The dissociation equilibrium of a sparingly soluble salt say $\left(A_p B_q\right)$ can be represented as follow. $\mathrm{A}_{\mathrm{p}} \mathrm{B}_{\mathrm{q}} \text { ?? } \underset{[\mathrm{S}]}{\mathrm{pA}}+\underset{[\mathrm{q}]}{\mathrm{q} \mathrm{B}^{\mathrm{p}-}}$ Let $\mathrm{S}$ be the solubility in $\mathrm{mol} /$ liter the expression for the solubility product will be as given below. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{A}^{+\mathrm{q}}\right]^{\mathrm{p}} \cdot\left[\mathrm{B}^{\mathrm{p}-}\right]^{\mathrm{q}} \\ & =[\mathrm{p} \cdot \mathrm{S}]^{\mathrm{p}} \cdot[\mathrm{q} \cdot \mathrm{S}]^{\mathrm{q}} \\ \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}} & =\mathrm{p}^{\mathrm{p}} \cdot \mathrm{q}^{\mathrm{q}}(\mathrm{S})^{\mathrm{p}+\mathrm{q}} \end{aligned}$
BITSAT-2016
Ionic Equilibrium
229410
A saturated solution of $\mathrm{Ag}_2 \mathrm{SO}_4$ is $2.5 \times 10^{-2} \mathrm{M}$. The value of its solubility product is-
229412
When $\mathrm{H}_2 \mathrm{~S}$ is passed in acidic medium in solution having $\mathrm{CuS}$ and $\mathrm{ZnS}$, only $\mathrm{CuS}$ is precipitated because
4 $\mathrm{ZnS}$ has lower melting point than $\mathrm{CuS}$
Explanation:
Precipitation of salt AB will occur when. $\mathrm{K}_{\mathrm{sp}}<\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]$ $\mathrm{ZnS}$ is not precipitated by passing $\mathrm{H}_2 \mathrm{~S}$ in acidic medium but CuS precipitated because by analysing above equation it is clear that $\mathrm{K}_{\mathrm{sp}} \mathrm{CuS}<<\mathrm{K}_{\mathrm{sp}} \mathrm{ZnS} \text {. }$
BITSAT-2005
Ionic Equilibrium
229421
In which of the following solvents, AgBr will have the highest solubility?
1 $10^{-3} \mathrm{M} \mathrm{NaBr}$
2 $10^{-3} \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$
3 Purewater
4 $10^{-3} \mathrm{M} \mathrm{HBr}$
Explanation:
$\mathrm{AgBr}$ will form soluble complex with $\mathrm{NH}_4 \mathrm{OH}$, i.e. $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Br}$. The key to this problem is something called the " common ion effect" In essence, the solubility of ionic solute is much lower is a solvent that already contains one the ions of the solute.
The dissociation equilibrium of a sparingly soluble salt say $\left(A_p B_q\right)$ can be represented as follow. $\mathrm{A}_{\mathrm{p}} \mathrm{B}_{\mathrm{q}} \text { ?? } \underset{[\mathrm{S}]}{\mathrm{pA}}+\underset{[\mathrm{q}]}{\mathrm{q} \mathrm{B}^{\mathrm{p}-}}$ Let $\mathrm{S}$ be the solubility in $\mathrm{mol} /$ liter the expression for the solubility product will be as given below. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{A}^{+\mathrm{q}}\right]^{\mathrm{p}} \cdot\left[\mathrm{B}^{\mathrm{p}-}\right]^{\mathrm{q}} \\ & =[\mathrm{p} \cdot \mathrm{S}]^{\mathrm{p}} \cdot[\mathrm{q} \cdot \mathrm{S}]^{\mathrm{q}} \\ \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}} & =\mathrm{p}^{\mathrm{p}} \cdot \mathrm{q}^{\mathrm{q}}(\mathrm{S})^{\mathrm{p}+\mathrm{q}} \end{aligned}$
BITSAT-2016
Ionic Equilibrium
229410
A saturated solution of $\mathrm{Ag}_2 \mathrm{SO}_4$ is $2.5 \times 10^{-2} \mathrm{M}$. The value of its solubility product is-
229412
When $\mathrm{H}_2 \mathrm{~S}$ is passed in acidic medium in solution having $\mathrm{CuS}$ and $\mathrm{ZnS}$, only $\mathrm{CuS}$ is precipitated because
4 $\mathrm{ZnS}$ has lower melting point than $\mathrm{CuS}$
Explanation:
Precipitation of salt AB will occur when. $\mathrm{K}_{\mathrm{sp}}<\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]$ $\mathrm{ZnS}$ is not precipitated by passing $\mathrm{H}_2 \mathrm{~S}$ in acidic medium but CuS precipitated because by analysing above equation it is clear that $\mathrm{K}_{\mathrm{sp}} \mathrm{CuS}<<\mathrm{K}_{\mathrm{sp}} \mathrm{ZnS} \text {. }$
BITSAT-2005
Ionic Equilibrium
229421
In which of the following solvents, AgBr will have the highest solubility?
1 $10^{-3} \mathrm{M} \mathrm{NaBr}$
2 $10^{-3} \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$
3 Purewater
4 $10^{-3} \mathrm{M} \mathrm{HBr}$
Explanation:
$\mathrm{AgBr}$ will form soluble complex with $\mathrm{NH}_4 \mathrm{OH}$, i.e. $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Br}$. The key to this problem is something called the " common ion effect" In essence, the solubility of ionic solute is much lower is a solvent that already contains one the ions of the solute.
