229453
The solubility product of $\mathrm{PbBr}_2$ is $10.8 \times 10^{-5}$. It is $70 \%$ dissociated in saturated solution. The solubility of salt is:
1 $4.18 \times 10^{-2}$
2 $6.76 \times 10^{-3}$
3 $3.4 \times 10^{-4}$
4 $5.44 \times 10^{-2}$
Explanation:
Given, $\mathrm{K}_{\mathrm{sp}}=10.8 \times 10^{-5}$ $\therefore \quad \mathrm{K}_{\mathrm{sp}}^{\mathrm{a}-\mathrm{s}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(2 \mathrm{~S})^2$ or $\quad 10.8 \times 10^{-5}=4 \mathrm{~S}^3$ or $\quad \mathrm{S}=3 \times 10^{-2}$ This is the solubility for the $100 \%$ dissociation. According to the question, we calculate for $70 \%$. $\therefore$ For $70 \%$ dissociation $(\mathrm{S})=\frac{100}{70} \times 3 \times 10^{-2}$ $\mathrm{S} \approx 4.28 \times 10^{-2}$ $10 \%$ dissociated $=\frac{70}{100}=.7$ $\mathrm{PbBr}_2$ ?? $\mathrm{Pb}_{.7 \mathrm{~S}}^{2+}+\underset{2 \times .7 \mathrm{~S}}{2 \mathrm{Br}^{-}}$ $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right] .\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{spp}}=[.7 \mathrm{~S}] .(2 \times .7 \mathrm{~S})^2$ $10.8 \times 10^{-5} \stackrel{5 \mathrm{pp}}{=}(.7)^3 \cdot 4 \mathrm{~S}^3$ $\mathrm{S}=4.2 \times 10^{-2}$
JCECE - 2003
Ionic Equilibrium
229454
A saturated solution of $\mathrm{CaF}_2$ is $2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$. Its solubility product constant is:
229455
Solubility product of a salt $\mathrm{AB}$ is $1 \times 10^{-8} \mathrm{M}^2$ in a solution in which the concentration of $\mathrm{A}^{+}$ions is $10^{-3} \mathrm{M}$. The salt will precipitate when the concentration of $\mathrm{B}^{-}$ions is kept
1 between $10^{-8} \mathrm{M}$ to $10^{-7} \mathrm{M}$
2 between $10^{-7} \mathrm{M}$ to $10^{-8} \mathrm{M}$
3 $>10^{-5} \mathrm{M}$
4 $<10^{-8} \mathrm{M}$
Explanation:
Exp: Given, $\mathrm{K}_{\mathrm{sp}} \mathrm{AB}=1 \times 10^{-8} \mathrm{M}^2,\left[\mathrm{~A}^{+}\right]=10^{-3} \mathrm{M}$ The ionisation of AB salt is - $\begin{gathered} \mathrm{AB} \text { ? } \mathrm{A}^{+}+\mathrm{B}^{-} \\ \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right] \end{gathered}$ Salt will precipitate if ionic concentration $>\mathrm{K}_{\mathrm{sp}}$ therefore $\begin{aligned} & {\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{+}\right]>1 \times 10^{-8}} \\ & 1 \times 10^{-3}\left[\mathrm{~B}^{-}\right]>1 \times 10^{-8} \\ & {\left[\mathrm{~B}^{-}\right]>\frac{1 \times 10^{-8}}{1 \times 10^{-3}}} \\ & {\left[\mathrm{~B}^{-}\right]>1 \times 10^{-5} \mathrm{M}} \end{aligned}$ Hence, salt will precipitate when the concentration of $\mathrm{B}^{-}$have the value greater than the $10^{-5} \mathrm{M}$.
JCECE - 2009
Ionic Equilibrium
229456
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{2-}$ ions in that solution?
