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Ionic Equilibrium

229424 The solubility product of $Cr (OH)_{3}$ at $298 k$ is $6.0 \times 10^{- 31}$ . The concentration of hydroxide ions in a saturated solution of $Cr (OH)_{3}$ will be

1

{(2.22 \times 10^{- 31})}^{1 / 4}

2

{(18 \times 10^{- 31})}^{1 / 4}

3

{(4.86 \times 10^{- 29})}^{1 / 4}

4

{(18 \times 10^{- 31})}^{1 / 2}

Explanation:

$Cr (OH)_{3}$ ? ? ${Cr}^{3 +} + 3 {OH}^{-}$
$\begin{aligned} S 3 S \\ K_{sp} = [{Cr}^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \\ 6 \times 10^{- 31} = 27 S^{4} \\ S^{4} = \frac{6 \times 10^{- 31}}{27} \\ S = {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = 3 S \\ [{OH}^{-}] = 3 \times {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = {(18 \times 10^{- 31})}^{1 / 4} \end{aligned}$

JEE Main 2020

Ionic Equilibrium

229426 What is the molar solubility of $Al (OH)_{3}$ in 0.2 M NaOH solution? Given that, solubility product of $Al (OH)_{3} = 2.4 \times 10^{- 24}$

1

3 \times 10^{- 19}

2

12 \times 10^{- 21}

3

3 \times 10^{- 22}

4

12 \times 10^{- 23}

Explanation:

$Al (OH)_{3} ◻ {Al}_{[S]}^{+ 3} + 3 {OH}_{[3 S]}^{-}$
$\begin{aligned} K_{sp} = [{Al}^{3 +}] \cdot [OH]^{3} \\ [{OH}^{-}] = 3 (S) + 0.2 m \approx 0.2 m \\ K_{sp} = S \times (0.2)^{3} \\ \begin{matrix} 2.4 \times 10^{- 24} = S \times [0.2]^{3} \end{matrix} \\ S = \frac{2.4 \times 10^{- 24}}{0.2 \times 0.2 \times 0.2} \\ S = 3 \times 10^{- 22} mol / L . \end{aligned}$

JEE Main 2019

Ionic Equilibrium

229427 If $K_{sp}$ of ${Ag}_{2} {CO}_{3}$ is $8 \times 10^{- 12}$ , the molar solubility of ${Ag}_{2} {CO}_{3}$ in $0.1 M AgNO O_{3}$ is

1

8 \times 10^{- 12} M

2

8 \times 10^{- 13} M

3

8 \times 10^{- 10} M

4

8 \times 10^{- 11} M

Explanation:

${Ag}_{2} {CO}_{3} ◻ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ \underset{0.1 M]}{{Ag}_{3}} {NO}_{CS]}^{{Ag}^{+}} + \underset{0.1 M}{{NO}_{3}^{-}} 0.1 M^{-} \end{aligned}$
[S]
Hence, total ${Ag}^{+}$ ion $= [0.1 M + S]$
$[0.1 M > S]$
$S$ is negligible.
$\begin{array}{r} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ = [0.1 + S]^{2} \cdot [S] \\ 8 \times 10^{- 12} = [0.1]^{2} \cdot [S] \\ S = \frac{8 \times 10^{- 12}}{0.1 \times 0.1} \\ S = 8 \times 10^{- 10} M \end{array}$

JEE Main 2019

Ionic Equilibrium

229428 An aqueous solution contains $0.10 M H_{2} S$ and $0.20 M$ HCI. If the equilibrium constants for the formation of ${HS}^{-}$ from $H_{2} S$ is $1.0 \times 10^{- 7}$ and that of $S^{2 -}$ from ${HS}^{-}$ ions is $1.2 \times 10^{- 13}$ then the concentration of $s^{2 -}$ ions in aqueous solution is

1

5 \times 10^{- 8}

2

3 \times 10^{- 20}

3

6 \times 10^{- 21}

4

5 \times 10^{- 19}

Explanation:

