229394
What is the value of $K_{s p}$ for Bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$, Which has solubility of $1.0 \times 10^{-15} \mathrm{~mol} / \mathrm{L}$ at $25^{\circ} \mathrm{C}$ ?
229395
What is the minimum volume of water required to dissolve $1 \mathrm{~g}$ of calcium sulphate at $298 \mathrm{~K}$ ? $\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{CaSO}_4$ is $9.0 \times 10^{-6}$ (Molar mass of $\mathrm{CaSO}_4=136 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229397
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ni}(\mathrm{OH})_2$ is $2.0 \times 10^{-15} \mathrm{M}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.10 \mathrm{M} \mathrm{NaOH}$ is
1 $2.0 \times 10^{-15} \mathrm{M}$
2 $2.0 \times 10^{-13} \mathrm{M}$
3 $2.0 \times 10^{-11} \mathrm{M}$
4 $2.0 \times 10^{-9} \mathrm{M}$
Explanation:
$\mathrm{K}_{\mathrm{sp}}=2.0 \times 10^{-15} \mathrm{M}$ $\begin{aligned} & \mathrm{Ni}(\mathrm{OH})_2 \text { ?? } \mathrm{Ni}_{\mathrm{S}}^{\mathrm{N}^{2+}}+2 \mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ni}^{-2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & =[s][0.10+2 \mathrm{~s}]^2 \\ & \end{aligned}$ ( because total concentration of $\left[\mathrm{OH}^{-}\right]$ions $=[0.10+2 \mathrm{~s}]$ As $\mathrm{K}_{\mathrm{sp}}$ is small $2 \mathrm{~s} \ll<0.10$ Thus, $(0.10+2 \mathrm{~s}) \approx 0.10$ Hence, $2.0 \times 10^{-15}=\mathrm{s}(0.10)^2$ $\mathrm{S}=2.0 \times 10^{-13} \mathrm{M}=\left[\mathrm{Ni}^{2+}\right]$
AMU-2013
Ionic Equilibrium
229398
If the solubility of $\mathrm{Pr}_2$ is $\mathrm{S}$ g-mole per litre, its solubility product, considering it to be $80 \%$ ionized, is
229394
What is the value of $K_{s p}$ for Bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$, Which has solubility of $1.0 \times 10^{-15} \mathrm{~mol} / \mathrm{L}$ at $25^{\circ} \mathrm{C}$ ?
229395
What is the minimum volume of water required to dissolve $1 \mathrm{~g}$ of calcium sulphate at $298 \mathrm{~K}$ ? $\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{CaSO}_4$ is $9.0 \times 10^{-6}$ (Molar mass of $\mathrm{CaSO}_4=136 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229397
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ni}(\mathrm{OH})_2$ is $2.0 \times 10^{-15} \mathrm{M}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.10 \mathrm{M} \mathrm{NaOH}$ is
1 $2.0 \times 10^{-15} \mathrm{M}$
2 $2.0 \times 10^{-13} \mathrm{M}$
3 $2.0 \times 10^{-11} \mathrm{M}$
4 $2.0 \times 10^{-9} \mathrm{M}$
Explanation:
$\mathrm{K}_{\mathrm{sp}}=2.0 \times 10^{-15} \mathrm{M}$ $\begin{aligned} & \mathrm{Ni}(\mathrm{OH})_2 \text { ?? } \mathrm{Ni}_{\mathrm{S}}^{\mathrm{N}^{2+}}+2 \mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ni}^{-2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & =[s][0.10+2 \mathrm{~s}]^2 \\ & \end{aligned}$ ( because total concentration of $\left[\mathrm{OH}^{-}\right]$ions $=[0.10+2 \mathrm{~s}]$ As $\mathrm{K}_{\mathrm{sp}}$ is small $2 \mathrm{~s} \ll<0.10$ Thus, $(0.10+2 \mathrm{~s}) \approx 0.10$ Hence, $2.0 \times 10^{-15}=\mathrm{s}(0.10)^2$ $\mathrm{S}=2.0 \times 10^{-13} \mathrm{M}=\left[\mathrm{Ni}^{2+}\right]$
AMU-2013
Ionic Equilibrium
229398
If the solubility of $\mathrm{Pr}_2$ is $\mathrm{S}$ g-mole per litre, its solubility product, considering it to be $80 \%$ ionized, is
229394
What is the value of $K_{s p}$ for Bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$, Which has solubility of $1.0 \times 10^{-15} \mathrm{~mol} / \mathrm{L}$ at $25^{\circ} \mathrm{C}$ ?
