01. Solubility and Solubility Product Constant
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Ionic Equilibrium

229474 Consider the electrochemical reaction between $\mathrm{Ag}$ (s) and $\mathrm{Cl}_2(\mathrm{~g})$ electrodes in $1 \mathrm{~L}$ of $0.1 \mathrm{M}$ $\mathrm{KCl}$ aqueous solution. Solubility product of $\mathrm{AgCl}$ is $1.8 \times 10^{-10}$ and $\mathrm{F}=96500 \mathrm{C} / \mathrm{mol}$ At $1 \geq$ $10^{-6} \mathrm{~A}$ current, calculate the time required to start observing the $\mathrm{AgCl}$ precipitation in the galvanic cell

1 $173 \mathrm{~s}$
2 $346 \mathrm{~s}$
3 $125 \times 10^6 \mathrm{~s}$
4 $1.25 \times 10^5 \mathrm{~s}$
5 $101 \mathrm{~s}$
Kerala-CEE-2018
Ionic Equilibrium

229480 The solubility of $\mathrm{AgCl}$ is $1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$. Its solubility in 0.1 molar sodium chloride solution is

1 $1 \times 10^{-10}$
2 $1 \times 10^{-5}$
3 $1 \times 10^{-9}$
4 $1 \times 10^{-4}$
MHT CET-2007
Ionic Equilibrium

229481 In qualitative analysis the metals of group 1 can be separated from other ions by precipitating them as chloride salts. A solutions initially contains $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentrations of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium ?
$\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}$ for $\mathrm{PbCl}_2$ $=1.7 \times 10^{-5}$ )

1 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
2 $\left[\mathrm{Ag}_{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=8.5 \times 10^{-5} \mathrm{M}$
3 $\left[\mathrm{Ag}_{-}\right]=1.8 \times 10^{-9} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
4 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=1.7 \times 10^{-4} \mathrm{M}$
NEET-Mains 2011
Ionic Equilibrium

229483 The solubility product of a sparingly soluble salt $\mathrm{AX}_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles $\mathrm{L}$ ) is

1 $5.6 \times 10^{-6}$
2 $3.1 \times 10^{-4}$
3 $2 \times 10^{-4}$
4 $4 \times 10^{-4}$
NEET-2004
Ionic Equilibrium

229484 The solubility product of \(\mathrm{AgI}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-16} \mathrm{~mol}^{-1} \mathrm{~L}^{-1}\). The solubility of \(\mathrm{AgI}\) in \(10^{-4}\) \(\mathrm{N}\) solution of \(\mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) is approximately (in mol L \({ }^{-1}\) )

1 \(1.0 \times 10^{-16}\)
2 \(1.0 \times 10^{-12}\)
3 \(1.0 \times 10^{-10}\)
4 \(1.0 \times 10^{-8}\)
NEET-2003
NEET DIGITAL LIBRARY
Ionic Equilibrium

229474 Consider the electrochemical reaction between $\mathrm{Ag}$ (s) and $\mathrm{Cl}_2(\mathrm{~g})$ electrodes in $1 \mathrm{~L}$ of $0.1 \mathrm{M}$ $\mathrm{KCl}$ aqueous solution. Solubility product of $\mathrm{AgCl}$ is $1.8 \times 10^{-10}$ and $\mathrm{F}=96500 \mathrm{C} / \mathrm{mol}$ At $1 \geq$ $10^{-6} \mathrm{~A}$ current, calculate the time required to start observing the $\mathrm{AgCl}$ precipitation in the galvanic cell

1 $173 \mathrm{~s}$
2 $346 \mathrm{~s}$
3 $125 \times 10^6 \mathrm{~s}$
4 $1.25 \times 10^5 \mathrm{~s}$
5 $101 \mathrm{~s}$
Kerala-CEE-2018
Ionic Equilibrium

229480 The solubility of $\mathrm{AgCl}$ is $1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$. Its solubility in 0.1 molar sodium chloride solution is

1 $1 \times 10^{-10}$
2 $1 \times 10^{-5}$
3 $1 \times 10^{-9}$
4 $1 \times 10^{-4}$
MHT CET-2007
Ionic Equilibrium

229481 In qualitative analysis the metals of group 1 can be separated from other ions by precipitating them as chloride salts. A solutions initially contains $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentrations of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium ?
$\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}$ for $\mathrm{PbCl}_2$ $=1.7 \times 10^{-5}$ )

1 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
2 $\left[\mathrm{Ag}_{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=8.5 \times 10^{-5} \mathrm{M}$
3 $\left[\mathrm{Ag}_{-}\right]=1.8 \times 10^{-9} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
4 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=1.7 \times 10^{-4} \mathrm{M}$
NEET-Mains 2011
Ionic Equilibrium

229483 The solubility product of a sparingly soluble salt $\mathrm{AX}_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles $\mathrm{L}$ ) is

1 $5.6 \times 10^{-6}$
2 $3.1 \times 10^{-4}$
3 $2 \times 10^{-4}$
4 $4 \times 10^{-4}$
NEET-2004
Ionic Equilibrium

