229502
If the solubility of $\mathrm{BaSO}_4$ (mol. wt. 233) is 2.33 $\times 10^{-4} \mathrm{~g} / 100 \mathrm{ml}$, then the solubility product of $\mathrm{BaSO}_4$ is
229509
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CaSO}_4$ is $4 \times 10^{-12}$. $\mathrm{CaSO}_4$ is precipitated on mixing equal volumes of the following solutions.
Super Saturated Solution- It is defined as precipitate, which form by the mixture of two solution, if the product of ions is greater than product of solubility, then it is called super saturated solution. $\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{SO}_4^2\right]>\mathrm{K}_{5 \mathrm{p}}$
UPTU/UPSEE-2015
Ionic Equilibrium
229511
The solubility of AgI in NaI solution is less than that in pure water because
1 AgI forms complex with NaI
2 Of common ion effect
3 Solubility product of AgI is less then that of $\mathrm{NaI}$
4 The temperature of the solution decreases
Explanation:
The solubility of AgI in NaI Solution is less than in pure water because of common ion effect. $\mathrm{AgI} \square \mathrm{Ag}^{+}+\mathrm{I}^{-}$ $\mathrm{NaI} \square \mathrm{Na}^{+}+\mathrm{I}^{-}$
UPTU/UPSEE-2014
Ionic Equilibrium
229513
At a certain temperature, the solubility of the salt, $M_m \mathbf{A}_{\mathrm{n}}$ in water is $s \mathrm{~mol} / \mathrm{L}$. The solubility product of the salt is
let us consider solubility of any general salt $\mathrm{M}_{\mathrm{m}} \mathrm{A}_{\mathrm{n}-}$ $\mathrm{M}_m \mathrm{~A}_n \text { ?? }\left[\mathrm{mM}^{\mathrm{n}}\right]\left[\mathrm{nA}^{m-}\right]$ The solution which contains maximam possible amount of solute in it are called saturated solution for saturated solution, at equilibrium. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{M}^{\mathrm{n}+}\right]^{\mathrm{m}}\left[\mathrm{A}^{\mathrm{m}+}\right]^{\mathrm{n}} \\ & =[\mathrm{ms}]^{\mathrm{m}}[\mathrm{ns}]^{\mathrm{n}} \\ \mathrm{K}_{\mathrm{sp}} & =\mathrm{m}^{\mathrm{m}} \mathrm{n}^{\mathrm{n}}[\mathrm{S}]^{\mathrm{m}-\mathrm{n}} \end{aligned}$
229502
If the solubility of $\mathrm{BaSO}_4$ (mol. wt. 233) is 2.33 $\times 10^{-4} \mathrm{~g} / 100 \mathrm{ml}$, then the solubility product of $\mathrm{BaSO}_4$ is
229509
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CaSO}_4$ is $4 \times 10^{-12}$. $\mathrm{CaSO}_4$ is precipitated on mixing equal volumes of the following solutions.
Super Saturated Solution- It is defined as precipitate, which form by the mixture of two solution, if the product of ions is greater than product of solubility, then it is called super saturated solution. $\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{SO}_4^2\right]>\mathrm{K}_{5 \mathrm{p}}$
UPTU/UPSEE-2015
Ionic Equilibrium
229511
The solubility of AgI in NaI solution is less than that in pure water because
1 AgI forms complex with NaI
2 Of common ion effect
3 Solubility product of AgI is less then that of $\mathrm{NaI}$
4 The temperature of the solution decreases
Explanation:
The solubility of AgI in NaI Solution is less than in pure water because of common ion effect. $\mathrm{AgI} \square \mathrm{Ag}^{+}+\mathrm{I}^{-}$ $\mathrm{NaI} \square \mathrm{Na}^{+}+\mathrm{I}^{-}$
UPTU/UPSEE-2014
Ionic Equilibrium
229513
At a certain temperature, the solubility of the salt, $M_m \mathbf{A}_{\mathrm{n}}$ in water is $s \mathrm{~mol} / \mathrm{L}$. The solubility product of the salt is
let us consider solubility of any general salt $\mathrm{M}_{\mathrm{m}} \mathrm{A}_{\mathrm{n}-}$ $\mathrm{M}_m \mathrm{~A}_n \text { ?? }\left[\mathrm{mM}^{\mathrm{n}}\right]\left[\mathrm{nA}^{m-}\right]$ The solution which contains maximam possible amount of solute in it are called saturated solution for saturated solution, at equilibrium. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{M}^{\mathrm{n}+}\right]^{\mathrm{m}}\left[\mathrm{A}^{\mathrm{m}+}\right]^{\mathrm{n}} \\ & =[\mathrm{ms}]^{\mathrm{m}}[\mathrm{ns}]^{\mathrm{n}} \\ \mathrm{K}_{\mathrm{sp}} & =\mathrm{m}^{\mathrm{m}} \mathrm{n}^{\mathrm{n}}[\mathrm{S}]^{\mathrm{m}-\mathrm{n}} \end{aligned}$
229502
If the solubility of $\mathrm{BaSO}_4$ (mol. wt. 233) is 2.33 $\times 10^{-4} \mathrm{~g} / 100 \mathrm{ml}$, then the solubility product of $\mathrm{BaSO}_4$ is
229509
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CaSO}_4$ is $4 \times 10^{-12}$. $\mathrm{CaSO}_4$ is precipitated on mixing equal volumes of the following solutions.