Formation of $H \overset{―}{S}$ from $H_{2} \overset{―}{S}$
$H_{2} S ? H H^{+} (aq) + {HS}^{-} (aq) - - - (i)$
from of $S^{2 -}$ from $H^{S -}$
${HS}^{-} aq ?? H^{+} (aq) + S^{2 -} (aq) - - - (ii)$
Where in- ${Ka}_{1} = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]}$
${Ka}_{2} = \frac{[H^{+}] \cdot [S^{2 -}]}{[{HS}^{- 1}]}$
Add eq ${eq}^{n}$ (i) \& (2), we get -
$H_{2} S (aq) ?$
$K = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]} \times \frac{[H^{+}] [S^{2 -}]}{[{HS}^{-}]}$
$\begin{aligned} K = \frac{{[H^{+}]}^{2} [S^{2 -}]}{[H_{2} S]} \\ = 10^{- 7} \times 1.2 \times 10^{- 13} \\ K = 1.2 \times 10^{- 20} \\ at. t = 00.10 m, 0.20 m \\ K = \frac{(0.2)^{2} (x)}{0.1} \\ 1.2 \times 10^{- 20} = \frac{0.2 \times 0.2 \times x}{0.1} \\ x = 3 \times 10^{- 20} \end{aligned}$

JEE Main-2018

Ionic Equilibrium

229424 The solubility product of $Cr (OH)_{3}$ at $298 k$ is $6.0 \times 10^{- 31}$ . The concentration of hydroxide ions in a saturated solution of $Cr (OH)_{3}$ will be

1

{(2.22 \times 10^{- 31})}^{1 / 4}

2

{(18 \times 10^{- 31})}^{1 / 4}

3

{(4.86 \times 10^{- 29})}^{1 / 4}

4

{(18 \times 10^{- 31})}^{1 / 2}

Explanation:

$Cr (OH)_{3}$ ? ? ${Cr}^{3 +} + 3 {OH}^{-}$
$\begin{aligned} S 3 S \\ K_{sp} = [{Cr}^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \\ 6 \times 10^{- 31} = 27 S^{4} \\ S^{4} = \frac{6 \times 10^{- 31}}{27} \\ S = {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = 3 S \\ [{OH}^{-}] = 3 \times {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = {(18 \times 10^{- 31})}^{1 / 4} \end{aligned}$

JEE Main 2020

Ionic Equilibrium

229426 What is the molar solubility of $Al (OH)_{3}$ in 0.2 M NaOH solution? Given that, solubility product of $Al (OH)_{3} = 2.4 \times 10^{- 24}$

1

3 \times 10^{- 19}

2

12 \times 10^{- 21}

3

3 \times 10^{- 22}

4

12 \times 10^{- 23}

Explanation:

$Al (OH)_{3} ◻ {Al}_{[S]}^{+ 3} + 3 {OH}_{[3 S]}^{-}$
$\begin{aligned} K_{sp} = [{Al}^{3 +}] \cdot [OH]^{3} \\ [{OH}^{-}] = 3 (S) + 0.2 m \approx 0.2 m \\ K_{sp} = S \times (0.2)^{3} \\ \begin{matrix} 2.4 \times 10^{- 24} = S \times [0.2]^{3} \end{matrix} \\ S = \frac{2.4 \times 10^{- 24}}{0.2 \times 0.2 \times 0.2} \\ S = 3 \times 10^{- 22} mol / L . \end{aligned}$

JEE Main 2019

Ionic Equilibrium

229427 If $K_{sp}$ of ${Ag}_{2} {CO}_{3}$ is $8 \times 10^{- 12}$ , the molar solubility of ${Ag}_{2} {CO}_{3}$ in $0.1 M AgNO O_{3}$ is

1

8 \times 10^{- 12} M

2

8 \times 10^{- 13} M

3

8 \times 10^{- 10} M

4

8 \times 10^{- 11} M

Explanation:

${Ag}_{2} {CO}_{3} ◻ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ \underset{0.1 M]}{{Ag}_{3}} {NO}_{CS]}^{{Ag}^{+}} + \underset{0.1 M}{{NO}_{3}^{-}} 0.1 M^{-} \end{aligned}$
[S]
Hence, total ${Ag}^{+}$ ion $= [0.1 M + S]$
$[0.1 M > S]$
$S$ is negligible.
$\begin{array}{r} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ = [0.1 + S]^{2} \cdot [S] \\ 8 \times 10^{- 12} = [0.1]^{2} \cdot [S] \\ S = \frac{8 \times 10^{- 12}}{0.1 \times 0.1} \\ S = 8 \times 10^{- 10} M \end{array}$

