01. Solubility and Solubility Product Constant
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Ionic Equilibrium

229424 The solubility product of $\mathrm{Cr}(\mathrm{OH})_3$ at $298 \mathrm{k}$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\mathrm{Cr}(\mathrm{OH})_3$ will be

1 $\left(2.22 \times 10^{-31}\right)^{1 / 4}$
2 $\left(18 \times 10^{-31}\right)^{1 / 4}$
3 $\left(4.86 \times 10^{-29}\right)^{1 / 4}$
4 $\left(18 \times 10^{-31}\right)^{1 / 2}$
JEE Main 2020
Ionic Equilibrium

229426 What is the molar solubility of $\mathrm{Al}(\mathrm{OH})_3$ in 0.2 M NaOH solution? Given that, solubility product of $\mathrm{Al}(\mathrm{OH})_3=2.4 \times 10^{-24}$

1 $3 \times 10^{-19}$
2 $12 \times 10^{-21}$
3 $3 \times 10^{-22}$
4 $12 \times 10^{-23}$
JEE Main 2019
Ionic Equilibrium

229427 If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $8 \times 10^{-12}$, the molar solubility of $\mathrm{Ag}_2 \mathrm{CO}_3$ in $0.1 \mathrm{M} \mathrm{AgNO} \mathrm{O}_3$ is

1 $8 \times 10^{-12} \mathrm{M}$
2 $8 \times 10^{-13} \mathrm{M}$
3 $8 \times 10^{-10} \mathrm{M}$
4 $8 \times 10^{-11} \mathrm{M}$
JEE Main 2019
Ionic Equilibrium

229428 An aqueous solution contains $0.10 \mathrm{M} \mathrm{H}_2 \mathrm{~S}$ and $0.20 M$ HCI. If the equilibrium constants for the formation of $\mathrm{HS}^{-}$from $\mathrm{H}_2 \mathrm{~S}$ is $1.0 \times 10^{-7}$ and that of $\mathrm{S}^{2-}$ from $\mathrm{HS}^{-}$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 $5 \times 10^{-19}$
JEE Main-2018
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Ionic Equilibrium

229424 The solubility product of $\mathrm{Cr}(\mathrm{OH})_3$ at $298 \mathrm{k}$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\mathrm{Cr}(\mathrm{OH})_3$ will be

1 $\left(2.22 \times 10^{-31}\right)^{1 / 4}$
2 $\left(18 \times 10^{-31}\right)^{1 / 4}$
3 $\left(4.86 \times 10^{-29}\right)^{1 / 4}$
4 $\left(18 \times 10^{-31}\right)^{1 / 2}$
JEE Main 2020
Ionic Equilibrium

229426 What is the molar solubility of $\mathrm{Al}(\mathrm{OH})_3$ in 0.2 M NaOH solution? Given that, solubility product of $\mathrm{Al}(\mathrm{OH})_3=2.4 \times 10^{-24}$

1 $3 \times 10^{-19}$
2 $12 \times 10^{-21}$
3 $3 \times 10^{-22}$
4 $12 \times 10^{-23}$
JEE Main 2019
Ionic Equilibrium

229427 If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $8 \times 10^{-12}$, the molar solubility of $\mathrm{Ag}_2 \mathrm{CO}_3$ in $0.1 \mathrm{M} \mathrm{AgNO} \mathrm{O}_3$ is

1 $8 \times 10^{-12} \mathrm{M}$
2 $8 \times 10^{-13} \mathrm{M}$
3 $8 \times 10^{-10} \mathrm{M}$
4 $8 \times 10^{-11} \mathrm{M}$
JEE Main 2019
Ionic Equilibrium

229428 An aqueous solution contains $0.10 \mathrm{M} \mathrm{H}_2 \mathrm{~S}$ and $0.20 M$ HCI. If the equilibrium constants for the formation of $\mathrm{HS}^{-}$from $\mathrm{H}_2 \mathrm{~S}$ is $1.0 \times 10^{-7}$ and that of $\mathrm{S}^{2-}$ from $\mathrm{HS}^{-}$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 $5 \times 10^{-19}$
JEE Main-2018
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Ionic Equilibrium