229453
The solubility product of $\mathrm{PbBr}_2$ is $10.8 \times 10^{-5}$. It is $70 \%$ dissociated in saturated solution. The solubility of salt is:
1 $4.18 \times 10^{-2}$
2 $6.76 \times 10^{-3}$
3 $3.4 \times 10^{-4}$
4 $5.44 \times 10^{-2}$
Explanation:
Given, $\mathrm{K}_{\mathrm{sp}}=10.8 \times 10^{-5}$ $\therefore \quad \mathrm{K}_{\mathrm{sp}}^{\mathrm{a}-\mathrm{s}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(2 \mathrm{~S})^2$ or $\quad 10.8 \times 10^{-5}=4 \mathrm{~S}^3$ or $\quad \mathrm{S}=3 \times 10^{-2}$ This is the solubility for the $100 \%$ dissociation. According to the question, we calculate for $70 \%$. $\therefore$ For $70 \%$ dissociation $(\mathrm{S})=\frac{100}{70} \times 3 \times 10^{-2}$ $\mathrm{S} \approx 4.28 \times 10^{-2}$ $10 \%$ dissociated $=\frac{70}{100}=.7$ $\mathrm{PbBr}_2$ ?? $\mathrm{Pb}_{.7 \mathrm{~S}}^{2+}+\underset{2 \times .7 \mathrm{~S}}{2 \mathrm{Br}^{-}}$ $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right] .\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{spp}}=[.7 \mathrm{~S}] .(2 \times .7 \mathrm{~S})^2$ $10.8 \times 10^{-5} \stackrel{5 \mathrm{pp}}{=}(.7)^3 \cdot 4 \mathrm{~S}^3$ $\mathrm{S}=4.2 \times 10^{-2}$
JCECE - 2003
Ionic Equilibrium
229454
A saturated solution of $\mathrm{CaF}_2$ is $2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$. Its solubility product constant is:
229455
Solubility product of a salt $\mathrm{AB}$ is $1 \times 10^{-8} \mathrm{M}^2$ in a solution in which the concentration of $\mathrm{A}^{+}$ions is $10^{-3} \mathrm{M}$. The salt will precipitate when the concentration of $\mathrm{B}^{-}$ions is kept
1 between $10^{-8} \mathrm{M}$ to $10^{-7} \mathrm{M}$
2 between $10^{-7} \mathrm{M}$ to $10^{-8} \mathrm{M}$
3 $>10^{-5} \mathrm{M}$
4 $<10^{-8} \mathrm{M}$
Explanation:
Exp: Given, $\mathrm{K}_{\mathrm{sp}} \mathrm{AB}=1 \times 10^{-8} \mathrm{M}^2,\left[\mathrm{~A}^{+}\right]=10^{-3} \mathrm{M}$ The ionisation of AB salt is - $\begin{gathered} \mathrm{AB} \text { ? } \mathrm{A}^{+}+\mathrm{B}^{-} \\ \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right] \end{gathered}$ Salt will precipitate if ionic concentration $>\mathrm{K}_{\mathrm{sp}}$ therefore $\begin{aligned} & {\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{+}\right]>1 \times 10^{-8}} \\ & 1 \times 10^{-3}\left[\mathrm{~B}^{-}\right]>1 \times 10^{-8} \\ & {\left[\mathrm{~B}^{-}\right]>\frac{1 \times 10^{-8}}{1 \times 10^{-3}}} \\ & {\left[\mathrm{~B}^{-}\right]>1 \times 10^{-5} \mathrm{M}} \end{aligned}$ Hence, salt will precipitate when the concentration of $\mathrm{B}^{-}$have the value greater than the $10^{-5} \mathrm{M}$.
JCECE - 2009
Ionic Equilibrium
229456
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{2-}$ ions in that solution?
229453
The solubility product of $\mathrm{PbBr}_2$ is $10.8 \times 10^{-5}$. It is $70 \%$ dissociated in saturated solution. The solubility of salt is:
1 $4.18 \times 10^{-2}$
2 $6.76 \times 10^{-3}$
3 $3.4 \times 10^{-4}$
4 $5.44 \times 10^{-2}$
Explanation:
Given, $\mathrm{K}_{\mathrm{sp}}=10.8 \times 10^{-5}$ $\therefore \quad \mathrm{K}_{\mathrm{sp}}^{\mathrm{a}-\mathrm{s}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(2 \mathrm{~S})^2$ or $\quad 10.8 \times 10^{-5}=4 \mathrm{~S}^3$ or $\quad \mathrm{S}=3 \times 10^{-2}$ This is the solubility for the $100 \%$ dissociation. According to the question, we calculate for $70 \%$. $\therefore$ For $70 \%$ dissociation $(\mathrm{S})=\frac{100}{70} \times 3 \times 10^{-2}$ $\mathrm{S} \approx 4.28 \times 10^{-2}$ $10 \%$ dissociated $=\frac{70}{100}=.7$ $\mathrm{PbBr}_2$ ?? $\mathrm{Pb}_{.7 \mathrm{~S}}^{2+}+\underset{2 \times .7 \mathrm{~S}}{2 \mathrm{Br}^{-}}$ $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right] .\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{spp}}=[.7 \mathrm{~S}] .(2 \times .7 \mathrm{~S})^2$ $10.8 \times 10^{-5} \stackrel{5 \mathrm{pp}}{=}(.7)^3 \cdot 4 \mathrm{~S}^3$ $\mathrm{S}=4.2 \times 10^{-2}$
JCECE - 2003
Ionic Equilibrium
229454
A saturated solution of $\mathrm{CaF}_2$ is $2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$. Its solubility product constant is:
