1.0 \times 10^{-15}

S = \sqrt[3]{\frac{1.0 \times 10^{-15}}{4}} = 6.3 \times 10^{-6} \, \text{mol/L}

For

\mathrm{Hg_2Cl_2}

(1:2 salt):

K_{sp} = 4S^3 = 1.3 \times 10^{-18}

S = \sqrt[3]{\frac{1.3 \times 10^{-18}}{4}} = 6.9 \times 10^{-7} \, \text{mol/L}

Comparing the molar solubilities:

6.3 \times 10^{-6} > 7.07 \times 10^{-7} > 6.9 \times 10^{-7}

Thus order is

\mathrm{Zn(OH)_2} > \mathrm{AgBr} > \mathrm{Hg_2Cl_2}

Answer:

\mathrm{Zn(OH)_2}>\mathrm{AgBr}>\mathrm{Hg_2Cl_2}

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