229395
What is the minimum volume of water required to dissolve $1 \mathrm{~g}$ of calcium sulphate at $298 \mathrm{~K}$ ? $\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{CaSO}_4$ is $9.0 \times 10^{-6}$ (Molar mass of $\mathrm{CaSO}_4=136 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229397
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ni}(\mathrm{OH})_2$ is $2.0 \times 10^{-15} \mathrm{M}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.10 \mathrm{M} \mathrm{NaOH}$ is
1 $2.0 \times 10^{-15} \mathrm{M}$
2 $2.0 \times 10^{-13} \mathrm{M}$
3 $2.0 \times 10^{-11} \mathrm{M}$
4 $2.0 \times 10^{-9} \mathrm{M}$
Explanation:
$\mathrm{K}_{\mathrm{sp}}=2.0 \times 10^{-15} \mathrm{M}$ $\begin{aligned} & \mathrm{Ni}(\mathrm{OH})_2 \text { ?? } \mathrm{Ni}_{\mathrm{S}}^{\mathrm{N}^{2+}}+2 \mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ni}^{-2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & =[s][0.10+2 \mathrm{~s}]^2 \\ & \end{aligned}$ ( because total concentration of $\left[\mathrm{OH}^{-}\right]$ions $=[0.10+2 \mathrm{~s}]$ As $\mathrm{K}_{\mathrm{sp}}$ is small $2 \mathrm{~s} \ll<0.10$ Thus, $(0.10+2 \mathrm{~s}) \approx 0.10$ Hence, $2.0 \times 10^{-15}=\mathrm{s}(0.10)^2$ $\mathrm{S}=2.0 \times 10^{-13} \mathrm{M}=\left[\mathrm{Ni}^{2+}\right]$
AMU-2013
Ionic Equilibrium
229398
If the solubility of $\mathrm{Pr}_2$ is $\mathrm{S}$ g-mole per litre, its solubility product, considering it to be $80 \%$ ionized, is
229394
What is the value of $K_{s p}$ for Bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$, Which has solubility of $1.0 \times 10^{-15} \mathrm{~mol} / \mathrm{L}$ at $25^{\circ} \mathrm{C}$ ?
229395
What is the minimum volume of water required to dissolve $1 \mathrm{~g}$ of calcium sulphate at $298 \mathrm{~K}$ ? $\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{CaSO}_4$ is $9.0 \times 10^{-6}$ (Molar mass of $\mathrm{CaSO}_4=136 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229397
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ni}(\mathrm{OH})_2$ is $2.0 \times 10^{-15} \mathrm{M}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.10 \mathrm{M} \mathrm{NaOH}$ is
1 $2.0 \times 10^{-15} \mathrm{M}$
2 $2.0 \times 10^{-13} \mathrm{M}$
3 $2.0 \times 10^{-11} \mathrm{M}$
4 $2.0 \times 10^{-9} \mathrm{M}$
Explanation:
$\mathrm{K}_{\mathrm{sp}}=2.0 \times 10^{-15} \mathrm{M}$ $\begin{aligned} & \mathrm{Ni}(\mathrm{OH})_2 \text { ?? } \mathrm{Ni}_{\mathrm{S}}^{\mathrm{N}^{2+}}+2 \mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ni}^{-2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & =[s][0.10+2 \mathrm{~s}]^2 \\ & \end{aligned}$ ( because total concentration of $\left[\mathrm{OH}^{-}\right]$ions $=[0.10+2 \mathrm{~s}]$ As $\mathrm{K}_{\mathrm{sp}}$ is small $2 \mathrm{~s} \ll<0.10$ Thus, $(0.10+2 \mathrm{~s}) \approx 0.10$ Hence, $2.0 \times 10^{-15}=\mathrm{s}(0.10)^2$ $\mathrm{S}=2.0 \times 10^{-13} \mathrm{M}=\left[\mathrm{Ni}^{2+}\right]$
AMU-2013
Ionic Equilibrium
229398
If the solubility of $\mathrm{Pr}_2$ is $\mathrm{S}$ g-mole per litre, its solubility product, considering it to be $80 \%$ ionized, is
229394
What is the value of $K_{s p}$ for Bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$, Which has solubility of $1.0 \times 10^{-15} \mathrm{~mol} / \mathrm{L}$ at $25^{\circ} \mathrm{C}$ ?
229395
What is the minimum volume of water required to dissolve $1 \mathrm{~g}$ of calcium sulphate at $298 \mathrm{~K}$ ? $\mathrm{K}_{\mathrm{sp}}$ for $\mathrm{CaSO}_4$ is $9.0 \times 10^{-6}$ (Molar mass of $\mathrm{CaSO}_4=136 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229397
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ni}(\mathrm{OH})_2$ is $2.0 \times 10^{-15} \mathrm{M}$, the molar solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.10 \mathrm{M} \mathrm{NaOH}$ is
1 $2.0 \times 10^{-15} \mathrm{M}$
2 $2.0 \times 10^{-13} \mathrm{M}$
3 $2.0 \times 10^{-11} \mathrm{M}$
4 $2.0 \times 10^{-9} \mathrm{M}$
Explanation:
$\mathrm{K}_{\mathrm{sp}}=2.0 \times 10^{-15} \mathrm{M}$ $\begin{aligned} & \mathrm{Ni}(\mathrm{OH})_2 \text { ?? } \mathrm{Ni}_{\mathrm{S}}^{\mathrm{N}^{2+}}+2 \mathrm{OH}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ni}^{-2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ & =[s][0.10+2 \mathrm{~s}]^2 \\ & \end{aligned}$ ( because total concentration of $\left[\mathrm{OH}^{-}\right]$ions $=[0.10+2 \mathrm{~s}]$ As $\mathrm{K}_{\mathrm{sp}}$ is small $2 \mathrm{~s} \ll<0.10$ Thus, $(0.10+2 \mathrm{~s}) \approx 0.10$ Hence, $2.0 \times 10^{-15}=\mathrm{s}(0.10)^2$ $\mathrm{S}=2.0 \times 10^{-13} \mathrm{M}=\left[\mathrm{Ni}^{2+}\right]$
AMU-2013
Ionic Equilibrium
229398
If the solubility of $\mathrm{Pr}_2$ is $\mathrm{S}$ g-mole per litre, its solubility product, considering it to be $80 \%$ ionized, is
