The solubilities of AgBr in water, in 0.01M CaBr_ 2 ,0.01MKBr and in 0.05M AgNO_ 3 are S _ 1 ,S_ 2 ,S_ 3 and S _ 4 respectively. Then the order of solubilities is

Correct answer: B. S _ 1 > S _ 3 > S _ 2 > S _ 4 Solubility of AgBr in 0.01M CaBr_ 2 ( S _ 2 ) ......(i) CaBr _ 2 is a strong electrolyte and dissociates completely to give common ion Br ^ − CaBr _ 2 _ 0.01 M → Ca ^ 2+ _ 0.01M + 2Br ^ − _ 0.02M AgBr(s) ⇌ Ag ^ + (aq)+Br ^ − (aq); K _ sp S _ 2 0 .02+S_ 2 ⇒ S _ 2 ×0.02 = K _ sp 0.02 or S _ 2 = 50 K _ sp Similarly, KBr gives common ion Br ^ − And AgNO _ 3 gives common ion Ag ^ + and we can calculate S _ 3 = 100 K_ sp ......(iii) And S _ 4 = 20 K _ sp ......(iv) respectively From (i), (ii), (iii) and (iv) S _ 1 > S _ 3 > S _ 2 > S _ 4 (Note: K _ sp >100 K _ sp ∵ K _ sp order - 10 ^ - 12 )

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Chemistry
Chapter: Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

The solubilities of AgBr in water, in 0.01M CaBr_ 2 ,0.01MKBr and...

The solubilities of

\text{AgBr}

in water, in

\text{0}\text{.01}\text{M}{\text{CaBr}_{2} \text{,}\text{0}\text{.01}\text{M}\text{KBr}

and in

\text{0}\text{.05}\text{M}{\text{AgNO}_{3}

are

{S}_{1} {\text{,S}_{2} {\text{,S}_{3}

and

{S}_{4}

respectively. Then the order of solubilities is

Options

A ${S}_{1} \text{>}{S}_{2} \text{>}{S}_{3} \text{>}{S}_{4}$
B ${S}_{1} \text{> }{S}_{3} \text{> }{S}_{2} \text{> }{S}_{4}$
C ${S}_{2} \text{>}{S}_{1} \text{>}{S}_{3} \text{>}{S}_{4}$
D ${S}_{4} \text{>}{S}_{3} \text{>}{S}_{1} \text{>}{S}_{2}$

Correct answer

B. ${S}_{1} \text{> }{S}_{3} \text{> }{S}_{2} \text{> }{S}_{4}$

Step-by-step solution

Solubility of

\text{AgBr}

\text{0}\text{.01}\text{M}{\text{CaBr}_{2} \left({S}_{2}\right) ......(i)<br /><br />

{

\text{CaBr}

}_{2}

is a strong electrolyte and dissociates completely to give common ion

{

\text{Br}

}^{−}

<br /><br />

\mathop{{

\text{CaBr}

}_{2}}_{0.01

\text{M}

} \to \mathop{{

\text{Ca}

}^{

\text{2+}

}}_{

\text{0}\text{.01}\text{M}

} + \mathop{{

\text{2Br}

}^{−}}_{

\text{0}\text{.02}\text{M}

}

<br /><br />\text{AgBr(s)}

⇌ {

\text{Ag}

}^{+} {

\text{(aq)+Br}

}^{−}

\text{(aq);}{K}_{\text{sp}}<br /><br />

{S}_{2}

\text{0}{\text{.02+S}_{2}

⇒ {S}_{2}

\times0.02 = {K}_{\text{sp}}

<br /><br /> 0.02

{S}_{2} = 50{K}_{\text{sp}}

<br /><br />Similarly, \text{KBr} gives common ion

{

\text{Br}

}^{−}

<br /><br />And

{

\text{AgNO}

}_{3}

gives common ion

{

\text{Ag}

}^{+}

and we can calculate<br /><br />

{S}_{3}

\text{=}{\text{100 K}_{\text{sp}} ......(iii)<br /><br />And

{S}_{4} = 20{K}_{

\text{sp}

}

......(iv) respectively<br /><br />From (i), (ii), (iii) and (iv)<br /><br />

{S}_{1}

\text{>}{S}_{3} \text{>}{S}_{2} \text{>}{S}_{4}<br /><br />(Note: \sqrt{{K}_{\text{sp}}} \text{>}\text{100}{K}_{\text{sp}}

∵ {K}_{

\text{sp}

}

order

- {10}^{- 12}

)

Practice Chemical Equilibrium on Quantrex Academy →

More from Chemical Equilibrium

Consider the following reactions in which all the reactants and products are present in… JEE Main 2026 Solid carbon, CaO and CaCO _3 are mixed and allowed to attain equilibrium at T K.… JEE Main 2026 In a closed flask at 600 K, one mole of X _2 Y _4 (g) attains equilibrium as given below… JEE Main 2026 One mole each of He and A(g) are taken in a 10 L closed flask and heated to 400 K to… JEE Main 2026 The reaction A(g) leftharpoons B(g) + C(g) was initiated with the amount ' a ' of A(g) .… JEE Main 2026 The values of pressure equilibrium constant recorded at different temperatures for the… JEE Main 2026 For the following reaction at 50° C and at 2 atm pressure, 2N_2O_5(g) leftharpoons… JEE Main 2026 At T(K) , the equilibrium constant of A_2(g) + B_2(g) leftharpoons C(g) is 2.7 × 10^ -5 .… JEE Main 2026 Full Chemical Equilibrium list All NTA Abhyas JEE Main PYQs