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Ionic Equilibrium

229463 Conductivity of a saturated solution of a sparingly soluble salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is $1.85 \times 10^{-6}$ $\mathrm{Sm}^{-1}$. Solubility product of the salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is
Given, $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{AB})=140 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$

1 $5.7 \times 10^{-12}$

2 $1.32 \times 10^{-12}$

3 $7.5 \times 10^{-12}$

4 $1.74 \times 10^{-12}$

Karnataka-CET-2014

Ionic Equilibrium

229465 $1 \mathrm{dm}^3$ solution containing $10^{-5}$ moles each of $\mathrm{Cl}^{-}$ ions and $\mathrm{CrO}_4^{2-}$ ions is treated with $10^{-4}$ moles of silver nitrate. Which one of the following observations is made?
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{Ag}_2 \mathrm{CrO}_4=4 \times 10^{-12}\right]$
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{AgCl}=1 \times 10^{-10}\right]$

1 Precipitation does not occur

2 Silver chromate gets precipitated first

3 Silver chloride gets precipitated first

4 Both silver chromate and silver chloride start precipitating simultaneously

Karnataka-CET-2009

Ionic Equilibrium

229468 Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^5 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is 0.8 . The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is

1 $4.0 \times 10^{-4}$

2 $4.0 \times 10^{-5}$

3 $5.0 \times 10^{-4}$

4 $4.0 \times 10^{-6}$

Karnataka-CET-2021

Ionic Equilibrium

229469 For which of the following sparingly soluble salt, the solubility (s) and solubility product $\left(K_{\mathrm{sp}}\right)$ are related by the expression $s=$ $\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{1 / 3} ?$

1 $\mathrm{BaSO}_4$

2 $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$

3 $\mathrm{Hg}_2 \mathrm{Cl}_2$

4 $\mathrm{Ag}_3 \mathrm{PO}_4$

5 $\mathrm{CuS}$

Explanation:

(3) $\mathrm{Hg}_2 \mathrm{Cl}_2$ ? $\mathrm{r}_{\mathrm{s}} \mathrm{Hg}_2^{2+}+\underset{2 \mathrm{~s}}{2 \mathrm{Cl}^{-}}$
$\begin{gathered}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Hg}_2^{2-}\right] \cdot\left[\mathrm{Cl}^{-}\right]^2 \\
\mathrm{~K}_{\mathrm{sp}}=(\mathrm{S}) \cdot(2 \mathrm{~S})^2 \\
\mathrm{~K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\
\mathrm{~S}=3 \sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}
\end{gathered}$
(2) $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right]^2 \\
& \mathrm{~K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(2 \mathrm{~S})^3 \\
& \mathrm{~K}_{\mathrm{sp}}=108 \mathrm{~S}^5 \\
& \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5} \\
&
\end{aligned}$
(1) $\mathrm{BaSO}_4$
The ionisation of $\mathrm{BaSO}_4$ is-
$\begin{aligned}
& \mathrm{BaSO}_4 \text { ?? } \mathrm{Ba}^{2+}+\mathrm{SO}_4{ }^{2-} \\
& \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right] \cdot\left[\mathrm{SO}_4^{-2}\right] \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S} . \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
(4) $\mathrm{Ag}_3 \mathrm{PO}_4$ ?? $3 \mathrm{Ag}^{+}+\mathrm{PO}_4^{3-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(\mathrm{~S}) \\
& \mathrm{K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\
& \mathrm{~S}=\sqrt[4]{\frac{\mathrm{K}_{\mathrm{sp}}}{27}}
\end{aligned}$
(5) $\mathrm{CuS} ?$ ? $\mathrm{Cu}^{2+}+\mathrm{S}_{\mathrm{s}}^{2-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{S}^{2-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(\mathrm{S}) \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
The solubility product and solubility of $\mathrm{Hg}_2 \mathrm{Cl}_2$ has the relation :
$\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{1 / 3}$

Kerala-CEE-2006

Ionic Equilibrium

229463 Conductivity of a saturated solution of a sparingly soluble salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is $1.85 \times 10^{-6}$ $\mathrm{Sm}^{-1}$. Solubility product of the salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is
Given, $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{AB})=140 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$