229484 The solubility product of \(\mathrm{AgI}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-16} \mathrm{~mol}^{-1} \mathrm{~L}^{-1}\). The solubility of \(\mathrm{AgI}\) in \(10^{-4}\) \(\mathrm{N}\) solution of \(\mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) is approximately (in mol L \({ }^{-1}\) )

1 \(1.0 \times 10^{-16}\)
2 \(1.0 \times 10^{-12}\)
3 \(1.0 \times 10^{-10}\)
4 \(1.0 \times 10^{-8}\)
NEET-2003
Join With Us for More Updates
Ionic Equilibrium

229474 Consider the electrochemical reaction between $\mathrm{Ag}$ (s) and $\mathrm{Cl}_2(\mathrm{~g})$ electrodes in $1 \mathrm{~L}$ of $0.1 \mathrm{M}$ $\mathrm{KCl}$ aqueous solution. Solubility product of $\mathrm{AgCl}$ is $1.8 \times 10^{-10}$ and $\mathrm{F}=96500 \mathrm{C} / \mathrm{mol}$ At $1 \geq$ $10^{-6} \mathrm{~A}$ current, calculate the time required to start observing the $\mathrm{AgCl}$ precipitation in the galvanic cell

1 $173 \mathrm{~s}$
2 $346 \mathrm{~s}$
3 $125 \times 10^6 \mathrm{~s}$
4 $1.25 \times 10^5 \mathrm{~s}$
5 $101 \mathrm{~s}$
Kerala-CEE-2018
Ionic Equilibrium

229480 The solubility of $\mathrm{AgCl}$ is $1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$. Its solubility in 0.1 molar sodium chloride solution is

1 $1 \times 10^{-10}$
2 $1 \times 10^{-5}$
3 $1 \times 10^{-9}$
4 $1 \times 10^{-4}$
MHT CET-2007
Ionic Equilibrium

229481 In qualitative analysis the metals of group 1 can be separated from other ions by precipitating them as chloride salts. A solutions initially contains $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentrations of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium ?
$\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}$ for $\mathrm{PbCl}_2$ $=1.7 \times 10^{-5}$ )

1 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
2 $\left[\mathrm{Ag}_{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=8.5 \times 10^{-5} \mathrm{M}$
3 $\left[\mathrm{Ag}_{-}\right]=1.8 \times 10^{-9} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
4 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=1.7 \times 10^{-4} \mathrm{M}$
NEET-Mains 2011
Ionic Equilibrium

229483 The solubility product of a sparingly soluble salt $\mathrm{AX}_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles $\mathrm{L}$ ) is

1 $5.6 \times 10^{-6}$
2 $3.1 \times 10^{-4}$
3 $2 \times 10^{-4}$
4 $4 \times 10^{-4}$
NEET-2004
Ionic Equilibrium

229484 The solubility product of \(\mathrm{AgI}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-16} \mathrm{~mol}^{-1} \mathrm{~L}^{-1}\). The solubility of \(\mathrm{AgI}\) in \(10^{-4}\) \(\mathrm{N}\) solution of \(\mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) is approximately (in mol L \({ }^{-1}\) )

1 \(1.0 \times 10^{-16}\)
2 \(1.0 \times 10^{-12}\)
3 \(1.0 \times 10^{-10}\)
4 \(1.0 \times 10^{-8}\)
NEET-2003
For More Updates join with Us
Ionic Equilibrium

229474 Consider the electrochemical reaction between $\mathrm{Ag}$ (s) and $\mathrm{Cl}_2(\mathrm{~g})$ electrodes in $1 \mathrm{~L}$ of $0.1 \mathrm{M}$ $\mathrm{KCl}$ aqueous solution. Solubility product of $\mathrm{AgCl}$ is $1.8 \times 10^{-10}$ and $\mathrm{F}=96500 \mathrm{C} / \mathrm{mol}$ At $1 \geq$ $10^{-6} \mathrm{~A}$ current, calculate the time required to start observing the $\mathrm{AgCl}$ precipitation in the galvanic cell

1 $173 \mathrm{~s}$
2 $346 \mathrm{~s}$
3 $125 \times 10^6 \mathrm{~s}$
4 $1.25 \times 10^5 \mathrm{~s}$
5 $101 \mathrm{~s}$
Kerala-CEE-2018
Ionic Equilibrium

229480 The solubility of $\mathrm{AgCl}$ is $1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$. Its solubility in 0.1 molar sodium chloride solution is

1 $1 \times 10^{-10}$
2 $1 \times 10^{-5}$
3 $1 \times 10^{-9}$
4 $1 \times 10^{-4}$
MHT CET-2007
Ionic Equilibrium

229481 In qualitative analysis the metals of group 1 can be separated from other ions by precipitating them as chloride salts. A solutions initially contains $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentrations of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium ?
$\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}$ for $\mathrm{PbCl}_2$ $=1.7 \times 10^{-5}$ )

