229413
The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}$ are $1^{-31}, 1^{-44}$ and $10^{-54}$ respectively. The solubility of these sulphides are in the order :
229415
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{~S}$ is $10^{-17}$, the solubility of $\mathrm{Ag}_2 \mathrm{~S}$ in $0.1 \mathrm{M}$ solution of $\mathrm{Na}_2 \mathrm{~S}$ will be
229416
If solubility of $\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{S}$, its solubility product will be:
1 $108 \mathrm{~S}^5$
2 $27 \mathrm{~S}^3$
3 $4 \mathrm{~S}^4$
4 $27 \mathrm{~S}^4$
Explanation:
$\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{AB}_3$ Type salt its solubility product will be- $\begin{aligned} & \begin{array}{r} \mathrm{M}(\mathrm{OH})_3 \text { ?? } \\ \mathrm{M}^{3+}+3 \mathrm{OH}^{-} \\ \mathrm{S} \end{array} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{3+}\right] \cdot\left[\mathrm{OH}^{-}\right]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=[\mathrm{S}] \cdot[3 \mathrm{~S}]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\ & \end{aligned}$
CG PET-2019
Ionic Equilibrium
229417
The product of $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$of water will be
1 $\mathrm{K}_{\mathrm{w}}=10^{-12}$
2 $\mathrm{K}_{\mathrm{w}}=10^{-14}$
3 $\mathrm{K}_{\mathrm{w}}=10^{-11}$
4 $\mathrm{K}_{\mathrm{w}}=10^{-10}$
Explanation:
The product of the molar concentration of $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water. At $298 \mathrm{~K}$, For pure water $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-7} \mathrm{~mol} / \mathrm{dm}^{-3}} \\ & \mathrm{~K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ & =\left[1 \times 10^{-7}\right] .\left[1 \times 10^{-7}\right] \\ & \mathrm{K}_{\mathrm{w}}=1 \times 10^{-14} \\ & \end{aligned}$
229413
The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}$ are $1^{-31}, 1^{-44}$ and $10^{-54}$ respectively. The solubility of these sulphides are in the order :
229415
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{~S}$ is $10^{-17}$, the solubility of $\mathrm{Ag}_2 \mathrm{~S}$ in $0.1 \mathrm{M}$ solution of $\mathrm{Na}_2 \mathrm{~S}$ will be
229416
If solubility of $\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{S}$, its solubility product will be:
1 $108 \mathrm{~S}^5$
2 $27 \mathrm{~S}^3$
3 $4 \mathrm{~S}^4$
4 $27 \mathrm{~S}^4$
Explanation:
$\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{AB}_3$ Type salt its solubility product will be- $\begin{aligned} & \begin{array}{r} \mathrm{M}(\mathrm{OH})_3 \text { ?? } \\ \mathrm{M}^{3+}+3 \mathrm{OH}^{-} \\ \mathrm{S} \end{array} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{3+}\right] \cdot\left[\mathrm{OH}^{-}\right]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=[\mathrm{S}] \cdot[3 \mathrm{~S}]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\ & \end{aligned}$
CG PET-2019
Ionic Equilibrium
229417
The product of $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$of water will be
1 $\mathrm{K}_{\mathrm{w}}=10^{-12}$
2 $\mathrm{K}_{\mathrm{w}}=10^{-14}$
3 $\mathrm{K}_{\mathrm{w}}=10^{-11}$
4 $\mathrm{K}_{\mathrm{w}}=10^{-10}$
Explanation:
The product of the molar concentration of $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water. At $298 \mathrm{~K}$, For pure water $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-7} \mathrm{~mol} / \mathrm{dm}^{-3}} \\ & \mathrm{~K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ & =\left[1 \times 10^{-7}\right] .\left[1 \times 10^{-7}\right] \\ & \mathrm{K}_{\mathrm{w}}=1 \times 10^{-14} \\ & \end{aligned}$
229413