1 $5.7 \times 10^{-12}$

2 $1.32 \times 10^{-12}$

3 $7.5 \times 10^{-12}$

4 $1.74 \times 10^{-12}$

Karnataka-CET-2014

Ionic Equilibrium

229465 $1 \mathrm{dm}^3$ solution containing $10^{-5}$ moles each of $\mathrm{Cl}^{-}$ ions and $\mathrm{CrO}_4^{2-}$ ions is treated with $10^{-4}$ moles of silver nitrate. Which one of the following observations is made?
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{Ag}_2 \mathrm{CrO}_4=4 \times 10^{-12}\right]$
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{AgCl}=1 \times 10^{-10}\right]$

1 Precipitation does not occur

2 Silver chromate gets precipitated first

3 Silver chloride gets precipitated first

4 Both silver chromate and silver chloride start precipitating simultaneously

Karnataka-CET-2009

Ionic Equilibrium

229468 Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^5 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is 0.8 . The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is

1 $4.0 \times 10^{-4}$

2 $4.0 \times 10^{-5}$

3 $5.0 \times 10^{-4}$

4 $4.0 \times 10^{-6}$

Karnataka-CET-2021

Ionic Equilibrium

229469 For which of the following sparingly soluble salt, the solubility (s) and solubility product $\left(K_{\mathrm{sp}}\right)$ are related by the expression $s=$ $\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{1 / 3} ?$

1 $\mathrm{BaSO}_4$

2 $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$

3 $\mathrm{Hg}_2 \mathrm{Cl}_2$

4 $\mathrm{Ag}_3 \mathrm{PO}_4$

5 $\mathrm{CuS}$

Explanation:

(3) $\mathrm{Hg}_2 \mathrm{Cl}_2$ ? $\mathrm{r}_{\mathrm{s}} \mathrm{Hg}_2^{2+}+\underset{2 \mathrm{~s}}{2 \mathrm{Cl}^{-}}$
$\begin{gathered}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Hg}_2^{2-}\right] \cdot\left[\mathrm{Cl}^{-}\right]^2 \\
\mathrm{~K}_{\mathrm{sp}}=(\mathrm{S}) \cdot(2 \mathrm{~S})^2 \\
\mathrm{~K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\
\mathrm{~S}=3 \sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}
\end{gathered}$
(2) $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right]^2 \\
& \mathrm{~K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(2 \mathrm{~S})^3 \\
& \mathrm{~K}_{\mathrm{sp}}=108 \mathrm{~S}^5 \\
& \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5} \\
&
\end{aligned}$
(1) $\mathrm{BaSO}_4$
The ionisation of $\mathrm{BaSO}_4$ is-
$\begin{aligned}
& \mathrm{BaSO}_4 \text { ?? } \mathrm{Ba}^{2+}+\mathrm{SO}_4{ }^{2-} \\
& \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right] \cdot\left[\mathrm{SO}_4^{-2}\right] \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S} . \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
(4) $\mathrm{Ag}_3 \mathrm{PO}_4$ ?? $3 \mathrm{Ag}^{+}+\mathrm{PO}_4^{3-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(\mathrm{~S}) \\
& \mathrm{K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\
& \mathrm{~S}=\sqrt[4]{\frac{\mathrm{K}_{\mathrm{sp}}}{27}}
\end{aligned}$
(5) $\mathrm{CuS} ?$ ? $\mathrm{Cu}^{2+}+\mathrm{S}_{\mathrm{s}}^{2-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{S}^{2-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(\mathrm{S}) \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
The solubility product and solubility of $\mathrm{Hg}_2 \mathrm{Cl}_2$ has the relation :
$\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{1 / 3}$

Kerala-CEE-2006

Ionic Equilibrium

229463 Conductivity of a saturated solution of a sparingly soluble salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is $1.85 \times 10^{-6}$ $\mathrm{Sm}^{-1}$. Solubility product of the salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is
Given, $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{AB})=140 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$