229424 The solubility product of $\mathrm{Cr}(\mathrm{OH})_3$ at $298 \mathrm{k}$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\mathrm{Cr}(\mathrm{OH})_3$ will be

1 $\left(2.22 \times 10^{-31}\right)^{1 / 4}$
2 $\left(18 \times 10^{-31}\right)^{1 / 4}$
3 $\left(4.86 \times 10^{-29}\right)^{1 / 4}$
4 $\left(18 \times 10^{-31}\right)^{1 / 2}$
JEE Main 2020
Ionic Equilibrium

229426 What is the molar solubility of $\mathrm{Al}(\mathrm{OH})_3$ in 0.2 M NaOH solution? Given that, solubility product of $\mathrm{Al}(\mathrm{OH})_3=2.4 \times 10^{-24}$

1 $3 \times 10^{-19}$
2 $12 \times 10^{-21}$
3 $3 \times 10^{-22}$
4 $12 \times 10^{-23}$
JEE Main 2019
Ionic Equilibrium

229427 If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $8 \times 10^{-12}$, the molar solubility of $\mathrm{Ag}_2 \mathrm{CO}_3$ in $0.1 \mathrm{M} \mathrm{AgNO} \mathrm{O}_3$ is

1 $8 \times 10^{-12} \mathrm{M}$
2 $8 \times 10^{-13} \mathrm{M}$
3 $8 \times 10^{-10} \mathrm{M}$
4 $8 \times 10^{-11} \mathrm{M}$
JEE Main 2019
Ionic Equilibrium

229428 An aqueous solution contains $0.10 \mathrm{M} \mathrm{H}_2 \mathrm{~S}$ and $0.20 M$ HCI. If the equilibrium constants for the formation of $\mathrm{HS}^{-}$from $\mathrm{H}_2 \mathrm{~S}$ is $1.0 \times 10^{-7}$ and that of $\mathrm{S}^{2-}$ from $\mathrm{HS}^{-}$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 $5 \times 10^{-19}$
JEE Main-2018
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Ionic Equilibrium

229424 The solubility product of $\mathrm{Cr}(\mathrm{OH})_3$ at $298 \mathrm{k}$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $\mathrm{Cr}(\mathrm{OH})_3$ will be

1 $\left(2.22 \times 10^{-31}\right)^{1 / 4}$
2 $\left(18 \times 10^{-31}\right)^{1 / 4}$
3 $\left(4.86 \times 10^{-29}\right)^{1 / 4}$
4 $\left(18 \times 10^{-31}\right)^{1 / 2}$
JEE Main 2020
Ionic Equilibrium

229426 What is the molar solubility of $\mathrm{Al}(\mathrm{OH})_3$ in 0.2 M NaOH solution? Given that, solubility product of $\mathrm{Al}(\mathrm{OH})_3=2.4 \times 10^{-24}$

1 $3 \times 10^{-19}$
2 $12 \times 10^{-21}$
3 $3 \times 10^{-22}$
4 $12 \times 10^{-23}$
JEE Main 2019
Ionic Equilibrium

229427 If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $8 \times 10^{-12}$, the molar solubility of $\mathrm{Ag}_2 \mathrm{CO}_3$ in $0.1 \mathrm{M} \mathrm{AgNO} \mathrm{O}_3$ is

1 $8 \times 10^{-12} \mathrm{M}$
2 $8 \times 10^{-13} \mathrm{M}$
3 $8 \times 10^{-10} \mathrm{M}$
4 $8 \times 10^{-11} \mathrm{M}$
JEE Main 2019
Ionic Equilibrium

229428 An aqueous solution contains $0.10 \mathrm{M} \mathrm{H}_2 \mathrm{~S}$ and $0.20 M$ HCI. If the equilibrium constants for the formation of $\mathrm{HS}^{-}$from $\mathrm{H}_2 \mathrm{~S}$ is $1.0 \times 10^{-7}$ and that of $\mathrm{S}^{2-}$ from $\mathrm{HS}^{-}$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 $5 \times 10^{-19}$
JEE Main-2018
For More Updates join with Us
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here