229365
The solubility of $A_2 X_3$ is ' $y^{\prime}$ M. Its solubility product is ........ M.
1 $6 y^4$
2 $64 y^4$
3 $36 y^5$
4 $108 y^5$
Explanation:
For a sparingly soluble salt of molecular formula $\mathrm{A}_2 \mathrm{X}_3$, if solubility in pure water is $\mathrm{y} M$ at a given temperature, $\mathrm{A}_2 \mathrm{X}_3 \rightleftharpoons \underset{2 \mathrm{y}}{2 \mathrm{~A}^{3+}}+3 \mathrm{X}^{2-}$ Solubility product, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3$ $\begin{aligned} & =(2 y)^2 \times(3 \mathrm{y})^3 \\ & =4 y^2 \times 27 y^3=108 y^5 \mathrm{M} \end{aligned}$
Shift-II
Ionic Equilibrium
229366
When equal volumes of $\mathrm{Ca}^{2+}$ and $\mathrm{F}^{-}$solutions are mixed, in which of the solutions precipitation will not occur ? $\left(\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{CaF}_2=1.6 \times 10^{-10}\right)$
Ionic product of $\left(\mathrm{CaF}_2\right)=\left[\mathrm{Ca}^{2+}\right] .\left[\mathrm{F}^{-}\right]^2$. Concentration of ions will be halved after mixing. Thus, ionic products will be 1.$\frac{10^{-2}}{2} \times\left(\frac{10^{-5}}{2}\right)^2=\frac{1}{8} \times 10^{-12}$ 2.$\frac{10^{-3}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-9}$ 3.$\frac{10^{-4}}{2} \times\left(\frac{10^{-2}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ 4.$\frac{10^{-2}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ In option (2), (3) and (4), ionic product $>\mathrm{K}_{\mathrm{sp}}$. Hence option (1) of the solution precipitation will not occur.
Shift-II
Ionic Equilibrium
229367
If the solubility product of $\mathrm{MgF}_2$ at a certain temperature is $1.08 \times 10^{-10}$, its solubility in mol $\mathbf{L}^{-1}$ is
1 $1.04 \times 10^{-5}$
2 $7.3 \times 10^{-4}$
3 $3.0 \times 10^{-5}$
4 $3.0 \times 10^{-4}$
Explanation:
The dissociation of $\mathrm{MgF}_2$ is given as $\mathrm{MgF}_2$ ?? $\mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility (S) of $\mathrm{Mg}^{+2}$ and $\mathrm{F}^{-}$is then from the formula of solubility product we get- $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\ \mathrm{~K}_{\mathrm{sp}} & =\mathrm{S} \times(2 \mathrm{~S})^2 \\ \mathrm{~S}^3 & =\frac{\mathrm{K}_{\mathrm{sp}}}{4} \\ \mathrm{~S}^3 & =\frac{1.08 \times 10^{-10}}{4} \\ \mathrm{~S} & =3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned}$
AP EAMCET (Medical) - 2013
Ionic Equilibrium
229369
The solubility of $\mathrm{AgBr}$ with solubility product $5.0 \times 10^{-13}$ at $298 \mathrm{~K}$ in $0.1 \mathrm{M}$ NaBr solution would be
1 $7 \times 10^{-6} \mathrm{M}$
2 $5 \times 10^{-12} \mathrm{M}$
3 $5 \times 10^{-14} \mathrm{M}$
4 $5 \times 10^{-6} \mathrm{M}$
Explanation:
Let, the solubility of $\mathrm{AgBr}$ be $\mathrm{S} \mathrm{mol} / \mathrm{L}$. $\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}$ Hence, $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=5 \times 10^{-13}$ Given that, $\left[\mathrm{Br}^{-}\right]=0.1$ It means solubility in $\mathrm{NaBr}$ is $5 \times 10^{-12}$.
229365
The solubility of $A_2 X_3$ is ' $y^{\prime}$ M. Its solubility product is ........ M.
