QBManager

CHXI07:EQUILIBRIUM

314901 $K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)$ . This means that

1 AgCl is more ionized than AgBr and

A g I

.

2 Both AgBr and

A g I

are less soluble than AgCl

3

A g I

is most soluble.

4 AgBr is more soluble than AgCl but less soluble than

A g I

.

Explanation:

Solubility $\propto K_{sp} .$

CHXI07:EQUILIBRIUM

314902 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $10^{- 31}, 10^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$F o r C u S, s o l u b i l i t y = {(1 0^{- 31})}^{1 / 2};$
$F o r A g_{2} S = {(\frac{K_{s p}}{4})}^{\frac{1}{3}} = {(\frac{1 0^{- 44}}{4})}^{\frac{1}{3}} a n d$
$f o r H g S = {(1 0^{- 54})}^{\frac{1}{2}}$

CHXI07:EQUILIBRIUM

314903 Assertion :
To precipitate the cations of fourth group in qualitative analysis, medium is made alkaline before passing $H_{2} S$ gas.
Reason :
This is done to suppress the ionisation of $H_{2} S$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$H_{2} S ⇌ 2 H^{+} + S^{2 -}$
${NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-}$
$H^{+}$ combines with ${OH}^{-}$ to form water and thus the first equilibrium shifts in the forward direction to increase the ionisation of $H_{2} S$ . So option (3) is correct.

CHXI07:EQUILIBRIUM

314904 The minimum volume of water required for 0.1 $g$ lead (II) chloride to get a saturated solution $(K_{s p}$ of $P b C l_{2} = 3.2 \times 10^{- 8};$ atomic mass of $P b = 207 u)$

1

1.798 L

2

0.36 L

3

17.95 L

4

0.18 L

Explanation:

${(K_{s p})}_{P b C_{2}} = 32 \times 10^{- 9}$
$\begin{matrix} PbC l_{2} ⇌ P b^{2 +} + 2C l^{-} \\ s 2s \end{matrix}$
$K_{sp} = [P b^{2 +}] {[C l^{-}]}^{2}$
$K_{sp} = 4 s^{2} = 32 \times 1 0^{- 9}$
$s^{3} = 8 \times 1 0^{- 9}$
$s = 2 \times 1 0^{- 3} M$
$\frac{w}{M \times w .} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$\frac{0 .1}{278} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$V_{L} = \frac{0 .1 \times 1000}{278 \times 2} = 0 .18 L$

JEE - 2018

CHXI07:EQUILIBRIUM

314905 Why only $A s^{+ 3}$ gets precipitated as $A s_{2} S_{3}$ and not ${Zn}^{2 +}$ as $ZnS$ when $H_{2} S$ is passed through an acidic solution containing ${As}^{+ 3}$ and ${Zn}^{+ 2}$ ?

1 Solubility product of

A s_{2} S_{3}

is less than that of

ZnS

2 Enough

{As}^{+ 3}

are present in acidic medium

3 Zinc salt does not ionise in acidic medium

4 Solubility product changes in presence of an acid.

CHXI07:EQUILIBRIUM

314901 $K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)$ . This means that

1 AgCl is more ionized than AgBr and

A g I

.

2 Both AgBr and

A g I

are less soluble than AgCl

3

A g I

is most soluble.

4 AgBr is more soluble than AgCl but less soluble than

A g I

.

Explanation:

Solubility $\propto K_{sp} .$

CHXI07:EQUILIBRIUM

314902 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $10^{- 31}, 10^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$F o r C u S, s o l u b i l i t y = {(1 0^{- 31})}^{1 / 2};$
$F o r A g_{2} S = {(\frac{K_{s p}}{4})}^{\frac{1}{3}} = {(\frac{1 0^{- 44}}{4})}^{\frac{1}{3}} a n d$
$f o r H g S = {(1 0^{- 54})}^{\frac{1}{2}}$

CHXI07:EQUILIBRIUM

314903 Assertion :
To precipitate the cations of fourth group in qualitative analysis, medium is made alkaline before passing $H_{2} S$ gas.
Reason :
This is done to suppress the ionisation of $H_{2} S$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$H_{2} S ⇌ 2 H^{+} + S^{2 -}$
${NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-}$
$H^{+}$ combines with ${OH}^{-}$ to form water and thus the first equilibrium shifts in the forward direction to increase the ionisation of $H_{2} S$ . So option (3) is correct.