1 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
2 $\left[\mathrm{Ag}_{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=8.5 \times 10^{-5} \mathrm{M}$
3 $\left[\mathrm{Ag}_{-}\right]=1.8 \times 10^{-9} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
4 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=1.7 \times 10^{-4} \mathrm{M}$
NEET-Mains 2011
Ionic Equilibrium

229483 The solubility product of a sparingly soluble salt $\mathrm{AX}_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles $\mathrm{L}$ ) is

1 $5.6 \times 10^{-6}$
2 $3.1 \times 10^{-4}$
3 $2 \times 10^{-4}$
4 $4 \times 10^{-4}$
NEET-2004
Ionic Equilibrium

229484 The solubility product of \(\mathrm{AgI}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-16} \mathrm{~mol}^{-1} \mathrm{~L}^{-1}\). The solubility of \(\mathrm{AgI}\) in \(10^{-4}\) \(\mathrm{N}\) solution of \(\mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) is approximately (in mol L \({ }^{-1}\) )

1 \(1.0 \times 10^{-16}\)
2 \(1.0 \times 10^{-12}\)
3 \(1.0 \times 10^{-10}\)
4 \(1.0 \times 10^{-8}\)
NEET-2003
NEET DIGITAL LIBRARY
Ionic Equilibrium

229474 Consider the electrochemical reaction between $\mathrm{Ag}$ (s) and $\mathrm{Cl}_2(\mathrm{~g})$ electrodes in $1 \mathrm{~L}$ of $0.1 \mathrm{M}$ $\mathrm{KCl}$ aqueous solution. Solubility product of $\mathrm{AgCl}$ is $1.8 \times 10^{-10}$ and $\mathrm{F}=96500 \mathrm{C} / \mathrm{mol}$ At $1 \geq$ $10^{-6} \mathrm{~A}$ current, calculate the time required to start observing the $\mathrm{AgCl}$ precipitation in the galvanic cell

1 $173 \mathrm{~s}$
2 $346 \mathrm{~s}$
3 $125 \times 10^6 \mathrm{~s}$
4 $1.25 \times 10^5 \mathrm{~s}$
5 $101 \mathrm{~s}$
Kerala-CEE-2018
Ionic Equilibrium

229480 The solubility of $\mathrm{AgCl}$ is $1 \times 10^{-5} \mathrm{~mol} / \mathrm{L}$. Its solubility in 0.1 molar sodium chloride solution is

1 $1 \times 10^{-10}$
2 $1 \times 10^{-5}$
3 $1 \times 10^{-9}$
4 $1 \times 10^{-4}$
MHT CET-2007
Ionic Equilibrium

229481 In qualitative analysis the metals of group 1 can be separated from other ions by precipitating them as chloride salts. A solutions initially contains $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ at a concentration of $0.10 \mathrm{M}$. Aqueous $\mathrm{HCl}$ is added to this solution until the $\mathrm{Cl}^{-}$concentrations of $\mathrm{Ag}^{+}$and $\mathrm{Pb}^{2+}$ be at equilibrium ?
$\left(\mathrm{K}_{\mathrm{sp}}\right.$ for $\mathrm{AgCl}=1.8 \times 10^{-10}, \mathrm{~K}_{\mathrm{sp}}$ for $\mathrm{PbCl}_2$ $=1.7 \times 10^{-5}$ )

1 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-7} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-6} \mathrm{M}$
2 $\left[\mathrm{Ag}_{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=8.5 \times 10^{-5} \mathrm{M}$
3 $\left[\mathrm{Ag}_{-}\right]=1.8 \times 10^{-9} \mathrm{M},\left[\mathrm{Pb}^{2+}\right]=1.7 \times 10^{-3} \mathrm{M}$
4 $\left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-11} \mathrm{M},\left[\mathrm{Pb}^{2-}\right]=1.7 \times 10^{-4} \mathrm{M}$
NEET-Mains 2011
Ionic Equilibrium

229483 The solubility product of a sparingly soluble salt $\mathrm{AX}_2$ is $3.2 \times 10^{-11}$. Its solubility (in moles $\mathrm{L}$ ) is

1 $5.6 \times 10^{-6}$
2 $3.1 \times 10^{-4}$
3 $2 \times 10^{-4}$
4 $4 \times 10^{-4}$
NEET-2004
Ionic Equilibrium

229484 The solubility product of \(\mathrm{AgI}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-16} \mathrm{~mol}^{-1} \mathrm{~L}^{-1}\). The solubility of \(\mathrm{AgI}\) in \(10^{-4}\) \(\mathrm{N}\) solution of \(\mathrm{KI}\) at \(25^{\circ} \mathrm{C}\) is approximately (in mol L \({ }^{-1}\) )

1 \(1.0 \times 10^{-16}\)
2 \(1.0 \times 10^{-12}\)
3 \(1.0 \times 10^{-10}\)
4 \(1.0 \times 10^{-8}\)
NEET-2003