229374
If the concentration of $\mathrm{Ag}^{+}$ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$. then find the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$.
1 $5.30 \times 10^{-12}$
2 $4.50 \times 10^{-11}$
3 $2.66 \times 10^{-12}$
4 $6.90 \times 10^{-12}$
Explanation:
Given that, $\left[\mathrm{Ag}^{+}\right]=1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$ Silver carbonate dissociates as follows - $\mathrm{Ag}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{Ag}^{+}+\mathrm{CO}_3^{2-}$ Let the solubility is $\mathrm{S}$. The solubility of silver will be $2 \mathrm{~s}$ as two moles ion are dissociated - $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \end{aligned}$ $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4 \times\left(1.20 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=6.91 \times 10^{-12} \end{aligned}$
Shift-I
Ionic Equilibrium
229376
The solubility of $\mathrm{AgCl}$ at $20^{\circ} \mathrm{C}$ is $1.435 \times 10^{-5} \mathrm{gL}^{-1}$. The solubility product of $\mathrm{AgCl}$ is
1 $2 \times 10^{-16}$
2 $108 \times 10^{-3}$
3 $1.0 \times 10^{-14}$
4 $1.035 \times 10^{-5}$
Explanation:
Solubility of $\mathrm{AgCl}=1.435 \times 10^{-5} \mathrm{gL}^{-1}$ [Molecular weight of $\mathrm{AgCl}=143.5$ ] We know that solubility of $\mathrm{AgCl}$ in moles $\begin{aligned} & \mathrm{AgCl} \square \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(\mathrm{s})^2 \\ \because \quad & \mathrm{S}=\frac{\left(1.435 \times 10^{-5}\right)}{143.5}=1 \times 10^{-7} \end{aligned}$ Therefore solubility product of $\mathrm{AgCl}=\mathrm{S}^2$
COMEDK-2018
Ionic Equilibrium
229377
The solubility of $\mathrm{AgCN}$ in a buffer solution of $p H=3$ is $x$. The value of $x$ is : [Assume : No cyano complex is found; $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ and $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times$ $\left.10^{-10}\right]$
1 $0.625 \times 10^{-6}$
2 $1.6 \times 10^{-6}$
3 $2.2 \times 10^{-16}$
4 $1.9 \times 10^{-5}$
Explanation:
Given $: \mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times 10^{-10}$ The solubility of $\mathrm{AgCN}$ in a buffer solution of $\mathrm{pH}=3$ Let, solubility is $S$ then, $\mathrm{AgCN} \text { ? } \mathrm{Ag}^{+}+\mathrm{CN}^{-}$ $\mathrm{K}_{\mathrm{sp}}=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2$ $\mathrm{H}^{+}+\mathrm{CN}^{-}$?? $\mathrm{HCN}, \quad \mathrm{K}=\frac{1}{\mathrm{~K}_{\mathrm{a}}}=\frac{1}{6.2 \times 10^{-10}}$ Therefore, $\mathrm{K}_{\mathrm{sp}} \times \frac{1}{\mathrm{~K}_{\mathrm{a}}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right] \frac{[\mathrm{HCN}]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}$ or $2.2 \times 10^{-16} \times \frac{1}{6.2 \times 10^{-10}}=\frac{\mathrm{S} \times \mathrm{S}}{10^{-3}}$ $\begin{array}{ll} \hline & S^2=\frac{2.2 \times 10^{-16} \times 10^{-3}}{6.2 \times 10^{-10}}=0.354 \times 10^{-9} \\ \text { or } & S^2=3.54 \times 10^{-10} \\ \text { or } & S=1.9 \times 10^{-5} \end{array}$
Shift-I
Ionic Equilibrium
229378
From the graph, the value of Henry's constant for the solubility of $\mathrm{HCl}$ gas in cyclohexane is
1 $10 \mathrm{k}$ torr
2 100 torr