1 $6 y^4$
2 $64 y^4$
3 $36 y^5$
4 $108 y^5$
Explanation:
For a sparingly soluble salt of molecular formula $\mathrm{A}_2 \mathrm{X}_3$, if solubility in pure water is $\mathrm{y} M$ at a given temperature, $\mathrm{A}_2 \mathrm{X}_3 \rightleftharpoons \underset{2 \mathrm{y}}{2 \mathrm{~A}^{3+}}+3 \mathrm{X}^{2-}$ Solubility product, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3$ $\begin{aligned} & =(2 y)^2 \times(3 \mathrm{y})^3 \\ & =4 y^2 \times 27 y^3=108 y^5 \mathrm{M} \end{aligned}$
Shift-II
Ionic Equilibrium
229366
When equal volumes of $\mathrm{Ca}^{2+}$ and $\mathrm{F}^{-}$solutions are mixed, in which of the solutions precipitation will not occur ? $\left(\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{CaF}_2=1.6 \times 10^{-10}\right)$
Ionic product of $\left(\mathrm{CaF}_2\right)=\left[\mathrm{Ca}^{2+}\right] .\left[\mathrm{F}^{-}\right]^2$. Concentration of ions will be halved after mixing. Thus, ionic products will be 1.$\frac{10^{-2}}{2} \times\left(\frac{10^{-5}}{2}\right)^2=\frac{1}{8} \times 10^{-12}$ 2.$\frac{10^{-3}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-9}$ 3.$\frac{10^{-4}}{2} \times\left(\frac{10^{-2}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ 4.$\frac{10^{-2}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ In option (2), (3) and (4), ionic product $>\mathrm{K}_{\mathrm{sp}}$. Hence option (1) of the solution precipitation will not occur.
Shift-II
Ionic Equilibrium
229367
If the solubility product of $\mathrm{MgF}_2$ at a certain temperature is $1.08 \times 10^{-10}$, its solubility in mol $\mathbf{L}^{-1}$ is
1 $1.04 \times 10^{-5}$
2 $7.3 \times 10^{-4}$
3 $3.0 \times 10^{-5}$
4 $3.0 \times 10^{-4}$
Explanation:
The dissociation of $\mathrm{MgF}_2$ is given as $\mathrm{MgF}_2$ ?? $\mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility (S) of $\mathrm{Mg}^{+2}$ and $\mathrm{F}^{-}$is then from the formula of solubility product we get- $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\ \mathrm{~K}_{\mathrm{sp}} & =\mathrm{S} \times(2 \mathrm{~S})^2 \\ \mathrm{~S}^3 & =\frac{\mathrm{K}_{\mathrm{sp}}}{4} \\ \mathrm{~S}^3 & =\frac{1.08 \times 10^{-10}}{4} \\ \mathrm{~S} & =3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned}$
AP EAMCET (Medical) - 2013
Ionic Equilibrium
229369
The solubility of $\mathrm{AgBr}$ with solubility product $5.0 \times 10^{-13}$ at $298 \mathrm{~K}$ in $0.1 \mathrm{M}$ NaBr solution would be
1 $7 \times 10^{-6} \mathrm{M}$
2 $5 \times 10^{-12} \mathrm{M}$
3 $5 \times 10^{-14} \mathrm{M}$
4 $5 \times 10^{-6} \mathrm{M}$
Explanation:
Let, the solubility of $\mathrm{AgBr}$ be $\mathrm{S} \mathrm{mol} / \mathrm{L}$. $\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}$ Hence, $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=5 \times 10^{-13}$ Given that, $\left[\mathrm{Br}^{-}\right]=0.1$ It means solubility in $\mathrm{NaBr}$ is $5 \times 10^{-12}$.
229365
The solubility of $A_2 X_3$ is ' $y^{\prime}$ M. Its solubility product is ........ M.