The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}$ are $1^{-31}, 1^{-44}$ and $10^{-54}$ respectively. The solubility of these sulphides are in the order :
229415
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{~S}$ is $10^{-17}$, the solubility of $\mathrm{Ag}_2 \mathrm{~S}$ in $0.1 \mathrm{M}$ solution of $\mathrm{Na}_2 \mathrm{~S}$ will be
229416
If solubility of $\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{S}$, its solubility product will be:
1 $108 \mathrm{~S}^5$
2 $27 \mathrm{~S}^3$
3 $4 \mathrm{~S}^4$
4 $27 \mathrm{~S}^4$
Explanation:
$\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{AB}_3$ Type salt its solubility product will be- $\begin{aligned} & \begin{array}{r} \mathrm{M}(\mathrm{OH})_3 \text { ?? } \\ \mathrm{M}^{3+}+3 \mathrm{OH}^{-} \\ \mathrm{S} \end{array} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{3+}\right] \cdot\left[\mathrm{OH}^{-}\right]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=[\mathrm{S}] \cdot[3 \mathrm{~S}]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\ & \end{aligned}$
CG PET-2019
Ionic Equilibrium
229417
The product of $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$of water will be
1 $\mathrm{K}_{\mathrm{w}}=10^{-12}$
2 $\mathrm{K}_{\mathrm{w}}=10^{-14}$
3 $\mathrm{K}_{\mathrm{w}}=10^{-11}$
4 $\mathrm{K}_{\mathrm{w}}=10^{-10}$
Explanation:
The product of the molar concentration of $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water. At $298 \mathrm{~K}$, For pure water $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-7} \mathrm{~mol} / \mathrm{dm}^{-3}} \\ & \mathrm{~K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ & =\left[1 \times 10^{-7}\right] .\left[1 \times 10^{-7}\right] \\ & \mathrm{K}_{\mathrm{w}}=1 \times 10^{-14} \\ & \end{aligned}$
229413
The $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{CuS}, \mathrm{Ag}_2 \mathrm{~S}$ and $\mathrm{HgS}$ are $1^{-31}, 1^{-44}$ and $10^{-54}$ respectively. The solubility of these sulphides are in the order :
229415
If $\mathrm{K}_{\mathrm{sp}}$ of $\mathrm{Ag}_2 \mathrm{~S}$ is $10^{-17}$, the solubility of $\mathrm{Ag}_2 \mathrm{~S}$ in $0.1 \mathrm{M}$ solution of $\mathrm{Na}_2 \mathrm{~S}$ will be
229416
If solubility of $\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{S}$, its solubility product will be:
1 $108 \mathrm{~S}^5$
2 $27 \mathrm{~S}^3$
3 $4 \mathrm{~S}^4$
4 $27 \mathrm{~S}^4$
Explanation:
$\mathrm{M}(\mathrm{OH})_3$ is $\mathrm{AB}_3$ Type salt its solubility product will be- $\begin{aligned} & \begin{array}{r} \mathrm{M}(\mathrm{OH})_3 \text { ?? } \\ \mathrm{M}^{3+}+3 \mathrm{OH}^{-} \\ \mathrm{S} \end{array} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{M}^{3+}\right] \cdot\left[\mathrm{OH}^{-}\right]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=[\mathrm{S}] \cdot[3 \mathrm{~S}]^3 \\ & \mathrm{~K}_{\mathrm{sp}}=27 \mathrm{~S}^4 \\ & \end{aligned}$
CG PET-2019
Ionic Equilibrium
229417
The product of $\mathrm{H}^{+}$and $\mathrm{OH}^{-}$of water will be
1 $\mathrm{K}_{\mathrm{w}}=10^{-12}$
2 $\mathrm{K}_{\mathrm{w}}=10^{-14}$
3 $\mathrm{K}_{\mathrm{w}}=10^{-11}$
4 $\mathrm{K}_{\mathrm{w}}=10^{-10}$
Explanation:
The product of the molar concentration of $\mathrm{H}^{+}$ and $\mathrm{OH}^{-}$ion pure water or an aqueous solution at constant temperature is constant which is called the ionic product of water. At $298 \mathrm{~K}$, For pure water $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-7} \mathrm{~mol} / \mathrm{dm}^{-3}} \\ & \mathrm{~K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ & =\left[1 \times 10^{-7}\right] .\left[1 \times 10^{-7}\right] \\ & \mathrm{K}_{\mathrm{w}}=1 \times 10^{-14} \\ & \end{aligned}$