3 50 torr
4 $2.4 \times 10^2$ torr
Explanation:
According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(\mathrm{X})$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law. $\mathrm{P}_{\text {solute }} \propto \mathrm{X}_{\text {solute }}$ in solution $\mathrm{P}_{\text {solute }}=\mathrm{k}_{\mathrm{H}} \mathrm{X}_{\text {solute }}$ in solution Where, $\mathrm{k}_{\mathrm{H}}=$ Henry's constant The slope of the straight line gives the value of $\mathrm{k}_{\mathrm{H}}$. $\begin{aligned} \therefore \quad \tan \theta & =\mathrm{k}_{\mathrm{H}}=\frac{50}{0.005} \\ & =\frac{50000}{5}=10000 \\ \mathrm{k}_{\mathrm{H}} & =10 \mathrm{k} \text { torr } \end{aligned}$
Shift-I
Ionic Equilibrium
229379
The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is : [Given : The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water $\left.=5.5 \times 10^{-6}\right]$
1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
Explanation:
Given: $\mathrm{K}_{\mathrm{sp}}=5.5 \times 10^{-6}$ Let, the solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is $\mathrm{S}$, $\begin{aligned} & \mathrm{Ca}(\mathrm{OH})_2 \square \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ & \text {(S) }(2 \mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{OH}^{-}\right]^2 \\ & \mathrm{~K}_{\mathrm{sp}}=(\mathrm{S})^1 \times(2 \mathrm{~S})^2 \\ & 5.5 \times 10^{-6}=4 \mathrm{~S}^3 \\ & S=\left(\frac{5.5 \times 10^{-6}}{4}\right)^{1 / 3} \\ & \mathrm{~S}=1.11 \times 10^{-2} \\ & \end{aligned}$
229374
If the concentration of $\mathrm{Ag}^{+}$ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$. then find the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$.
1 $5.30 \times 10^{-12}$
2 $4.50 \times 10^{-11}$
3 $2.66 \times 10^{-12}$
4 $6.90 \times 10^{-12}$
Explanation:
Given that, $\left[\mathrm{Ag}^{+}\right]=1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$ Silver carbonate dissociates as follows - $\mathrm{Ag}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{Ag}^{+}+\mathrm{CO}_3^{2-}$ Let the solubility is $\mathrm{S}$. The solubility of silver will be $2 \mathrm{~s}$ as two moles ion are dissociated - $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \end{aligned}$ $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4 \times\left(1.20 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=6.91 \times 10^{-12} \end{aligned}$
Shift-I
Ionic Equilibrium
229376
The solubility of $\mathrm{AgCl}$ at $20^{\circ} \mathrm{C}$ is $1.435 \times 10^{-5} \mathrm{gL}^{-1}$. The solubility product of $\mathrm{AgCl}$ is
1 $2 \times 10^{-16}$
2 $108 \times 10^{-3}$
3 $1.0 \times 10^{-14}$
4 $1.035 \times 10^{-5}$
Explanation:
Solubility of $\mathrm{AgCl}=1.435 \times 10^{-5} \mathrm{gL}^{-1}$ [Molecular weight of $\mathrm{AgCl}=143.5$ ] We know that solubility of $\mathrm{AgCl}$ in moles $\begin{aligned} & \mathrm{AgCl} \square \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(\mathrm{s})^2 \\ \because \quad & \mathrm{S}=\frac{\left(1.435 \times 10^{-5}\right)}{143.5}=1 \times 10^{-7} \end{aligned}$ Therefore solubility product of $\mathrm{AgCl}=\mathrm{S}^2$
COMEDK-2018
Ionic Equilibrium
229377