Super Saturated Solution- It is defined as precipitate, which form by the mixture of two solution, if the product of ions is greater than product of solubility, then it is called super saturated solution. $\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{SO}_4^2\right]>\mathrm{K}_{5 \mathrm{p}}$
UPTU/UPSEE-2015
Ionic Equilibrium
229511
The solubility of AgI in NaI solution is less than that in pure water because
1 AgI forms complex with NaI
2 Of common ion effect
3 Solubility product of AgI is less then that of $\mathrm{NaI}$
4 The temperature of the solution decreases
Explanation:
The solubility of AgI in NaI Solution is less than in pure water because of common ion effect. $\mathrm{AgI} \square \mathrm{Ag}^{+}+\mathrm{I}^{-}$ $\mathrm{NaI} \square \mathrm{Na}^{+}+\mathrm{I}^{-}$
UPTU/UPSEE-2014
Ionic Equilibrium
229513
At a certain temperature, the solubility of the salt, $M_m \mathbf{A}_{\mathrm{n}}$ in water is $s \mathrm{~mol} / \mathrm{L}$. The solubility product of the salt is
let us consider solubility of any general salt $\mathrm{M}_{\mathrm{m}} \mathrm{A}_{\mathrm{n}-}$ $\mathrm{M}_m \mathrm{~A}_n \text { ?? }\left[\mathrm{mM}^{\mathrm{n}}\right]\left[\mathrm{nA}^{m-}\right]$ The solution which contains maximam possible amount of solute in it are called saturated solution for saturated solution, at equilibrium. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{M}^{\mathrm{n}+}\right]^{\mathrm{m}}\left[\mathrm{A}^{\mathrm{m}+}\right]^{\mathrm{n}} \\ & =[\mathrm{ms}]^{\mathrm{m}}[\mathrm{ns}]^{\mathrm{n}} \\ \mathrm{K}_{\mathrm{sp}} & =\mathrm{m}^{\mathrm{m}} \mathrm{n}^{\mathrm{n}}[\mathrm{S}]^{\mathrm{m}-\mathrm{n}} \end{aligned}$
229502
If the solubility of $\mathrm{BaSO}_4$ (mol. wt. 233) is 2.33 $\times 10^{-4} \mathrm{~g} / 100 \mathrm{ml}$, then the solubility product of $\mathrm{BaSO}_4$ is
229509
$\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CaSO}_4$ is $4 \times 10^{-12}$. $\mathrm{CaSO}_4$ is precipitated on mixing equal volumes of the following solutions.
Super Saturated Solution- It is defined as precipitate, which form by the mixture of two solution, if the product of ions is greater than product of solubility, then it is called super saturated solution. $\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{SO}_4^2\right]>\mathrm{K}_{5 \mathrm{p}}$
UPTU/UPSEE-2015
Ionic Equilibrium
229511
The solubility of AgI in NaI solution is less than that in pure water because
1 AgI forms complex with NaI
2 Of common ion effect
3 Solubility product of AgI is less then that of $\mathrm{NaI}$
4 The temperature of the solution decreases
Explanation:
The solubility of AgI in NaI Solution is less than in pure water because of common ion effect. $\mathrm{AgI} \square \mathrm{Ag}^{+}+\mathrm{I}^{-}$ $\mathrm{NaI} \square \mathrm{Na}^{+}+\mathrm{I}^{-}$
UPTU/UPSEE-2014
Ionic Equilibrium
229513
At a certain temperature, the solubility of the salt, $M_m \mathbf{A}_{\mathrm{n}}$ in water is $s \mathrm{~mol} / \mathrm{L}$. The solubility product of the salt is
let us consider solubility of any general salt $\mathrm{M}_{\mathrm{m}} \mathrm{A}_{\mathrm{n}-}$ $\mathrm{M}_m \mathrm{~A}_n \text { ?? }\left[\mathrm{mM}^{\mathrm{n}}\right]\left[\mathrm{nA}^{m-}\right]$ The solution which contains maximam possible amount of solute in it are called saturated solution for saturated solution, at equilibrium. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{M}^{\mathrm{n}+}\right]^{\mathrm{m}}\left[\mathrm{A}^{\mathrm{m}+}\right]^{\mathrm{n}} \\ & =[\mathrm{ms}]^{\mathrm{m}}[\mathrm{ns}]^{\mathrm{n}} \\ \mathrm{K}_{\mathrm{sp}} & =\mathrm{m}^{\mathrm{m}} \mathrm{n}^{\mathrm{n}}[\mathrm{S}]^{\mathrm{m}-\mathrm{n}} \end{aligned}$