1 $6 y^4$
2 $64 y^4$
3 $36 y^5$
4 $108 y^5$
Explanation:
For a sparingly soluble salt of molecular formula $\mathrm{A}_2 \mathrm{X}_3$, if solubility in pure water is $\mathrm{y} M$ at a given temperature, $\mathrm{A}_2 \mathrm{X}_3 \rightleftharpoons \underset{2 \mathrm{y}}{2 \mathrm{~A}^{3+}}+3 \mathrm{X}^{2-}$ Solubility product, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3$ $\begin{aligned} & =(2 y)^2 \times(3 \mathrm{y})^3 \\ & =4 y^2 \times 27 y^3=108 y^5 \mathrm{M} \end{aligned}$
Shift-II
Ionic Equilibrium
229366
When equal volumes of $\mathrm{Ca}^{2+}$ and $\mathrm{F}^{-}$solutions are mixed, in which of the solutions precipitation will not occur ? $\left(\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{CaF}_2=1.6 \times 10^{-10}\right)$
Ionic product of $\left(\mathrm{CaF}_2\right)=\left[\mathrm{Ca}^{2+}\right] .\left[\mathrm{F}^{-}\right]^2$. Concentration of ions will be halved after mixing. Thus, ionic products will be 1.$\frac{10^{-2}}{2} \times\left(\frac{10^{-5}}{2}\right)^2=\frac{1}{8} \times 10^{-12}$ 2.$\frac{10^{-3}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-9}$ 3.$\frac{10^{-4}}{2} \times\left(\frac{10^{-2}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ 4.$\frac{10^{-2}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ In option (2), (3) and (4), ionic product $>\mathrm{K}_{\mathrm{sp}}$. Hence option (1) of the solution precipitation will not occur.
Shift-II
Ionic Equilibrium
229367
If the solubility product of $\mathrm{MgF}_2$ at a certain temperature is $1.08 \times 10^{-10}$, its solubility in mol $\mathbf{L}^{-1}$ is
1 $1.04 \times 10^{-5}$
2 $7.3 \times 10^{-4}$
3 $3.0 \times 10^{-5}$
4 $3.0 \times 10^{-4}$
Explanation:
The dissociation of $\mathrm{MgF}_2$ is given as $\mathrm{MgF}_2$ ?? $\mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility (S) of $\mathrm{Mg}^{+2}$ and $\mathrm{F}^{-}$is then from the formula of solubility product we get- $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\ \mathrm{~K}_{\mathrm{sp}} & =\mathrm{S} \times(2 \mathrm{~S})^2 \\ \mathrm{~S}^3 & =\frac{\mathrm{K}_{\mathrm{sp}}}{4} \\ \mathrm{~S}^3 & =\frac{1.08 \times 10^{-10}}{4} \\ \mathrm{~S} & =3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned}$
AP EAMCET (Medical) - 2013
Ionic Equilibrium
229369
The solubility of $\mathrm{AgBr}$ with solubility product $5.0 \times 10^{-13}$ at $298 \mathrm{~K}$ in $0.1 \mathrm{M}$ NaBr solution would be
1 $7 \times 10^{-6} \mathrm{M}$
2 $5 \times 10^{-12} \mathrm{M}$
3 $5 \times 10^{-14} \mathrm{M}$
4 $5 \times 10^{-6} \mathrm{M}$
Explanation:
Let, the solubility of $\mathrm{AgBr}$ be $\mathrm{S} \mathrm{mol} / \mathrm{L}$. $\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}$ Hence, $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=5 \times 10^{-13}$ Given that, $\left[\mathrm{Br}^{-}\right]=0.1$ It means solubility in $\mathrm{NaBr}$ is $5 \times 10^{-12}$.
229365
The solubility of $A_2 X_3$ is ' $y^{\prime}$ M. Its solubility product is ........ M.