229455
Solubility product of a salt $\mathrm{AB}$ is $1 \times 10^{-8} \mathrm{M}^2$ in a solution in which the concentration of $\mathrm{A}^{+}$ions is $10^{-3} \mathrm{M}$. The salt will precipitate when the concentration of $\mathrm{B}^{-}$ions is kept
1 between $10^{-8} \mathrm{M}$ to $10^{-7} \mathrm{M}$
2 between $10^{-7} \mathrm{M}$ to $10^{-8} \mathrm{M}$
3 $>10^{-5} \mathrm{M}$
4 $<10^{-8} \mathrm{M}$
Explanation:
Exp: Given, $\mathrm{K}_{\mathrm{sp}} \mathrm{AB}=1 \times 10^{-8} \mathrm{M}^2,\left[\mathrm{~A}^{+}\right]=10^{-3} \mathrm{M}$ The ionisation of AB salt is - $\begin{gathered} \mathrm{AB} \text { ? } \mathrm{A}^{+}+\mathrm{B}^{-} \\ \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right] \end{gathered}$ Salt will precipitate if ionic concentration $>\mathrm{K}_{\mathrm{sp}}$ therefore $\begin{aligned} & {\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{+}\right]>1 \times 10^{-8}} \\ & 1 \times 10^{-3}\left[\mathrm{~B}^{-}\right]>1 \times 10^{-8} \\ & {\left[\mathrm{~B}^{-}\right]>\frac{1 \times 10^{-8}}{1 \times 10^{-3}}} \\ & {\left[\mathrm{~B}^{-}\right]>1 \times 10^{-5} \mathrm{M}} \end{aligned}$ Hence, salt will precipitate when the concentration of $\mathrm{B}^{-}$have the value greater than the $10^{-5} \mathrm{M}$.
JCECE - 2009
Ionic Equilibrium
229456
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{2-}$ ions in that solution?
229453
The solubility product of $\mathrm{PbBr}_2$ is $10.8 \times 10^{-5}$. It is $70 \%$ dissociated in saturated solution. The solubility of salt is:
1 $4.18 \times 10^{-2}$
2 $6.76 \times 10^{-3}$
3 $3.4 \times 10^{-4}$
4 $5.44 \times 10^{-2}$
Explanation:
Given, $\mathrm{K}_{\mathrm{sp}}=10.8 \times 10^{-5}$ $\therefore \quad \mathrm{K}_{\mathrm{sp}}^{\mathrm{a}-\mathrm{s}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(2 \mathrm{~S})^2$ or $\quad 10.8 \times 10^{-5}=4 \mathrm{~S}^3$ or $\quad \mathrm{S}=3 \times 10^{-2}$ This is the solubility for the $100 \%$ dissociation. According to the question, we calculate for $70 \%$. $\therefore$ For $70 \%$ dissociation $(\mathrm{S})=\frac{100}{70} \times 3 \times 10^{-2}$ $\mathrm{S} \approx 4.28 \times 10^{-2}$ $10 \%$ dissociated $=\frac{70}{100}=.7$ $\mathrm{PbBr}_2$ ?? $\mathrm{Pb}_{.7 \mathrm{~S}}^{2+}+\underset{2 \times .7 \mathrm{~S}}{2 \mathrm{Br}^{-}}$ $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right] .\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{spp}}=[.7 \mathrm{~S}] .(2 \times .7 \mathrm{~S})^2$ $10.8 \times 10^{-5} \stackrel{5 \mathrm{pp}}{=}(.7)^3 \cdot 4 \mathrm{~S}^3$ $\mathrm{S}=4.2 \times 10^{-2}$
JCECE - 2003
Ionic Equilibrium
229454
A saturated solution of $\mathrm{CaF}_2$ is $2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$. Its solubility product constant is:
229455
Solubility product of a salt $\mathrm{AB}$ is $1 \times 10^{-8} \mathrm{M}^2$ in a solution in which the concentration of $\mathrm{A}^{+}$ions is $10^{-3} \mathrm{M}$. The salt will precipitate when the concentration of $\mathrm{B}^{-}$ions is kept
1 between $10^{-8} \mathrm{M}$ to $10^{-7} \mathrm{M}$
2 between $10^{-7} \mathrm{M}$ to $10^{-8} \mathrm{M}$
3 $>10^{-5} \mathrm{M}$
4 $<10^{-8} \mathrm{M}$
Explanation:
Exp: Given, $\mathrm{K}_{\mathrm{sp}} \mathrm{AB}=1 \times 10^{-8} \mathrm{M}^2,\left[\mathrm{~A}^{+}\right]=10^{-3} \mathrm{M}$ The ionisation of AB salt is - $\begin{gathered} \mathrm{AB} \text { ? } \mathrm{A}^{+}+\mathrm{B}^{-} \\ \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right] \end{gathered}$ Salt will precipitate if ionic concentration $>\mathrm{K}_{\mathrm{sp}}$ therefore $\begin{aligned} & {\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{+}\right]>1 \times 10^{-8}} \\ & 1 \times 10^{-3}\left[\mathrm{~B}^{-}\right]>1 \times 10^{-8} \\ & {\left[\mathrm{~B}^{-}\right]>\frac{1 \times 10^{-8}}{1 \times 10^{-3}}} \\ & {\left[\mathrm{~B}^{-}\right]>1 \times 10^{-5} \mathrm{M}} \end{aligned}$ Hence, salt will precipitate when the concentration of $\mathrm{B}^{-}$have the value greater than the $10^{-5} \mathrm{M}$.