The solubility of $\mathrm{AgCN}$ in a buffer solution of $p H=3$ is $x$. The value of $x$ is : [Assume : No cyano complex is found; $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ and $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times$ $\left.10^{-10}\right]$
1 $0.625 \times 10^{-6}$
2 $1.6 \times 10^{-6}$
3 $2.2 \times 10^{-16}$
4 $1.9 \times 10^{-5}$
Explanation:
Given $: \mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times 10^{-10}$ The solubility of $\mathrm{AgCN}$ in a buffer solution of $\mathrm{pH}=3$ Let, solubility is $S$ then, $\mathrm{AgCN} \text { ? } \mathrm{Ag}^{+}+\mathrm{CN}^{-}$ $\mathrm{K}_{\mathrm{sp}}=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2$ $\mathrm{H}^{+}+\mathrm{CN}^{-}$?? $\mathrm{HCN}, \quad \mathrm{K}=\frac{1}{\mathrm{~K}_{\mathrm{a}}}=\frac{1}{6.2 \times 10^{-10}}$ Therefore, $\mathrm{K}_{\mathrm{sp}} \times \frac{1}{\mathrm{~K}_{\mathrm{a}}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right] \frac{[\mathrm{HCN}]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}$ or $2.2 \times 10^{-16} \times \frac{1}{6.2 \times 10^{-10}}=\frac{\mathrm{S} \times \mathrm{S}}{10^{-3}}$ $\begin{array}{ll} \hline & S^2=\frac{2.2 \times 10^{-16} \times 10^{-3}}{6.2 \times 10^{-10}}=0.354 \times 10^{-9} \\ \text { or } & S^2=3.54 \times 10^{-10} \\ \text { or } & S=1.9 \times 10^{-5} \end{array}$
Shift-I
Ionic Equilibrium
229378
From the graph, the value of Henry's constant for the solubility of $\mathrm{HCl}$ gas in cyclohexane is
1 $10 \mathrm{k}$ torr
2 100 torr
3 50 torr
4 $2.4 \times 10^2$ torr
Explanation:
According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(\mathrm{X})$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law. $\mathrm{P}_{\text {solute }} \propto \mathrm{X}_{\text {solute }}$ in solution $\mathrm{P}_{\text {solute }}=\mathrm{k}_{\mathrm{H}} \mathrm{X}_{\text {solute }}$ in solution Where, $\mathrm{k}_{\mathrm{H}}=$ Henry's constant The slope of the straight line gives the value of $\mathrm{k}_{\mathrm{H}}$. $\begin{aligned} \therefore \quad \tan \theta & =\mathrm{k}_{\mathrm{H}}=\frac{50}{0.005} \\ & =\frac{50000}{5}=10000 \\ \mathrm{k}_{\mathrm{H}} & =10 \mathrm{k} \text { torr } \end{aligned}$
Shift-I
Ionic Equilibrium
229379
The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is : [Given : The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water $\left.=5.5 \times 10^{-6}\right]$
1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
Explanation:
Given: $\mathrm{K}_{\mathrm{sp}}=5.5 \times 10^{-6}$ Let, the solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is $\mathrm{S}$, $\begin{aligned} & \mathrm{Ca}(\mathrm{OH})_2 \square \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ & \text {(S) }(2 \mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{OH}^{-}\right]^2 \\ & \mathrm{~K}_{\mathrm{sp}}=(\mathrm{S})^1 \times(2 \mathrm{~S})^2 \\ & 5.5 \times 10^{-6}=4 \mathrm{~S}^3 \\ & S=\left(\frac{5.5 \times 10^{-6}}{4}\right)^{1 / 3} \\ & \mathrm{~S}=1.11 \times 10^{-2} \\ & \end{aligned}$
229374
If the concentration of $\mathrm{Ag}^{+}$ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$. then find the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$.
