229485
Solubility of $\mathbf{M X}_2$ type electrolytes is $0.5 \times 10^{-4}$ $\mathbf{m o l} /$ lit, then find out $\mathbf{K}_{\text {sp }}$ of electrolytes.
1 $5 \times 10^{-12}$
2 $25 \times 10^{-10}$
3 $1 \times 10^{-13}$
4 $5 \times 10^{-13}$
Explanation:
Exp: An electrolyte MX dissocitation andergoes aollows:- {|l|l|l|l|} | Concentration | $_2$ | $^{-2}$ | $^{-}$ | |---|---|---|---| |${l} { Initial } | | { concentration }$ | 1 | 0 | 0 | |${l} { Concentration } | | { at Equilibrium }$ | $1-$ | $$ | $2 $ | |Thus from the above condition we can say that, $\mathrm{K}_{\mathrm{sp}}=$ $\mathrm{s} \times(2 \mathrm{~s})^2=4 \times(s)^3$ Here, $s$ (the solubility) is $0.5 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$. $\begin{aligned} & \therefore \mathrm{K}_{55}=4 \times\left(0.5 \times 10^{-4}\right)^{-3} \\ & \mathrm{~K}_{\mathrm{sp}}=5 \times 10^{-13} \end{aligned}$
NEET-2002
Ionic Equilibrium
229486
Solubility of $M_2 \mathrm{~S}$ salt is $3.5 \times 10^{-6}$ then find out solubility product,
1 $1.7 \times 10^{-6}$
2 $1.7 \times 10^{-16}$
3 $1.7 \times 10^{-18}$
4 $1.7 \times 10^{-12}$
Explanation:
Let's be the solubility of salt $\mathrm{M}_2 \mathrm{~S}$ which undergoes dissociation as fallow:at aqulibrium $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{+}\right]_2^2\left[\mathrm{~S}^{-2}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & K_{\text {5p }}=4 \times\left(3.5 \times 10^{-6}\right)^3 \\ & K_{5 p}=1.7 \times 10^{-16} \\ & \end{aligned}$
NEET-2001
Ionic Equilibrium
229488
The solubility product of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}^{\mathrm{S}}$ are $10^{-31}, 10^{-f 4}$ and $10^{-54}$ respectively. The solubilities of these sulphides are in the order
229489
$\mathrm{pH}$ of a saturated solution if $\mathrm{Ba}(\mathrm{OH})_2$ is 12. The value of solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ba}(\mathrm{OH})_2$ is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229485
Solubility of $\mathbf{M X}_2$ type electrolytes is $0.5 \times 10^{-4}$ $\mathbf{m o l} /$ lit, then find out $\mathbf{K}_{\text {sp }}$ of electrolytes.
1 $5 \times 10^{-12}$
2 $25 \times 10^{-10}$
3 $1 \times 10^{-13}$
4 $5 \times 10^{-13}$
Explanation:
Exp: An electrolyte MX dissocitation andergoes aollows:- {|l|l|l|l|} | Concentration | $_2$ | $^{-2}$ | $^{-}$ | |---|---|---|---| |${l} { Initial } | | { concentration }$ | 1 | 0 | 0 | |${l} { Concentration } | | { at Equilibrium }$ | $1-$ | $$ | $2 $ | |Thus from the above condition we can say that, $\mathrm{K}_{\mathrm{sp}}=$ $\mathrm{s} \times(2 \mathrm{~s})^2=4 \times(s)^3$ Here, $s$ (the solubility) is $0.5 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$. $\begin{aligned} & \therefore \mathrm{K}_{55}=4 \times\left(0.5 \times 10^{-4}\right)^{-3} \\ & \mathrm{~K}_{\mathrm{sp}}=5 \times 10^{-13} \end{aligned}$
NEET-2002
Ionic Equilibrium
229486
Solubility of $M_2 \mathrm{~S}$ salt is $3.5 \times 10^{-6}$ then find out solubility product,
1 $1.7 \times 10^{-6}$
2 $1.7 \times 10^{-16}$
3 $1.7 \times 10^{-18}$
4 $1.7 \times 10^{-12}$
Explanation:
Let's be the solubility of salt $\mathrm{M}_2 \mathrm{~S}$ which undergoes dissociation as fallow:at aqulibrium $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{+}\right]_2^2\left[\mathrm{~S}^{-2}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & K_{\text {5p }}=4 \times\left(3.5 \times 10^{-6}\right)^3 \\ & K_{5 p}=1.7 \times 10^{-16} \\ & \end{aligned}$
NEET-2001
Ionic Equilibrium
229488
The solubility product of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}^{\mathrm{S}}$ are $10^{-31}, 10^{-f 4}$ and $10^{-54}$ respectively. The solubilities of these sulphides are in the order
229489
$\mathrm{pH}$ of a saturated solution if $\mathrm{Ba}(\mathrm{OH})_2$ is 12. The value of solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ba}(\mathrm{OH})_2$ is
229485
Solubility of $\mathbf{M X}_2$ type electrolytes is $0.5 \times 10^{-4}$ $\mathbf{m o l} /$ lit, then find out $\mathbf{K}_{\text {sp }}$ of electrolytes.