1 $6 y^4$
2 $64 y^4$
3 $36 y^5$
4 $108 y^5$
Explanation:
For a sparingly soluble salt of molecular formula $\mathrm{A}_2 \mathrm{X}_3$, if solubility in pure water is $\mathrm{y} M$ at a given temperature, $\mathrm{A}_2 \mathrm{X}_3 \rightleftharpoons \underset{2 \mathrm{y}}{2 \mathrm{~A}^{3+}}+3 \mathrm{X}^{2-}$ Solubility product, $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{X}^{2-}\right]^3$ $\begin{aligned} & =(2 y)^2 \times(3 \mathrm{y})^3 \\ & =4 y^2 \times 27 y^3=108 y^5 \mathrm{M} \end{aligned}$
Shift-II
Ionic Equilibrium
229366
When equal volumes of $\mathrm{Ca}^{2+}$ and $\mathrm{F}^{-}$solutions are mixed, in which of the solutions precipitation will not occur ? $\left(\mathrm{K}_{\mathrm{sp}}\right.$ of $\left.\mathrm{CaF}_2=1.6 \times 10^{-10}\right)$
Ionic product of $\left(\mathrm{CaF}_2\right)=\left[\mathrm{Ca}^{2+}\right] .\left[\mathrm{F}^{-}\right]^2$. Concentration of ions will be halved after mixing. Thus, ionic products will be 1.$\frac{10^{-2}}{2} \times\left(\frac{10^{-5}}{2}\right)^2=\frac{1}{8} \times 10^{-12}$ 2.$\frac{10^{-3}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-9}$ 3.$\frac{10^{-4}}{2} \times\left(\frac{10^{-2}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ 4.$\frac{10^{-2}}{2} \times\left(\frac{10^{-3}}{2}\right)^2=\frac{1}{8} \times 10^{-8}$ In option (2), (3) and (4), ionic product $>\mathrm{K}_{\mathrm{sp}}$. Hence option (1) of the solution precipitation will not occur.
Shift-II
Ionic Equilibrium
229367
If the solubility product of $\mathrm{MgF}_2$ at a certain temperature is $1.08 \times 10^{-10}$, its solubility in mol $\mathbf{L}^{-1}$ is
1 $1.04 \times 10^{-5}$
2 $7.3 \times 10^{-4}$
3 $3.0 \times 10^{-5}$
4 $3.0 \times 10^{-4}$
Explanation:
The dissociation of $\mathrm{MgF}_2$ is given as $\mathrm{MgF}_2$ ?? $\mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility (S) of $\mathrm{Mg}^{+2}$ and $\mathrm{F}^{-}$is then from the formula of solubility product we get- $\begin{aligned} \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{F}^{-}\right]^2 \\ \mathrm{~K}_{\mathrm{sp}} & =\mathrm{S} \times(2 \mathrm{~S})^2 \\ \mathrm{~S}^3 & =\frac{\mathrm{K}_{\mathrm{sp}}}{4} \\ \mathrm{~S}^3 & =\frac{1.08 \times 10^{-10}}{4} \\ \mathrm{~S} & =3 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \end{aligned}$
AP EAMCET (Medical) - 2013
Ionic Equilibrium
229369
The solubility of $\mathrm{AgBr}$ with solubility product $5.0 \times 10^{-13}$ at $298 \mathrm{~K}$ in $0.1 \mathrm{M}$ NaBr solution would be
1 $7 \times 10^{-6} \mathrm{M}$
2 $5 \times 10^{-12} \mathrm{M}$
3 $5 \times 10^{-14} \mathrm{M}$
4 $5 \times 10^{-6} \mathrm{M}$
Explanation:
Let, the solubility of $\mathrm{AgBr}$ be $\mathrm{S} \mathrm{mol} / \mathrm{L}$. $\mathrm{AgBr} \rightleftharpoons \mathrm{Ag}^{+}+\mathrm{Br}^{-}$ Hence, $\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Br}^{-}\right]=5 \times 10^{-13}$ Given that, $\left[\mathrm{Br}^{-}\right]=0.1$ It means solubility in $\mathrm{NaBr}$ is $5 \times 10^{-12}$.