CHXI07:EQUILIBRIUM

314904 The minimum volume of water required for 0.1 $g$ lead (II) chloride to get a saturated solution $(K_{s p}$ of $P b C l_{2} = 3.2 \times 10^{- 8};$ atomic mass of $P b = 207 u)$

1

1.798 L

2

0.36 L

3

17.95 L

4

0.18 L

Explanation:

${(K_{s p})}_{P b C_{2}} = 32 \times 10^{- 9}$
$\begin{matrix} PbC l_{2} ⇌ P b^{2 +} + 2C l^{-} \\ s 2s \end{matrix}$
$K_{sp} = [P b^{2 +}] {[C l^{-}]}^{2}$
$K_{sp} = 4 s^{2} = 32 \times 1 0^{- 9}$
$s^{3} = 8 \times 1 0^{- 9}$
$s = 2 \times 1 0^{- 3} M$
$\frac{w}{M \times w .} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$\frac{0 .1}{278} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$V_{L} = \frac{0 .1 \times 1000}{278 \times 2} = 0 .18 L$

JEE - 2018

CHXI07:EQUILIBRIUM

314905 Why only $A s^{+ 3}$ gets precipitated as $A s_{2} S_{3}$ and not ${Zn}^{2 +}$ as $ZnS$ when $H_{2} S$ is passed through an acidic solution containing ${As}^{+ 3}$ and ${Zn}^{+ 2}$ ?

1 Solubility product of

A s_{2} S_{3}

is less than that of

ZnS

2 Enough

{As}^{+ 3}

are present in acidic medium

3 Zinc salt does not ionise in acidic medium

4 Solubility product changes in presence of an acid.

CHXI07:EQUILIBRIUM

314901 $K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)$ . This means that

1 AgCl is more ionized than AgBr and

A g I

.

2 Both AgBr and

A g I

are less soluble than AgCl

3

A g I

is most soluble.

4 AgBr is more soluble than AgCl but less soluble than

A g I

.

Explanation:

Solubility $\propto K_{sp} .$

CHXI07:EQUILIBRIUM

314902 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $10^{- 31}, 10^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$F o r C u S, s o l u b i l i t y = {(1 0^{- 31})}^{1 / 2};$
$F o r A g_{2} S = {(\frac{K_{s p}}{4})}^{\frac{1}{3}} = {(\frac{1 0^{- 44}}{4})}^{\frac{1}{3}} a n d$
$f o r H g S = {(1 0^{- 54})}^{\frac{1}{2}}$

CHXI07:EQUILIBRIUM

314903 Assertion :
To precipitate the cations of fourth group in qualitative analysis, medium is made alkaline before passing $H_{2} S$ gas.
Reason :
This is done to suppress the ionisation of $H_{2} S$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$H_{2} S ⇌ 2 H^{+} + S^{2 -}$
${NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-}$
$H^{+}$ combines with ${OH}^{-}$ to form water and thus the first equilibrium shifts in the forward direction to increase the ionisation of $H_{2} S$ . So option (3) is correct.