JEE Main 2019

Ionic Equilibrium

229428 An aqueous solution contains $0.10 M H_{2} S$ and $0.20 M$ HCI. If the equilibrium constants for the formation of ${HS}^{-}$ from $H_{2} S$ is $1.0 \times 10^{- 7}$ and that of $S^{2 -}$ from ${HS}^{-}$ ions is $1.2 \times 10^{- 13}$ then the concentration of $s^{2 -}$ ions in aqueous solution is

1

5 \times 10^{- 8}

2

3 \times 10^{- 20}

3

6 \times 10^{- 21}

4

5 \times 10^{- 19}

Explanation:

Formation of $H \overset{―}{S}$ from $H_{2} \overset{―}{S}$
$H_{2} S ? H H^{+} (aq) + {HS}^{-} (aq) - - - (i)$
from of $S^{2 -}$ from $H^{S -}$
${HS}^{-} aq ?? H^{+} (aq) + S^{2 -} (aq) - - - (ii)$
Where in- ${Ka}_{1} = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]}$
${Ka}_{2} = \frac{[H^{+}] \cdot [S^{2 -}]}{[{HS}^{- 1}]}$
Add eq ${eq}^{n}$ (i) \& (2), we get -
$H_{2} S (aq) ?$
$K = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]} \times \frac{[H^{+}] [S^{2 -}]}{[{HS}^{-}]}$
$\begin{aligned} K = \frac{{[H^{+}]}^{2} [S^{2 -}]}{[H_{2} S]} \\ = 10^{- 7} \times 1.2 \times 10^{- 13} \\ K = 1.2 \times 10^{- 20} \\ at. t = 00.10 m, 0.20 m \\ K = \frac{(0.2)^{2} (x)}{0.1} \\ 1.2 \times 10^{- 20} = \frac{0.2 \times 0.2 \times x}{0.1} \\ x = 3 \times 10^{- 20} \end{aligned}$

JEE Main-2018

Ionic Equilibrium

229424 The solubility product of $Cr (OH)_{3}$ at $298 k$ is $6.0 \times 10^{- 31}$ . The concentration of hydroxide ions in a saturated solution of $Cr (OH)_{3}$ will be

1

{(2.22 \times 10^{- 31})}^{1 / 4}

2

{(18 \times 10^{- 31})}^{1 / 4}

3

{(4.86 \times 10^{- 29})}^{1 / 4}

4

{(18 \times 10^{- 31})}^{1 / 2}

Explanation:

$Cr (OH)_{3}$ ? ? ${Cr}^{3 +} + 3 {OH}^{-}$
$\begin{aligned} S 3 S \\ K_{sp} = [{Cr}^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \\ 6 \times 10^{- 31} = 27 S^{4} \\ S^{4} = \frac{6 \times 10^{- 31}}{27} \\ S = {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = 3 S \\ [{OH}^{-}] = 3 \times {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = {(18 \times 10^{- 31})}^{1 / 4} \end{aligned}$

JEE Main 2020

Ionic Equilibrium

229426 What is the molar solubility of $Al (OH)_{3}$ in 0.2 M NaOH solution? Given that, solubility product of $Al (OH)_{3} = 2.4 \times 10^{- 24}$

1

3 \times 10^{- 19}

2

12 \times 10^{- 21}

3

3 \times 10^{- 22}

4

12 \times 10^{- 23}

Explanation:

$Al (OH)_{3} ◻ {Al}_{[S]}^{+ 3} + 3 {OH}_{[3 S]}^{-}$
$\begin{aligned} K_{sp} = [{Al}^{3 +}] \cdot [OH]^{3} \\ [{OH}^{-}] = 3 (S) + 0.2 m \approx 0.2 m \\ K_{sp} = S \times (0.2)^{3} \\ \begin{matrix} 2.4 \times 10^{- 24} = S \times [0.2]^{3} \end{matrix} \\ S = \frac{2.4 \times 10^{- 24}}{0.2 \times 0.2 \times 0.2} \\ S = 3 \times 10^{- 22} mol / L . \end{aligned}$

JEE Main 2019

Ionic Equilibrium

229427 If $K_{sp}$ of ${Ag}_{2} {CO}_{3}$ is $8 \times 10^{- 12}$ , the molar solubility of ${Ag}_{2} {CO}_{3}$ in $0.1 M AgNO O_{3}$ is

