NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229370
If the molar solubility (in mol.L ${ }^{-1}$ ) of a sparingly soluble salt $\mathrm{AB}_4$ is ' $S$ ', and the corresponding solubility product is ' $K_{\mathrm{sp}}$ ', then $S$ in terms of $K_{s p}$ is given by the relation
$\mathrm{AB}_4(\mathrm{aq}) \square \mathrm{A}^{+4}+4 \mathrm{~B}^{-}$ The molar solubility (in $\mathrm{mol} / lit.)$of a sparingly soluble salt $\mathrm{AB}_4$ is ' $\mathrm{S}$ '. The corresponding solubility product is $\mathrm{K}_{\mathrm{sp}}$ $\begin{array}{ll} \because & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+4}\right]\left[\mathrm{B}^{-}\right]^4 \\ & \mathrm{~K}_{\mathrm{sp}}=\mathrm{S}(4 \mathrm{~S})^4=256 \mathrm{~S}^5 \end{array}$ Hence, the molar solubility is $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$
Shift-II
Ionic Equilibrium
229371
The solubility product of $\mathrm{Ni}(\mathrm{OH})_2$ at $298 \mathrm{~K}$ is $2 \times 10^{-15} \mathrm{~mol}^3 \mathrm{dm}^{-9}$. The pH value if, its aqueous and saturated solution is
229372
The $\left[\mathrm{Ag}^{+}\right]$in a saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $1.5 \times 10^{-4}$ M. What is the solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
229370
If the molar solubility (in mol.L ${ }^{-1}$ ) of a sparingly soluble salt $\mathrm{AB}_4$ is ' $S$ ', and the corresponding solubility product is ' $K_{\mathrm{sp}}$ ', then $S$ in terms of $K_{s p}$ is given by the relation
$\mathrm{AB}_4(\mathrm{aq}) \square \mathrm{A}^{+4}+4 \mathrm{~B}^{-}$ The molar solubility (in $\mathrm{mol} / lit.)$of a sparingly soluble salt $\mathrm{AB}_4$ is ' $\mathrm{S}$ '. The corresponding solubility product is $\mathrm{K}_{\mathrm{sp}}$ $\begin{array}{ll} \because & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+4}\right]\left[\mathrm{B}^{-}\right]^4 \\ & \mathrm{~K}_{\mathrm{sp}}=\mathrm{S}(4 \mathrm{~S})^4=256 \mathrm{~S}^5 \end{array}$ Hence, the molar solubility is $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$
Shift-II
Ionic Equilibrium
229371
The solubility product of $\mathrm{Ni}(\mathrm{OH})_2$ at $298 \mathrm{~K}$ is $2 \times 10^{-15} \mathrm{~mol}^3 \mathrm{dm}^{-9}$. The pH value if, its aqueous and saturated solution is
229372
The $\left[\mathrm{Ag}^{+}\right]$in a saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $1.5 \times 10^{-4}$ M. What is the solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
229370
If the molar solubility (in mol.L ${ }^{-1}$ ) of a sparingly soluble salt $\mathrm{AB}_4$ is ' $S$ ', and the corresponding solubility product is ' $K_{\mathrm{sp}}$ ', then $S$ in terms of $K_{s p}$ is given by the relation
$\mathrm{AB}_4(\mathrm{aq}) \square \mathrm{A}^{+4}+4 \mathrm{~B}^{-}$ The molar solubility (in $\mathrm{mol} / lit.)$of a sparingly soluble salt $\mathrm{AB}_4$ is ' $\mathrm{S}$ '. The corresponding solubility product is $\mathrm{K}_{\mathrm{sp}}$ $\begin{array}{ll} \because & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+4}\right]\left[\mathrm{B}^{-}\right]^4 \\ & \mathrm{~K}_{\mathrm{sp}}=\mathrm{S}(4 \mathrm{~S})^4=256 \mathrm{~S}^5 \end{array}$ Hence, the molar solubility is $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$
Shift-II
Ionic Equilibrium
229371
The solubility product of $\mathrm{Ni}(\mathrm{OH})_2$ at $298 \mathrm{~K}$ is $2 \times 10^{-15} \mathrm{~mol}^3 \mathrm{dm}^{-9}$. The pH value if, its aqueous and saturated solution is
229372
The $\left[\mathrm{Ag}^{+}\right]$in a saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $1.5 \times 10^{-4}$ M. What is the solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
229370
If the molar solubility (in mol.L ${ }^{-1}$ ) of a sparingly soluble salt $\mathrm{AB}_4$ is ' $S$ ', and the corresponding solubility product is ' $K_{\mathrm{sp}}$ ', then $S$ in terms of $K_{s p}$ is given by the relation
$\mathrm{AB}_4(\mathrm{aq}) \square \mathrm{A}^{+4}+4 \mathrm{~B}^{-}$ The molar solubility (in $\mathrm{mol} / lit.)$of a sparingly soluble salt $\mathrm{AB}_4$ is ' $\mathrm{S}$ '. The corresponding solubility product is $\mathrm{K}_{\mathrm{sp}}$ $\begin{array}{ll} \because & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+4}\right]\left[\mathrm{B}^{-}\right]^4 \\ & \mathrm{~K}_{\mathrm{sp}}=\mathrm{S}(4 \mathrm{~S})^4=256 \mathrm{~S}^5 \end{array}$ Hence, the molar solubility is $\mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{256}\right)^{1 / 5}$
Shift-II
Ionic Equilibrium
229371
The solubility product of $\mathrm{Ni}(\mathrm{OH})_2$ at $298 \mathrm{~K}$ is $2 \times 10^{-15} \mathrm{~mol}^3 \mathrm{dm}^{-9}$. The pH value if, its aqueous and saturated solution is
229372
The $\left[\mathrm{Ag}^{+}\right]$in a saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $1.5 \times 10^{-4}$ M. What is the solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ ?