The dissociation equilibrium of a sparingly soluble salt say $\left(A_p B_q\right)$ can be represented as follow. $\mathrm{A}_{\mathrm{p}} \mathrm{B}_{\mathrm{q}} \text { ?? } \underset{[\mathrm{S}]}{\mathrm{pA}}+\underset{[\mathrm{q}]}{\mathrm{q} \mathrm{B}^{\mathrm{p}-}}$ Let $\mathrm{S}$ be the solubility in $\mathrm{mol} /$ liter the expression for the solubility product will be as given below. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{A}^{+\mathrm{q}}\right]^{\mathrm{p}} \cdot\left[\mathrm{B}^{\mathrm{p}-}\right]^{\mathrm{q}} \\ & =[\mathrm{p} \cdot \mathrm{S}]^{\mathrm{p}} \cdot[\mathrm{q} \cdot \mathrm{S}]^{\mathrm{q}} \\ \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}} & =\mathrm{p}^{\mathrm{p}} \cdot \mathrm{q}^{\mathrm{q}}(\mathrm{S})^{\mathrm{p}+\mathrm{q}} \end{aligned}$
BITSAT-2016
Ionic Equilibrium
229410
A saturated solution of $\mathrm{Ag}_2 \mathrm{SO}_4$ is $2.5 \times 10^{-2} \mathrm{M}$. The value of its solubility product is-
229412
When $\mathrm{H}_2 \mathrm{~S}$ is passed in acidic medium in solution having $\mathrm{CuS}$ and $\mathrm{ZnS}$, only $\mathrm{CuS}$ is precipitated because
4 $\mathrm{ZnS}$ has lower melting point than $\mathrm{CuS}$
Explanation:
Precipitation of salt AB will occur when. $\mathrm{K}_{\mathrm{sp}}<\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]$ $\mathrm{ZnS}$ is not precipitated by passing $\mathrm{H}_2 \mathrm{~S}$ in acidic medium but CuS precipitated because by analysing above equation it is clear that $\mathrm{K}_{\mathrm{sp}} \mathrm{CuS}<<\mathrm{K}_{\mathrm{sp}} \mathrm{ZnS} \text {. }$
BITSAT-2005
Ionic Equilibrium
229421
In which of the following solvents, AgBr will have the highest solubility?
1 $10^{-3} \mathrm{M} \mathrm{NaBr}$
2 $10^{-3} \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$
3 Purewater
4 $10^{-3} \mathrm{M} \mathrm{HBr}$
Explanation:
$\mathrm{AgBr}$ will form soluble complex with $\mathrm{NH}_4 \mathrm{OH}$, i.e. $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Br}$. The key to this problem is something called the " common ion effect" In essence, the solubility of ionic solute is much lower is a solvent that already contains one the ions of the solute.
The dissociation equilibrium of a sparingly soluble salt say $\left(A_p B_q\right)$ can be represented as follow. $\mathrm{A}_{\mathrm{p}} \mathrm{B}_{\mathrm{q}} \text { ?? } \underset{[\mathrm{S}]}{\mathrm{pA}}+\underset{[\mathrm{q}]}{\mathrm{q} \mathrm{B}^{\mathrm{p}-}}$ Let $\mathrm{S}$ be the solubility in $\mathrm{mol} /$ liter the expression for the solubility product will be as given below. $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{A}^{+\mathrm{q}}\right]^{\mathrm{p}} \cdot\left[\mathrm{B}^{\mathrm{p}-}\right]^{\mathrm{q}} \\ & =[\mathrm{p} \cdot \mathrm{S}]^{\mathrm{p}} \cdot[\mathrm{q} \cdot \mathrm{S}]^{\mathrm{q}} \\ \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}} & =\mathrm{p}^{\mathrm{p}} \cdot \mathrm{q}^{\mathrm{q}}(\mathrm{S})^{\mathrm{p}+\mathrm{q}} \end{aligned}$
BITSAT-2016
Ionic Equilibrium
229410
A saturated solution of $\mathrm{Ag}_2 \mathrm{SO}_4$ is $2.5 \times 10^{-2} \mathrm{M}$. The value of its solubility product is-
229412
When $\mathrm{H}_2 \mathrm{~S}$ is passed in acidic medium in solution having $\mathrm{CuS}$ and $\mathrm{ZnS}$, only $\mathrm{CuS}$ is precipitated because
4 $\mathrm{ZnS}$ has lower melting point than $\mathrm{CuS}$
Explanation:
Precipitation of salt AB will occur when. $\mathrm{K}_{\mathrm{sp}}<\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right]$ $\mathrm{ZnS}$ is not precipitated by passing $\mathrm{H}_2 \mathrm{~S}$ in acidic medium but CuS precipitated because by analysing above equation it is clear that $\mathrm{K}_{\mathrm{sp}} \mathrm{CuS}<<\mathrm{K}_{\mathrm{sp}} \mathrm{ZnS} \text {. }$
BITSAT-2005
Ionic Equilibrium
229421
In which of the following solvents, AgBr will have the highest solubility?
1 $10^{-3} \mathrm{M} \mathrm{NaBr}$
2 $10^{-3} \mathrm{M} \mathrm{NH}_4 \mathrm{OH}$
3 Purewater
4 $10^{-3} \mathrm{M} \mathrm{HBr}$
Explanation:
$\mathrm{AgBr}$ will form soluble complex with $\mathrm{NH}_4 \mathrm{OH}$, i.e. $\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right] \mathrm{Br}$. The key to this problem is something called the " common ion effect" In essence, the solubility of ionic solute is much lower is a solvent that already contains one the ions of the solute.