1 $5.30 \times 10^{-12}$
2 $4.50 \times 10^{-11}$
3 $2.66 \times 10^{-12}$
4 $6.90 \times 10^{-12}$
Explanation:
Given that, $\left[\mathrm{Ag}^{+}\right]=1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$ Silver carbonate dissociates as follows - $\mathrm{Ag}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{Ag}^{+}+\mathrm{CO}_3^{2-}$ Let the solubility is $\mathrm{S}$. The solubility of silver will be $2 \mathrm{~s}$ as two moles ion are dissociated - $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \end{aligned}$ $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4 \times\left(1.20 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=6.91 \times 10^{-12} \end{aligned}$
Shift-I
Ionic Equilibrium
229376
The solubility of $\mathrm{AgCl}$ at $20^{\circ} \mathrm{C}$ is $1.435 \times 10^{-5} \mathrm{gL}^{-1}$. The solubility product of $\mathrm{AgCl}$ is
1 $2 \times 10^{-16}$
2 $108 \times 10^{-3}$
3 $1.0 \times 10^{-14}$
4 $1.035 \times 10^{-5}$
Explanation:
Solubility of $\mathrm{AgCl}=1.435 \times 10^{-5} \mathrm{gL}^{-1}$ [Molecular weight of $\mathrm{AgCl}=143.5$ ] We know that solubility of $\mathrm{AgCl}$ in moles $\begin{aligned} & \mathrm{AgCl} \square \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(\mathrm{s})^2 \\ \because \quad & \mathrm{S}=\frac{\left(1.435 \times 10^{-5}\right)}{143.5}=1 \times 10^{-7} \end{aligned}$ Therefore solubility product of $\mathrm{AgCl}=\mathrm{S}^2$
COMEDK-2018
Ionic Equilibrium
229377
The solubility of $\mathrm{AgCN}$ in a buffer solution of $p H=3$ is $x$. The value of $x$ is : [Assume : No cyano complex is found; $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ and $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times$ $\left.10^{-10}\right]$
1 $0.625 \times 10^{-6}$
2 $1.6 \times 10^{-6}$
3 $2.2 \times 10^{-16}$
4 $1.9 \times 10^{-5}$
Explanation:
Given $: \mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times 10^{-10}$ The solubility of $\mathrm{AgCN}$ in a buffer solution of $\mathrm{pH}=3$ Let, solubility is $S$ then, $\mathrm{AgCN} \text { ? } \mathrm{Ag}^{+}+\mathrm{CN}^{-}$ $\mathrm{K}_{\mathrm{sp}}=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2$ $\mathrm{H}^{+}+\mathrm{CN}^{-}$?? $\mathrm{HCN}, \quad \mathrm{K}=\frac{1}{\mathrm{~K}_{\mathrm{a}}}=\frac{1}{6.2 \times 10^{-10}}$ Therefore, $\mathrm{K}_{\mathrm{sp}} \times \frac{1}{\mathrm{~K}_{\mathrm{a}}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right] \frac{[\mathrm{HCN}]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}$ or $2.2 \times 10^{-16} \times \frac{1}{6.2 \times 10^{-10}}=\frac{\mathrm{S} \times \mathrm{S}}{10^{-3}}$ $\begin{array}{ll} \hline & S^2=\frac{2.2 \times 10^{-16} \times 10^{-3}}{6.2 \times 10^{-10}}=0.354 \times 10^{-9} \\ \text { or } & S^2=3.54 \times 10^{-10} \\ \text { or } & S=1.9 \times 10^{-5} \end{array}$
Shift-I
Ionic Equilibrium
229378
From the graph, the value of Henry's constant for the solubility of $\mathrm{HCl}$ gas in cyclohexane is
1 $10 \mathrm{k}$ torr
2 100 torr
3 50 torr
4 $2.4 \times 10^2$ torr
Explanation:
According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(\mathrm{X})$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law. $\mathrm{P}_{\text {solute }} \propto \mathrm{X}_{\text {solute }}$ in solution $\mathrm{P}_{\text {solute }}=\mathrm{k}_{\mathrm{H}} \mathrm{X}_{\text {solute }}$ in solution Where, $\mathrm{k}_{\mathrm{H}}=$ Henry's constant The slope of the straight line gives the value of $\mathrm{k}_{\mathrm{H}}$. $\begin{aligned} \therefore \quad \tan \theta & =\mathrm{k}_{\mathrm{H}}=\frac{50}{0.005} \\ & =\frac{50000}{5}=10000 \\ \mathrm{k}_{\mathrm{H}} & =10 \mathrm{k} \text { torr } \end{aligned}$
Shift-I
Ionic Equilibrium
229379
The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is : [Given : The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water $\left.=5.5 \times 10^{-6}\right]$
1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
Explanation:
Given: $\mathrm{K}_{\mathrm{sp}}=5.5 \times 10^{-6}$ Let, the solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is $\mathrm{S}$, $\begin{aligned} & \mathrm{Ca}(\mathrm{OH})_2 \square \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ & \text {(S) }(2 \mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{OH}^{-}\right]^2 \\ & \mathrm{~K}_{\mathrm{sp}}=(\mathrm{S})^1 \times(2 \mathrm{~S})^2 \\ & 5.5 \times 10^{-6}=4 \mathrm{~S}^3 \\ & S=\left(\frac{5.5 \times 10^{-6}}{4}\right)^{1 / 3} \\ & \mathrm{~S}=1.11 \times 10^{-2} \\ & \end{aligned}$
229374
If the concentration of $\mathrm{Ag}^{+}$ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$. then find the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$.