1 $5.7 \times 10^{-12}$

2 $1.32 \times 10^{-12}$

3 $7.5 \times 10^{-12}$

4 $1.74 \times 10^{-12}$

Karnataka-CET-2014

Ionic Equilibrium

229465 $1 \mathrm{dm}^3$ solution containing $10^{-5}$ moles each of $\mathrm{Cl}^{-}$ ions and $\mathrm{CrO}_4^{2-}$ ions is treated with $10^{-4}$ moles of silver nitrate. Which one of the following observations is made?
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{Ag}_2 \mathrm{CrO}_4=4 \times 10^{-12}\right]$
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{AgCl}=1 \times 10^{-10}\right]$

1 Precipitation does not occur

2 Silver chromate gets precipitated first

3 Silver chloride gets precipitated first

4 Both silver chromate and silver chloride start precipitating simultaneously

Karnataka-CET-2009

Ionic Equilibrium

229468 Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^5 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is 0.8 . The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is

1 $4.0 \times 10^{-4}$

2 $4.0 \times 10^{-5}$

3 $5.0 \times 10^{-4}$

4 $4.0 \times 10^{-6}$

Karnataka-CET-2021

Ionic Equilibrium

229469 For which of the following sparingly soluble salt, the solubility (s) and solubility product $\left(K_{\mathrm{sp}}\right)$ are related by the expression $s=$ $\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{1 / 3} ?$

1 $\mathrm{BaSO}_4$

2 $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$

3 $\mathrm{Hg}_2 \mathrm{Cl}_2$

4 $\mathrm{Ag}_3 \mathrm{PO}_4$

5 $\mathrm{CuS}$

Explanation:

(3) $\mathrm{Hg}_2 \mathrm{Cl}_2$ ? $\mathrm{r}_{\mathrm{s}} \mathrm{Hg}_2^{2+}+\underset{2 \mathrm{~s}}{2 \mathrm{Cl}^{-}}$
$\begin{gathered}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Hg}_2^{2-}\right] \cdot\left[\mathrm{Cl}^{-}\right]^2 \\
\mathrm{~K}_{\mathrm{sp}}=(\mathrm{S}) \cdot(2 \mathrm{~S})^2 \\
\mathrm{~K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\
\mathrm{~S}=3 \sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}
\end{gathered}$
(2) $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right]^2 \\
& \mathrm{~K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(2 \mathrm{~S})^3 \\
& \mathrm{~K}_{\mathrm{sp}}=108 \mathrm{~S}^5 \\
& \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5} \\
&
\end{aligned}$
(1) $\mathrm{BaSO}_4$
The ionisation of $\mathrm{BaSO}_4$ is-
$\begin{aligned}
& \mathrm{BaSO}_4 \text { ?? } \mathrm{Ba}^{2+}+\mathrm{SO}_4{ }^{2-} \\
& \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right] \cdot\left[\mathrm{SO}_4^{-2}\right] \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S} . \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
(4) $\mathrm{Ag}_3 \mathrm{PO}_4$ ?? $3 \mathrm{Ag}^{+}+\mathrm{PO}_4^{3-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(\mathrm{~S}) \\
& \mathrm{K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\
& \mathrm{~S}=\sqrt[4]{\frac{\mathrm{K}_{\mathrm{sp}}}{27}}
\end{aligned}$
(5) $\mathrm{CuS} ?$ ? $\mathrm{Cu}^{2+}+\mathrm{S}_{\mathrm{s}}^{2-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{S}^{2-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(\mathrm{S}) \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
The solubility product and solubility of $\mathrm{Hg}_2 \mathrm{Cl}_2$ has the relation :
$\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{1 / 3}$

Kerala-CEE-2006

Ionic Equilibrium

229463 Conductivity of a saturated solution of a sparingly soluble salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is $1.85 \times 10^{-6}$ $\mathrm{Sm}^{-1}$. Solubility product of the salt $\mathrm{AB}$ at $298 \mathrm{~K}$ is
Given, $\Lambda_{\mathrm{m}}^{\circ}(\mathrm{AB})=140 \times 10^{-4} \mathrm{Sm}^2 \mathrm{~mol}^{-1}$