CHXI07:EQUILIBRIUM

314904 The minimum volume of water required for 0.1 $g$ lead (II) chloride to get a saturated solution $(K_{s p}$ of $P b C l_{2} = 3.2 \times 10^{- 8};$ atomic mass of $P b = 207 u)$

1

1.798 L

2

0.36 L

3

17.95 L

4

0.18 L

Explanation:

${(K_{s p})}_{P b C_{2}} = 32 \times 10^{- 9}$
$\begin{matrix} PbC l_{2} ⇌ P b^{2 +} + 2C l^{-} \\ s 2s \end{matrix}$
$K_{sp} = [P b^{2 +}] {[C l^{-}]}^{2}$
$K_{sp} = 4 s^{2} = 32 \times 1 0^{- 9}$
$s^{3} = 8 \times 1 0^{- 9}$
$s = 2 \times 1 0^{- 3} M$
$\frac{w}{M \times w .} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$\frac{0 .1}{278} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$V_{L} = \frac{0 .1 \times 1000}{278 \times 2} = 0 .18 L$

JEE - 2018

CHXI07:EQUILIBRIUM

314905 Why only $A s^{+ 3}$ gets precipitated as $A s_{2} S_{3}$ and not ${Zn}^{2 +}$ as $ZnS$ when $H_{2} S$ is passed through an acidic solution containing ${As}^{+ 3}$ and ${Zn}^{+ 2}$ ?

1 Solubility product of

A s_{2} S_{3}

is less than that of

ZnS

2 Enough

{As}^{+ 3}

are present in acidic medium

3 Zinc salt does not ionise in acidic medium

4 Solubility product changes in presence of an acid.

CHXI07:EQUILIBRIUM

314901 $K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)$ . This means that

1 AgCl is more ionized than AgBr and

A g I

.

2 Both AgBr and

A g I

are less soluble than AgCl

3

A g I

is most soluble.

4 AgBr is more soluble than AgCl but less soluble than

A g I

.

Explanation:

Solubility $\propto K_{sp} .$

CHXI07:EQUILIBRIUM

314902 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $10^{- 31}, 10^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$F o r C u S, s o l u b i l i t y = {(1 0^{- 31})}^{1 / 2};$
$F o r A g_{2} S = {(\frac{K_{s p}}{4})}^{\frac{1}{3}} = {(\frac{1 0^{- 44}}{4})}^{\frac{1}{3}} a n d$
$f o r H g S = {(1 0^{- 54})}^{\frac{1}{2}}$

CHXI07:EQUILIBRIUM

314903 Assertion :
To precipitate the cations of fourth group in qualitative analysis, medium is made alkaline before passing $H_{2} S$ gas.
Reason :
This is done to suppress the ionisation of $H_{2} S$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$H_{2} S ⇌ 2 H^{+} + S^{2 -}$
${NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-}$
$H^{+}$ combines with ${OH}^{-}$ to form water and thus the first equilibrium shifts in the forward direction to increase the ionisation of $H_{2} S$ . So option (3) is correct.

CHXI07:EQUILIBRIUM

314904 The minimum volume of water required for 0.1 $g$ lead (II) chloride to get a saturated solution $(K_{s p}$ of $P b C l_{2} = 3.2 \times 10^{- 8};$ atomic mass of $P b = 207 u)$

1

1.798 L

2

0.36 L

3

17.95 L

4

0.18 L

Explanation:

${(K_{s p})}_{P b C_{2}} = 32 \times 10^{- 9}$
$\begin{matrix} PbC l_{2} ⇌ P b^{2 +} + 2C l^{-} \\ s 2s \end{matrix}$
$K_{sp} = [P b^{2 +}] {[C l^{-}]}^{2}$
$K_{sp} = 4 s^{2} = 32 \times 1 0^{- 9}$
$s^{3} = 8 \times 1 0^{- 9}$
$s = 2 \times 1 0^{- 3} M$
$\frac{w}{M \times w .} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$\frac{0 .1}{278} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$V_{L} = \frac{0 .1 \times 1000}{278 \times 2} = 0 .18 L$

JEE - 2018

CHXI07:EQUILIBRIUM

314905 Why only $A s^{+ 3}$ gets precipitated as $A s_{2} S_{3}$ and not ${Zn}^{2 +}$ as $ZnS$ when $H_{2} S$ is passed through an acidic solution containing ${As}^{+ 3}$ and ${Zn}^{+ 2}$ ?