1 $5.30 \times 10^{-12}$
2 $4.50 \times 10^{-11}$
3 $2.66 \times 10^{-12}$
4 $6.90 \times 10^{-12}$
Explanation:
Given that, $\left[\mathrm{Ag}^{+}\right]=1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$ Silver carbonate dissociates as follows - $\mathrm{Ag}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{Ag}^{+}+\mathrm{CO}_3^{2-}$ Let the solubility is $\mathrm{S}$. The solubility of silver will be $2 \mathrm{~s}$ as two moles ion are dissociated - $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \end{aligned}$ $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4 \times\left(1.20 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=6.91 \times 10^{-12} \end{aligned}$
Shift-I
Ionic Equilibrium
229376
The solubility of $\mathrm{AgCl}$ at $20^{\circ} \mathrm{C}$ is $1.435 \times 10^{-5} \mathrm{gL}^{-1}$. The solubility product of $\mathrm{AgCl}$ is
1 $2 \times 10^{-16}$
2 $108 \times 10^{-3}$
3 $1.0 \times 10^{-14}$
4 $1.035 \times 10^{-5}$
Explanation:
Solubility of $\mathrm{AgCl}=1.435 \times 10^{-5} \mathrm{gL}^{-1}$ [Molecular weight of $\mathrm{AgCl}=143.5$ ] We know that solubility of $\mathrm{AgCl}$ in moles $\begin{aligned} & \mathrm{AgCl} \square \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(\mathrm{s})^2 \\ \because \quad & \mathrm{S}=\frac{\left(1.435 \times 10^{-5}\right)}{143.5}=1 \times 10^{-7} \end{aligned}$ Therefore solubility product of $\mathrm{AgCl}=\mathrm{S}^2$
COMEDK-2018
Ionic Equilibrium
229377
The solubility of $\mathrm{AgCN}$ in a buffer solution of $p H=3$ is $x$. The value of $x$ is : [Assume : No cyano complex is found; $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ and $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times$ $\left.10^{-10}\right]$
1 $0.625 \times 10^{-6}$
2 $1.6 \times 10^{-6}$
3 $2.2 \times 10^{-16}$
4 $1.9 \times 10^{-5}$
Explanation:
Given $: \mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times 10^{-10}$ The solubility of $\mathrm{AgCN}$ in a buffer solution of $\mathrm{pH}=3$ Let, solubility is $S$ then, $\mathrm{AgCN} \text { ? } \mathrm{Ag}^{+}+\mathrm{CN}^{-}$ $\mathrm{K}_{\mathrm{sp}}=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2$ $\mathrm{H}^{+}+\mathrm{CN}^{-}$?? $\mathrm{HCN}, \quad \mathrm{K}=\frac{1}{\mathrm{~K}_{\mathrm{a}}}=\frac{1}{6.2 \times 10^{-10}}$ Therefore, $\mathrm{K}_{\mathrm{sp}} \times \frac{1}{\mathrm{~K}_{\mathrm{a}}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right] \frac{[\mathrm{HCN}]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}$ or $2.2 \times 10^{-16} \times \frac{1}{6.2 \times 10^{-10}}=\frac{\mathrm{S} \times \mathrm{S}}{10^{-3}}$ $\begin{array}{ll} \hline & S^2=\frac{2.2 \times 10^{-16} \times 10^{-3}}{6.2 \times 10^{-10}}=0.354 \times 10^{-9} \\ \text { or } & S^2=3.54 \times 10^{-10} \\ \text { or } & S=1.9 \times 10^{-5} \end{array}$
Shift-I
Ionic Equilibrium
229378
From the graph, the value of Henry's constant for the solubility of $\mathrm{HCl}$ gas in cyclohexane is
1 $10 \mathrm{k}$ torr
2 100 torr
3 50 torr
4 $2.4 \times 10^2$ torr
Explanation:
According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(\mathrm{X})$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law. $\mathrm{P}_{\text {solute }} \propto \mathrm{X}_{\text {solute }}$ in solution $\mathrm{P}_{\text {solute }}=\mathrm{k}_{\mathrm{H}} \mathrm{X}_{\text {solute }}$ in solution Where, $\mathrm{k}_{\mathrm{H}}=$ Henry's constant The slope of the straight line gives the value of $\mathrm{k}_{\mathrm{H}}$. $\begin{aligned} \therefore \quad \tan \theta & =\mathrm{k}_{\mathrm{H}}=\frac{50}{0.005} \\ & =\frac{50000}{5}=10000 \\ \mathrm{k}_{\mathrm{H}} & =10 \mathrm{k} \text { torr } \end{aligned}$
Shift-I
Ionic Equilibrium
229379
The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is : [Given : The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water $\left.=5.5 \times 10^{-6}\right]$
1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
Explanation:
Given: $\mathrm{K}_{\mathrm{sp}}=5.5 \times 10^{-6}$ Let, the solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is $\mathrm{S}$, $\begin{aligned} & \mathrm{Ca}(\mathrm{OH})_2 \square \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ & \text {(S) }(2 \mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{OH}^{-}\right]^2 \\ & \mathrm{~K}_{\mathrm{sp}}=(\mathrm{S})^1 \times(2 \mathrm{~S})^2 \\ & 5.5 \times 10^{-6}=4 \mathrm{~S}^3 \\ & S=\left(\frac{5.5 \times 10^{-6}}{4}\right)^{1 / 3} \\ & \mathrm{~S}=1.11 \times 10^{-2} \\ & \end{aligned}$
229374
If the concentration of $\mathrm{Ag}^{+}$ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CO}_3$ is $1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$. then find the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$.
1 $5.30 \times 10^{-12}$
2 $4.50 \times 10^{-11}$
3 $2.66 \times 10^{-12}$
4 $6.90 \times 10^{-12}$
Explanation:
Given that, $\left[\mathrm{Ag}^{+}\right]=1.20 \times 10^{-4} \mathrm{~mol} . \mathrm{L}^{-1}$ Silver carbonate dissociates as follows - $\mathrm{Ag}_2 \mathrm{CO}_3 \rightarrow 2 \mathrm{Ag}^{+}+\mathrm{CO}_3^{2-}$ Let the solubility is $\mathrm{S}$. The solubility of silver will be $2 \mathrm{~s}$ as two moles ion are dissociated - $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2\left[\mathrm{CO}_3^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \end{aligned}$ $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4 \times\left(1.20 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=6.91 \times 10^{-12} \end{aligned}$
Shift-I
Ionic Equilibrium
229376
The solubility of $\mathrm{AgCl}$ at $20^{\circ} \mathrm{C}$ is $1.435 \times 10^{-5} \mathrm{gL}^{-1}$. The solubility product of $\mathrm{AgCl}$ is
1 $2 \times 10^{-16}$
2 $108 \times 10^{-3}$
3 $1.0 \times 10^{-14}$
4 $1.035 \times 10^{-5}$
Explanation:
Solubility of $\mathrm{AgCl}=1.435 \times 10^{-5} \mathrm{gL}^{-1}$ [Molecular weight of $\mathrm{AgCl}=143.5$ ] We know that solubility of $\mathrm{AgCl}$ in moles $\begin{aligned} & \mathrm{AgCl} \square \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{Cl}^{-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(\mathrm{s})^2 \\ \because \quad & \mathrm{S}=\frac{\left(1.435 \times 10^{-5}\right)}{143.5}=1 \times 10^{-7} \end{aligned}$ Therefore solubility product of $\mathrm{AgCl}=\mathrm{S}^2$
COMEDK-2018
Ionic Equilibrium
229377