1 $5 \times 10^{-12}$
2 $25 \times 10^{-10}$
3 $1 \times 10^{-13}$
4 $5 \times 10^{-13}$
Explanation:
Exp: An electrolyte MX dissocitation andergoes aollows:- {|l|l|l|l|} | Concentration | $_2$ | $^{-2}$ | $^{-}$ | |---|---|---|---| |${l} { Initial } | | { concentration }$ | 1 | 0 | 0 | |${l} { Concentration } | | { at Equilibrium }$ | $1-$ | $$ | $2 $ | |Thus from the above condition we can say that, $\mathrm{K}_{\mathrm{sp}}=$ $\mathrm{s} \times(2 \mathrm{~s})^2=4 \times(s)^3$ Here, $s$ (the solubility) is $0.5 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$. $\begin{aligned} & \therefore \mathrm{K}_{55}=4 \times\left(0.5 \times 10^{-4}\right)^{-3} \\ & \mathrm{~K}_{\mathrm{sp}}=5 \times 10^{-13} \end{aligned}$
NEET-2002
Ionic Equilibrium
229486
Solubility of $M_2 \mathrm{~S}$ salt is $3.5 \times 10^{-6}$ then find out solubility product,
1 $1.7 \times 10^{-6}$
2 $1.7 \times 10^{-16}$
3 $1.7 \times 10^{-18}$
4 $1.7 \times 10^{-12}$
Explanation:
Let's be the solubility of salt $\mathrm{M}_2 \mathrm{~S}$ which undergoes dissociation as fallow:at aqulibrium $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{+}\right]_2^2\left[\mathrm{~S}^{-2}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & K_{\text {5p }}=4 \times\left(3.5 \times 10^{-6}\right)^3 \\ & K_{5 p}=1.7 \times 10^{-16} \\ & \end{aligned}$
NEET-2001
Ionic Equilibrium
229488
The solubility product of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}^{\mathrm{S}}$ are $10^{-31}, 10^{-f 4}$ and $10^{-54}$ respectively. The solubilities of these sulphides are in the order
229489
$\mathrm{pH}$ of a saturated solution if $\mathrm{Ba}(\mathrm{OH})_2$ is 12. The value of solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ba}(\mathrm{OH})_2$ is
229485
Solubility of $\mathbf{M X}_2$ type electrolytes is $0.5 \times 10^{-4}$ $\mathbf{m o l} /$ lit, then find out $\mathbf{K}_{\text {sp }}$ of electrolytes.
1 $5 \times 10^{-12}$
2 $25 \times 10^{-10}$
3 $1 \times 10^{-13}$
4 $5 \times 10^{-13}$
Explanation:
Exp: An electrolyte MX dissocitation andergoes aollows:- {|l|l|l|l|} | Concentration | $_2$ | $^{-2}$ | $^{-}$ | |---|---|---|---| |${l} { Initial } | | { concentration }$ | 1 | 0 | 0 | |${l} { Concentration } | | { at Equilibrium }$ | $1-$ | $$ | $2 $ | |Thus from the above condition we can say that, $\mathrm{K}_{\mathrm{sp}}=$ $\mathrm{s} \times(2 \mathrm{~s})^2=4 \times(s)^3$ Here, $s$ (the solubility) is $0.5 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}$. $\begin{aligned} & \therefore \mathrm{K}_{55}=4 \times\left(0.5 \times 10^{-4}\right)^{-3} \\ & \mathrm{~K}_{\mathrm{sp}}=5 \times 10^{-13} \end{aligned}$
NEET-2002
Ionic Equilibrium
229486
Solubility of $M_2 \mathrm{~S}$ salt is $3.5 \times 10^{-6}$ then find out solubility product,
1 $1.7 \times 10^{-6}$
2 $1.7 \times 10^{-16}$
3 $1.7 \times 10^{-18}$
4 $1.7 \times 10^{-12}$
Explanation:
Let's be the solubility of salt $\mathrm{M}_2 \mathrm{~S}$ which undergoes dissociation as fallow:at aqulibrium $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{+}\right]_2^2\left[\mathrm{~S}^{-2}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~S})^2(\mathrm{~S}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~S}^3 \\ & K_{\text {5p }}=4 \times\left(3.5 \times 10^{-6}\right)^3 \\ & K_{5 p}=1.7 \times 10^{-16} \\ & \end{aligned}$
NEET-2001
Ionic Equilibrium
229488
The solubility product of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}^{\mathrm{S}}$ are $10^{-31}, 10^{-f 4}$ and $10^{-54}$ respectively. The solubilities of these sulphides are in the order
229489
$\mathrm{pH}$ of a saturated solution if $\mathrm{Ba}(\mathrm{OH})_2$ is 12. The value of solubility product $\left(\mathrm{K}_{\mathrm{sp}}\right)$ of $\mathrm{Ba}(\mathrm{OH})_2$ is