1 Solubility product of

A s_{2} S_{3}

is less than that of

ZnS

2 Enough

{As}^{+ 3}

are present in acidic medium

3 Zinc salt does not ionise in acidic medium

4 Solubility product changes in presence of an acid.

CHXI07:EQUILIBRIUM

314901 $K_{sp} (AgCl) > K_{sp} (AgBr) > K_{sp} (AgI)$ . This means that

1 AgCl is more ionized than AgBr and

A g I

.

2 Both AgBr and

A g I

are less soluble than AgCl

3

A g I

is most soluble.

4 AgBr is more soluble than AgCl but less soluble than

A g I

.

Explanation:

Solubility $\propto K_{sp} .$

CHXI07:EQUILIBRIUM

314902 The $K_{sp}$ of $CuS, {Ag}_{2} S$ and $HgS$ are $10^{- 31}, 10^{- 44}$ and $10^{- 54}$ respectively. The solubility of these sulphides are in the order :

1

{Ag}_{2} S > CuS > HgS

2

AgS > HgS > CuS

3

HgS > {Ag}_{2} S > CuS

4

CuS > {Ag}_{2} S > HgS

Explanation:

$F o r C u S, s o l u b i l i t y = {(1 0^{- 31})}^{1 / 2};$
$F o r A g_{2} S = {(\frac{K_{s p}}{4})}^{\frac{1}{3}} = {(\frac{1 0^{- 44}}{4})}^{\frac{1}{3}} a n d$
$f o r H g S = {(1 0^{- 54})}^{\frac{1}{2}}$

CHXI07:EQUILIBRIUM

314903 Assertion :
To precipitate the cations of fourth group in qualitative analysis, medium is made alkaline before passing $H_{2} S$ gas.
Reason :
This is done to suppress the ionisation of $H_{2} S$ .

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but Reason is correct.

Explanation:

$H_{2} S ⇌ 2 H^{+} + S^{2 -}$
${NH}_{4} OH ⇌ {NH}_{4}^{+} + {OH}^{-}$
$H^{+}$ combines with ${OH}^{-}$ to form water and thus the first equilibrium shifts in the forward direction to increase the ionisation of $H_{2} S$ . So option (3) is correct.

CHXI07:EQUILIBRIUM

314904 The minimum volume of water required for 0.1 $g$ lead (II) chloride to get a saturated solution $(K_{s p}$ of $P b C l_{2} = 3.2 \times 10^{- 8};$ atomic mass of $P b = 207 u)$

1

1.798 L

2

0.36 L

3

17.95 L

4

0.18 L

Explanation:

${(K_{s p})}_{P b C_{2}} = 32 \times 10^{- 9}$
$\begin{matrix} PbC l_{2} ⇌ P b^{2 +} + 2C l^{-} \\ s 2s \end{matrix}$
$K_{sp} = [P b^{2 +}] {[C l^{-}]}^{2}$
$K_{sp} = 4 s^{2} = 32 \times 1 0^{- 9}$
$s^{3} = 8 \times 1 0^{- 9}$
$s = 2 \times 1 0^{- 3} M$
$\frac{w}{M \times w .} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$\frac{0 .1}{278} \times \frac{1}{V_{L}} = 2 \times 1 0^{- 3}$
$V_{L} = \frac{0 .1 \times 1000}{278 \times 2} = 0 .18 L$

JEE - 2018

CHXI07:EQUILIBRIUM

314905 Why only $A s^{+ 3}$ gets precipitated as $A s_{2} S_{3}$ and not ${Zn}^{2 +}$ as $ZnS$ when $H_{2} S$ is passed through an acidic solution containing ${As}^{+ 3}$ and ${Zn}^{+ 2}$ ?

1 Solubility product of

A s_{2} S_{3}

is less than that of

ZnS

2 Enough

{As}^{+ 3}

are present in acidic medium

3 Zinc salt does not ionise in acidic medium

4 Solubility product changes in presence of an acid.

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