The solubility of $\mathrm{AgCN}$ in a buffer solution of $p H=3$ is $x$. The value of $x$ is : [Assume : No cyano complex is found; $\mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ and $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times$ $\left.10^{-10}\right]$
1 $0.625 \times 10^{-6}$
2 $1.6 \times 10^{-6}$
3 $2.2 \times 10^{-16}$
4 $1.9 \times 10^{-5}$
Explanation:
Given $: \mathrm{K}_{\mathrm{sp}}(\mathrm{AgCN})=2.2 \times 10^{-16}$ $\mathrm{K}_{\mathrm{a}}(\mathrm{HCN})=6.2 \times 10^{-10}$ The solubility of $\mathrm{AgCN}$ in a buffer solution of $\mathrm{pH}=3$ Let, solubility is $S$ then, $\mathrm{AgCN} \text { ? } \mathrm{Ag}^{+}+\mathrm{CN}^{-}$ $\mathrm{K}_{\mathrm{sp}}=\mathrm{S} \times \mathrm{S}=\mathrm{S}^2$ $\mathrm{H}^{+}+\mathrm{CN}^{-}$?? $\mathrm{HCN}, \quad \mathrm{K}=\frac{1}{\mathrm{~K}_{\mathrm{a}}}=\frac{1}{6.2 \times 10^{-10}}$ Therefore, $\mathrm{K}_{\mathrm{sp}} \times \frac{1}{\mathrm{~K}_{\mathrm{a}}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CN}^{-}\right] \frac{[\mathrm{HCN}]}{\left[\mathrm{H}^{+}\right]\left[\mathrm{CN}^{-}\right]}$ or $2.2 \times 10^{-16} \times \frac{1}{6.2 \times 10^{-10}}=\frac{\mathrm{S} \times \mathrm{S}}{10^{-3}}$ $\begin{array}{ll} \hline & S^2=\frac{2.2 \times 10^{-16} \times 10^{-3}}{6.2 \times 10^{-10}}=0.354 \times 10^{-9} \\ \text { or } & S^2=3.54 \times 10^{-10} \\ \text { or } & S=1.9 \times 10^{-5} \end{array}$
Shift-I
Ionic Equilibrium
229378
From the graph, the value of Henry's constant for the solubility of $\mathrm{HCl}$ gas in cyclohexane is
1 $10 \mathrm{k}$ torr
2 100 torr
3 50 torr
4 $2.4 \times 10^2$ torr
Explanation:
According to Henry's law, "The partial pressure of the gas in vapour phase (vapour pressure of the solute) is directly proportional to the mole fraction $(\mathrm{X})$ of the gaseous solute in the solution at low concentration. This statement is known as Henry's law. $\mathrm{P}_{\text {solute }} \propto \mathrm{X}_{\text {solute }}$ in solution $\mathrm{P}_{\text {solute }}=\mathrm{k}_{\mathrm{H}} \mathrm{X}_{\text {solute }}$ in solution Where, $\mathrm{k}_{\mathrm{H}}=$ Henry's constant The slope of the straight line gives the value of $\mathrm{k}_{\mathrm{H}}$. $\begin{aligned} \therefore \quad \tan \theta & =\mathrm{k}_{\mathrm{H}}=\frac{50}{0.005} \\ & =\frac{50000}{5}=10000 \\ \mathrm{k}_{\mathrm{H}} & =10 \mathrm{k} \text { torr } \end{aligned}$
Shift-I
Ionic Equilibrium
229379
The solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is : [Given : The solubility product of $\mathrm{Ca}(\mathrm{OH})_2$ in water $\left.=5.5 \times 10^{-6}\right]$
1 $1.11 \times 10^{-2}$
2 $1.11 \times 10^{-6}$
3 $1.77 \times 10^{-2}$
4 $1.77 \times 10^{-6}$
Explanation:
Given: $\mathrm{K}_{\mathrm{sp}}=5.5 \times 10^{-6}$ Let, the solubility of $\mathrm{Ca}(\mathrm{OH})_2$ in water is $\mathrm{S}$, $\begin{aligned} & \mathrm{Ca}(\mathrm{OH})_2 \square \mathrm{Ca}^{2+}+2 \mathrm{OH}^{-} \\ & \text {(S) }(2 \mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{OH}^{-}\right]^2 \\ & \mathrm{~K}_{\mathrm{sp}}=(\mathrm{S})^1 \times(2 \mathrm{~S})^2 \\ & 5.5 \times 10^{-6}=4 \mathrm{~S}^3 \\ & S=\left(\frac{5.5 \times 10^{-6}}{4}\right)^{1 / 3} \\ & \mathrm{~S}=1.11 \times 10^{-2} \\ & \end{aligned}$
