229490
$\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_2$ is 9 . The solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ca}(\mathrm{OH})_2$ is
229491
The molar solubility of $\mathrm{CaF}_2\left(\mathrm{~K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of $\mathrm{NaF}$ will be
$\begin{aligned} \mathrm{CaF}_2(\mathrm{~s}) \square \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-} \text {(aq) } \\ \mathrm{NaF}(\mathrm{aq}) \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})\end{aligned}$ In solution $-[F]=\left(2 s^{\prime}+C\right)$ $[\mathrm{F}] \approx \mathrm{C}$ (due to common ion effect) $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=[\mathrm{Ca}+2] \cdot[\mathrm{F}]^2$ $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=\mathrm{s}^{\prime} . \mathrm{C}^2$ $s^{\prime}=\frac{5.3 \times 10^{-11}}{\left(10^{-1}\right)^2}$ $s^{\prime}=5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$
Odisha NEET-2019
Ionic Equilibrium
229492
The solubility of $\mathrm{BaSO}_4$ in water is $2.42 \times 10^{-3} \mathrm{~g}$ $\mathrm{L}^{-1}$ at $298 \mathrm{~K}$. The value of its solubility product $\left(K_{\mathrm{sp}}\right.$ ) will be (Given molar mass of $\mathrm{BaSO}_4=233 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229493
Concentration of the $\mathrm{Ag}^{+}$ions in a saturated solution of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2.2 \times 10^{-4}$ mol $\mathrm{L}^{-1}$. Solubility product of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is
229490
$\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_2$ is 9 . The solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ca}(\mathrm{OH})_2$ is
229491
The molar solubility of $\mathrm{CaF}_2\left(\mathrm{~K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of $\mathrm{NaF}$ will be
$\begin{aligned} \mathrm{CaF}_2(\mathrm{~s}) \square \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-} \text {(aq) } \\ \mathrm{NaF}(\mathrm{aq}) \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})\end{aligned}$ In solution $-[F]=\left(2 s^{\prime}+C\right)$ $[\mathrm{F}] \approx \mathrm{C}$ (due to common ion effect) $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=[\mathrm{Ca}+2] \cdot[\mathrm{F}]^2$ $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=\mathrm{s}^{\prime} . \mathrm{C}^2$ $s^{\prime}=\frac{5.3 \times 10^{-11}}{\left(10^{-1}\right)^2}$ $s^{\prime}=5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$
Odisha NEET-2019
Ionic Equilibrium
229492
The solubility of $\mathrm{BaSO}_4$ in water is $2.42 \times 10^{-3} \mathrm{~g}$ $\mathrm{L}^{-1}$ at $298 \mathrm{~K}$. The value of its solubility product $\left(K_{\mathrm{sp}}\right.$ ) will be (Given molar mass of $\mathrm{BaSO}_4=233 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229493
Concentration of the $\mathrm{Ag}^{+}$ions in a saturated solution of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2.2 \times 10^{-4}$ mol $\mathrm{L}^{-1}$. Solubility product of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is
229490
$\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_2$ is 9 . The solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ca}(\mathrm{OH})_2$ is
229491
The molar solubility of $\mathrm{CaF}_2\left(\mathrm{~K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of $\mathrm{NaF}$ will be
$\begin{aligned} \mathrm{CaF}_2(\mathrm{~s}) \square \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-} \text {(aq) } \\ \mathrm{NaF}(\mathrm{aq}) \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})\end{aligned}$ In solution $-[F]=\left(2 s^{\prime}+C\right)$ $[\mathrm{F}] \approx \mathrm{C}$ (due to common ion effect) $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=[\mathrm{Ca}+2] \cdot[\mathrm{F}]^2$ $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=\mathrm{s}^{\prime} . \mathrm{C}^2$ $s^{\prime}=\frac{5.3 \times 10^{-11}}{\left(10^{-1}\right)^2}$ $s^{\prime}=5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$
Odisha NEET-2019
Ionic Equilibrium
229492
The solubility of $\mathrm{BaSO}_4$ in water is $2.42 \times 10^{-3} \mathrm{~g}$ $\mathrm{L}^{-1}$ at $298 \mathrm{~K}$. The value of its solubility product $\left(K_{\mathrm{sp}}\right.$ ) will be (Given molar mass of $\mathrm{BaSO}_4=233 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229493
Concentration of the $\mathrm{Ag}^{+}$ions in a saturated solution of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2.2 \times 10^{-4}$ mol $\mathrm{L}^{-1}$. Solubility product of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is
229490
$\mathrm{pH}$ of a saturated solution of $\mathrm{Ca}(\mathrm{OH})_2$ is 9 . The solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ca}(\mathrm{OH})_2$ is
229491
The molar solubility of $\mathrm{CaF}_2\left(\mathrm{~K}_{\mathrm{sp}}=5.3 \times 10^{-11}\right)$ in $0.1 \mathrm{M}$ solution of $\mathrm{NaF}$ will be
$\begin{aligned} \mathrm{CaF}_2(\mathrm{~s}) \square \mathrm{Ca}^{2+}(\mathrm{aq})+2 \mathrm{~F}^{-} \text {(aq) } \\ \mathrm{NaF}(\mathrm{aq}) \rightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{F}^{-}(\mathrm{aq})\end{aligned}$ In solution $-[F]=\left(2 s^{\prime}+C\right)$ $[\mathrm{F}] \approx \mathrm{C}$ (due to common ion effect) $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=[\mathrm{Ca}+2] \cdot[\mathrm{F}]^2$ $\mathrm{K}_{\mathrm{sp}}\left(\mathrm{CaF}_2\right)=\mathrm{s}^{\prime} . \mathrm{C}^2$ $s^{\prime}=\frac{5.3 \times 10^{-11}}{\left(10^{-1}\right)^2}$ $s^{\prime}=5.3 \times 10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}$
Odisha NEET-2019
Ionic Equilibrium
229492
The solubility of $\mathrm{BaSO}_4$ in water is $2.42 \times 10^{-3} \mathrm{~g}$ $\mathrm{L}^{-1}$ at $298 \mathrm{~K}$. The value of its solubility product $\left(K_{\mathrm{sp}}\right.$ ) will be (Given molar mass of $\mathrm{BaSO}_4=233 \mathrm{~g} \mathrm{~mol}^{-1}$ )
229493
Concentration of the $\mathrm{Ag}^{+}$ions in a saturated solution of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is $2.2 \times 10^{-4}$ mol $\mathrm{L}^{-1}$. Solubility product of $\mathrm{Ag}_2 \mathrm{C}_2 \mathrm{O}_4$ is