JCECE - 2009
Ionic Equilibrium
229456
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{2-}$ ions in that solution?
229453
The solubility product of $\mathrm{PbBr}_2$ is $10.8 \times 10^{-5}$. It is $70 \%$ dissociated in saturated solution. The solubility of salt is:
1 $4.18 \times 10^{-2}$
2 $6.76 \times 10^{-3}$
3 $3.4 \times 10^{-4}$
4 $5.44 \times 10^{-2}$
Explanation:
Given, $\mathrm{K}_{\mathrm{sp}}=10.8 \times 10^{-5}$ $\therefore \quad \mathrm{K}_{\mathrm{sp}}^{\mathrm{a}-\mathrm{s}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{sp}}=(\mathrm{S})(2 \mathrm{~S})^2$ or $\quad 10.8 \times 10^{-5}=4 \mathrm{~S}^3$ or $\quad \mathrm{S}=3 \times 10^{-2}$ This is the solubility for the $100 \%$ dissociation. According to the question, we calculate for $70 \%$. $\therefore$ For $70 \%$ dissociation $(\mathrm{S})=\frac{100}{70} \times 3 \times 10^{-2}$ $\mathrm{S} \approx 4.28 \times 10^{-2}$ $10 \%$ dissociated $=\frac{70}{100}=.7$ $\mathrm{PbBr}_2$ ?? $\mathrm{Pb}_{.7 \mathrm{~S}}^{2+}+\underset{2 \times .7 \mathrm{~S}}{2 \mathrm{Br}^{-}}$ $\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right] .\left[\mathrm{Br}^{-}\right]^2$ $\mathrm{K}_{\mathrm{spp}}=[.7 \mathrm{~S}] .(2 \times .7 \mathrm{~S})^2$ $10.8 \times 10^{-5} \stackrel{5 \mathrm{pp}}{=}(.7)^3 \cdot 4 \mathrm{~S}^3$ $\mathrm{S}=4.2 \times 10^{-2}$
JCECE - 2003
Ionic Equilibrium
229454
A saturated solution of $\mathrm{CaF}_2$ is $2 \times 10^{-4} \mathrm{~mol} / \mathrm{L}$. Its solubility product constant is:
229455
Solubility product of a salt $\mathrm{AB}$ is $1 \times 10^{-8} \mathrm{M}^2$ in a solution in which the concentration of $\mathrm{A}^{+}$ions is $10^{-3} \mathrm{M}$. The salt will precipitate when the concentration of $\mathrm{B}^{-}$ions is kept
1 between $10^{-8} \mathrm{M}$ to $10^{-7} \mathrm{M}$
2 between $10^{-7} \mathrm{M}$ to $10^{-8} \mathrm{M}$
3 $>10^{-5} \mathrm{M}$
4 $<10^{-8} \mathrm{M}$
Explanation:
Exp: Given, $\mathrm{K}_{\mathrm{sp}} \mathrm{AB}=1 \times 10^{-8} \mathrm{M}^2,\left[\mathrm{~A}^{+}\right]=10^{-3} \mathrm{M}$ The ionisation of AB salt is - $\begin{gathered} \mathrm{AB} \text { ? } \mathrm{A}^{+}+\mathrm{B}^{-} \\ \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{-}\right] \end{gathered}$ Salt will precipitate if ionic concentration $>\mathrm{K}_{\mathrm{sp}}$ therefore $\begin{aligned} & {\left[\mathrm{A}^{+}\right]\left[\mathrm{B}^{+}\right]>1 \times 10^{-8}} \\ & 1 \times 10^{-3}\left[\mathrm{~B}^{-}\right]>1 \times 10^{-8} \\ & {\left[\mathrm{~B}^{-}\right]>\frac{1 \times 10^{-8}}{1 \times 10^{-3}}} \\ & {\left[\mathrm{~B}^{-}\right]>1 \times 10^{-5} \mathrm{M}} \end{aligned}$ Hence, salt will precipitate when the concentration of $\mathrm{B}^{-}$have the value greater than the $10^{-5} \mathrm{M}$.
JCECE - 2009
Ionic Equilibrium
229456
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $32 \times 10^{-12}$. What is the concentration of $\mathrm{CrO}_4^{2-}$ ions in that solution?