1

8 \times 10^{- 12} M

2

8 \times 10^{- 13} M

3

8 \times 10^{- 10} M

4

8 \times 10^{- 11} M

Explanation:

${Ag}_{2} {CO}_{3} ◻ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ \underset{0.1 M]}{{Ag}_{3}} {NO}_{CS]}^{{Ag}^{+}} + \underset{0.1 M}{{NO}_{3}^{-}} 0.1 M^{-} \end{aligned}$
[S]
Hence, total ${Ag}^{+}$ ion $= [0.1 M + S]$
$[0.1 M > S]$
$S$ is negligible.
$\begin{array}{r} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ = [0.1 + S]^{2} \cdot [S] \\ 8 \times 10^{- 12} = [0.1]^{2} \cdot [S] \\ S = \frac{8 \times 10^{- 12}}{0.1 \times 0.1} \\ S = 8 \times 10^{- 10} M \end{array}$

JEE Main 2019

Ionic Equilibrium

229428 An aqueous solution contains $0.10 M H_{2} S$ and $0.20 M$ HCI. If the equilibrium constants for the formation of ${HS}^{-}$ from $H_{2} S$ is $1.0 \times 10^{- 7}$ and that of $S^{2 -}$ from ${HS}^{-}$ ions is $1.2 \times 10^{- 13}$ then the concentration of $s^{2 -}$ ions in aqueous solution is

1

5 \times 10^{- 8}

2

3 \times 10^{- 20}

3

6 \times 10^{- 21}

4

5 \times 10^{- 19}

Explanation:

Formation of $H \overset{―}{S}$ from $H_{2} \overset{―}{S}$
$H_{2} S ? H H^{+} (aq) + {HS}^{-} (aq) - - - (i)$
from of $S^{2 -}$ from $H^{S -}$
${HS}^{-} aq ?? H^{+} (aq) + S^{2 -} (aq) - - - (ii)$
Where in- ${Ka}_{1} = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]}$
${Ka}_{2} = \frac{[H^{+}] \cdot [S^{2 -}]}{[{HS}^{- 1}]}$
Add eq ${eq}^{n}$ (i) \& (2), we get -
$H_{2} S (aq) ?$
$K = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]} \times \frac{[H^{+}] [S^{2 -}]}{[{HS}^{-}]}$
$\begin{aligned} K = \frac{{[H^{+}]}^{2} [S^{2 -}]}{[H_{2} S]} \\ = 10^{- 7} \times 1.2 \times 10^{- 13} \\ K = 1.2 \times 10^{- 20} \\ at. t = 00.10 m, 0.20 m \\ K = \frac{(0.2)^{2} (x)}{0.1} \\ 1.2 \times 10^{- 20} = \frac{0.2 \times 0.2 \times x}{0.1} \\ x = 3 \times 10^{- 20} \end{aligned}$

JEE Main-2018

Ionic Equilibrium

229424 The solubility product of $Cr (OH)_{3}$ at $298 k$ is $6.0 \times 10^{- 31}$ . The concentration of hydroxide ions in a saturated solution of $Cr (OH)_{3}$ will be

1

{(2.22 \times 10^{- 31})}^{1 / 4}

2

{(18 \times 10^{- 31})}^{1 / 4}

3

{(4.86 \times 10^{- 29})}^{1 / 4}

4

{(18 \times 10^{- 31})}^{1 / 2}

Explanation:

$Cr (OH)_{3}$ ? ? ${Cr}^{3 +} + 3 {OH}^{-}$
$\begin{aligned} S 3 S \\ K_{sp} = [{Cr}^{3 +}] \cdot {[{OH}^{-}]}^{3} \\ = [S] \cdot [3 S]^{3} \\ K_{sp} = 27 S^{4} \\ 6 \times 10^{- 31} = 27 S^{4} \\ S^{4} = \frac{6 \times 10^{- 31}}{27} \\ S = {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = 3 S \\ [{OH}^{-}] = 3 \times {(\frac{6 \times 10^{- 31}}{27})}^{1 / 4} \\ [{OH}^{-}] = {(18 \times 10^{- 31})}^{1 / 4} \end{aligned}$

JEE Main 2020

Ionic Equilibrium

229426 What is the molar solubility of $Al (OH)_{3}$ in 0.2 M NaOH solution? Given that, solubility product of $Al (OH)_{3} = 2.4 \times 10^{- 24}$