1 $5.7 \times 10^{-12}$

2 $1.32 \times 10^{-12}$

3 $7.5 \times 10^{-12}$

4 $1.74 \times 10^{-12}$

Karnataka-CET-2014

Ionic Equilibrium

229465 $1 \mathrm{dm}^3$ solution containing $10^{-5}$ moles each of $\mathrm{Cl}^{-}$ ions and $\mathrm{CrO}_4^{2-}$ ions is treated with $10^{-4}$ moles of silver nitrate. Which one of the following observations is made?
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{Ag}_2 \mathrm{CrO}_4=4 \times 10^{-12}\right]$
$\left[\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{AgCl}=1 \times 10^{-10}\right]$

1 Precipitation does not occur

2 Silver chromate gets precipitated first

3 Silver chloride gets precipitated first

4 Both silver chromate and silver chloride start precipitating simultaneously

Karnataka-CET-2009

Ionic Equilibrium

229468 Henry's law constant for the solubility of $\mathrm{N}_2$ gas in water at $298 \mathrm{~K}$ is $1.0 \times 10^5 \mathrm{~atm}$. The mole fraction of $\mathrm{N}_2$ in air is 0.8 . The number of moles of $\mathrm{N}_2$ from air dissolved in 10 moles of water at $298 \mathrm{~K}$ and $5 \mathrm{~atm}$ pressure is

1 $4.0 \times 10^{-4}$

2 $4.0 \times 10^{-5}$

3 $5.0 \times 10^{-4}$

4 $4.0 \times 10^{-6}$

Karnataka-CET-2021

Ionic Equilibrium

229469 For which of the following sparingly soluble salt, the solubility (s) and solubility product $\left(K_{\mathrm{sp}}\right)$ are related by the expression $s=$ $\left(\mathrm{K}_{\mathrm{sp}} / 4\right)^{1 / 3} ?$

1 $\mathrm{BaSO}_4$

2 $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$

3 $\mathrm{Hg}_2 \mathrm{Cl}_2$

4 $\mathrm{Ag}_3 \mathrm{PO}_4$

5 $\mathrm{CuS}$

Explanation:

(3) $\mathrm{Hg}_2 \mathrm{Cl}_2$ ? $\mathrm{r}_{\mathrm{s}} \mathrm{Hg}_2^{2+}+\underset{2 \mathrm{~s}}{2 \mathrm{Cl}^{-}}$
$\begin{gathered}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Hg}_2^{2-}\right] \cdot\left[\mathrm{Cl}^{-}\right]^2 \\
\mathrm{~K}_{\mathrm{sp}}=(\mathrm{S}) \cdot(2 \mathrm{~S})^2 \\
\mathrm{~K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\
\mathrm{~S}=3 \sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{4}}
\end{gathered}$
(2) $\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right]^2 \\
& \mathrm{~K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(2 \mathrm{~S})^3 \\
& \mathrm{~K}_{\mathrm{sp}}=108 \mathrm{~S}^5 \\
& \mathrm{~S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5} \\
&
\end{aligned}$
(1) $\mathrm{BaSO}_4$
The ionisation of $\mathrm{BaSO}_4$ is-
$\begin{aligned}
& \mathrm{BaSO}_4 \text { ?? } \mathrm{Ba}^{2+}+\mathrm{SO}_4{ }^{2-} \\
& \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ba}^{2+}\right] \cdot\left[\mathrm{SO}_4^{-2}\right] \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S} . \mathrm{S} \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
(4) $\mathrm{Ag}_3 \mathrm{PO}_4$ ?? $3 \mathrm{Ag}^{+}+\mathrm{PO}_4^{3-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^3 \cdot\left[\mathrm{PO}_4{ }^{3-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(3 \mathrm{~S})^3(\mathrm{~S}) \\
& \mathrm{K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\
& \mathrm{~S}=\sqrt[4]{\frac{\mathrm{K}_{\mathrm{sp}}}{27}}
\end{aligned}$
(5) $\mathrm{CuS} ?$ ? $\mathrm{Cu}^{2+}+\mathrm{S}_{\mathrm{s}}^{2-}$
$\begin{aligned}
& \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{S}^{2-}\right] \\
& \mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(\mathrm{S}) \\
& \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^2 \\
& \mathrm{~S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}
\end{aligned}$
The solubility product and solubility of $\mathrm{Hg}_2 \mathrm{Cl}_2$ has the relation :
$\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{4}\right)^{1 / 3}$

Kerala-CEE-2006

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