1

3 \times 10^{- 19}

2

12 \times 10^{- 21}

3

3 \times 10^{- 22}

4

12 \times 10^{- 23}

Explanation:

$Al (OH)_{3} ◻ {Al}_{[S]}^{+ 3} + 3 {OH}_{[3 S]}^{-}$
$\begin{aligned} K_{sp} = [{Al}^{3 +}] \cdot [OH]^{3} \\ [{OH}^{-}] = 3 (S) + 0.2 m \approx 0.2 m \\ K_{sp} = S \times (0.2)^{3} \\ \begin{matrix} 2.4 \times 10^{- 24} = S \times [0.2]^{3} \end{matrix} \\ S = \frac{2.4 \times 10^{- 24}}{0.2 \times 0.2 \times 0.2} \\ S = 3 \times 10^{- 22} mol / L . \end{aligned}$

JEE Main 2019

Ionic Equilibrium

229427 If $K_{sp}$ of ${Ag}_{2} {CO}_{3}$ is $8 \times 10^{- 12}$ , the molar solubility of ${Ag}_{2} {CO}_{3}$ in $0.1 M AgNO O_{3}$ is

1

8 \times 10^{- 12} M

2

8 \times 10^{- 13} M

3

8 \times 10^{- 10} M

4

8 \times 10^{- 11} M

Explanation:

${Ag}_{2} {CO}_{3} ◻ 2 {Ag}^{+} + {CO}_{3}^{2 -}$
$\begin{aligned} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ \underset{0.1 M]}{{Ag}_{3}} {NO}_{CS]}^{{Ag}^{+}} + \underset{0.1 M}{{NO}_{3}^{-}} 0.1 M^{-} \end{aligned}$
[S]
Hence, total ${Ag}^{+}$ ion $= [0.1 M + S]$
$[0.1 M > S]$
$S$ is negligible.
$\begin{array}{r} K_{sp} = [{Ag}^{+}] \cdot [{CO}_{3}^{2 -}] \\ = [0.1 + S]^{2} \cdot [S] \\ 8 \times 10^{- 12} = [0.1]^{2} \cdot [S] \\ S = \frac{8 \times 10^{- 12}}{0.1 \times 0.1} \\ S = 8 \times 10^{- 10} M \end{array}$

JEE Main 2019

Ionic Equilibrium

229428 An aqueous solution contains $0.10 M H_{2} S$ and $0.20 M$ HCI. If the equilibrium constants for the formation of ${HS}^{-}$ from $H_{2} S$ is $1.0 \times 10^{- 7}$ and that of $S^{2 -}$ from ${HS}^{-}$ ions is $1.2 \times 10^{- 13}$ then the concentration of $s^{2 -}$ ions in aqueous solution is

1

5 \times 10^{- 8}

2

3 \times 10^{- 20}

3

6 \times 10^{- 21}

4

5 \times 10^{- 19}

Explanation:

Formation of $H \overset{―}{S}$ from $H_{2} \overset{―}{S}$
$H_{2} S ? H H^{+} (aq) + {HS}^{-} (aq) - - - (i)$
from of $S^{2 -}$ from $H^{S -}$
${HS}^{-} aq ?? H^{+} (aq) + S^{2 -} (aq) - - - (ii)$
Where in- ${Ka}_{1} = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]}$
${Ka}_{2} = \frac{[H^{+}] \cdot [S^{2 -}]}{[{HS}^{- 1}]}$
Add eq ${eq}^{n}$ (i) \& (2), we get -
$H_{2} S (aq) ?$
$K = \frac{[H^{+}] [{HS}^{-}]}{[H_{2} S]} \times \frac{[H^{+}] [S^{2 -}]}{[{HS}^{-}]}$
$\begin{aligned} K = \frac{{[H^{+}]}^{2} [S^{2 -}]}{[H_{2} S]} \\ = 10^{- 7} \times 1.2 \times 10^{- 13} \\ K = 1.2 \times 10^{- 20} \\ at. t = 00.10 m, 0.20 m \\ K = \frac{(0.2)^{2} (x)}{0.1} \\ 1.2 \times 10^{- 20} = \frac{0.2 \times 0.2 \times x}{0.1} \\ x = 3 \times 10^{- 20} \end{aligned}$

JEE Main-2018

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