229437
Solubility of $\mathrm{Ca}(\mathrm{OH})_2$ is $\mathrm{S}$ mol $\mathrm{L}^{-1}$. The solubility product $\left(K_{s p}\right)$ under the same condition is
229400
The molar solubility (s) of the equilibrium, $\mathbf{A}_{\mathbf{x}} \cdot \mathbf{B}_{\mathbf{y}}$ (solid) ?? $\mathbf{x A}^{z^{+}}{ }_{(a q)}+\mathbf{y B}^{z^{-}}{ }_{(\mathrm{aq})}$ in terms of the solubility product $\left(K_{\mathrm{sp}}\right)$ will be
Let us consider the molar solubility of a salt $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ is $s \mathrm{~mol} \mathrm{~L}^{-1}$, then, $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ (solid) ??? $\mathrm{xA}_{(\mathrm{aq})}^{z+}+\mathrm{yB}_{(\mathrm{aq})}^{z-}$ $\therefore \mathrm{K}_{\mathrm{sp}}=(\mathrm{xs})^{\mathrm{x}} \cdot(\mathrm{ys})^{\mathrm{y}} \quad \mathrm{xs} \quad \text { ys }$ Hence, molar solubility can be given as : $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^z\right]^{\mathrm{x}} \cdot\left[\mathrm{B}^{z-}\right] \\ & \quad=[\mathrm{x} \cdot \mathrm{s}]^{\mathrm{x}}[\mathrm{y} \cdot \mathrm{s}]^{\mathrm{y}} \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{y}} . \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{y}} \cdot \\ & \mathrm{s}^{\mathrm{x}+\mathrm{y}}=\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}} \text { Thus } s=\left[\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}}\right]^{\frac{1}{\mathrm{x}+\mathrm{y}}} \end{aligned}$
AMU-2007
Ionic Equilibrium
229401
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in water at $298 \mathrm{~K}$ is $3.2 \times 10^{-11}$. What will be concentration of $\mathrm{CrO}_4^{2-}$ ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
229437
Solubility of $\mathrm{Ca}(\mathrm{OH})_2$ is $\mathrm{S}$ mol $\mathrm{L}^{-1}$. The solubility product $\left(K_{s p}\right)$ under the same condition is
229400
The molar solubility (s) of the equilibrium, $\mathbf{A}_{\mathbf{x}} \cdot \mathbf{B}_{\mathbf{y}}$ (solid) ?? $\mathbf{x A}^{z^{+}}{ }_{(a q)}+\mathbf{y B}^{z^{-}}{ }_{(\mathrm{aq})}$ in terms of the solubility product $\left(K_{\mathrm{sp}}\right)$ will be
Let us consider the molar solubility of a salt $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ is $s \mathrm{~mol} \mathrm{~L}^{-1}$, then, $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ (solid) ??? $\mathrm{xA}_{(\mathrm{aq})}^{z+}+\mathrm{yB}_{(\mathrm{aq})}^{z-}$ $\therefore \mathrm{K}_{\mathrm{sp}}=(\mathrm{xs})^{\mathrm{x}} \cdot(\mathrm{ys})^{\mathrm{y}} \quad \mathrm{xs} \quad \text { ys }$ Hence, molar solubility can be given as : $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^z\right]^{\mathrm{x}} \cdot\left[\mathrm{B}^{z-}\right] \\ & \quad=[\mathrm{x} \cdot \mathrm{s}]^{\mathrm{x}}[\mathrm{y} \cdot \mathrm{s}]^{\mathrm{y}} \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{y}} . \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{y}} \cdot \\ & \mathrm{s}^{\mathrm{x}+\mathrm{y}}=\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}} \text { Thus } s=\left[\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}}\right]^{\frac{1}{\mathrm{x}+\mathrm{y}}} \end{aligned}$
AMU-2007
Ionic Equilibrium
229401
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in water at $298 \mathrm{~K}$ is $3.2 \times 10^{-11}$. What will be concentration of $\mathrm{CrO}_4^{2-}$ ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$
229437
Solubility of $\mathrm{Ca}(\mathrm{OH})_2$ is $\mathrm{S}$ mol $\mathrm{L}^{-1}$. The solubility product $\left(K_{s p}\right)$ under the same condition is
229400
The molar solubility (s) of the equilibrium, $\mathbf{A}_{\mathbf{x}} \cdot \mathbf{B}_{\mathbf{y}}$ (solid) ?? $\mathbf{x A}^{z^{+}}{ }_{(a q)}+\mathbf{y B}^{z^{-}}{ }_{(\mathrm{aq})}$ in terms of the solubility product $\left(K_{\mathrm{sp}}\right)$ will be
Let us consider the molar solubility of a salt $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ is $s \mathrm{~mol} \mathrm{~L}^{-1}$, then, $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ (solid) ??? $\mathrm{xA}_{(\mathrm{aq})}^{z+}+\mathrm{yB}_{(\mathrm{aq})}^{z-}$ $\therefore \mathrm{K}_{\mathrm{sp}}=(\mathrm{xs})^{\mathrm{x}} \cdot(\mathrm{ys})^{\mathrm{y}} \quad \mathrm{xs} \quad \text { ys }$ Hence, molar solubility can be given as : $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^z\right]^{\mathrm{x}} \cdot\left[\mathrm{B}^{z-}\right] \\ & \quad=[\mathrm{x} \cdot \mathrm{s}]^{\mathrm{x}}[\mathrm{y} \cdot \mathrm{s}]^{\mathrm{y}} \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{y}} . \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{y}} \cdot \\ & \mathrm{s}^{\mathrm{x}+\mathrm{y}}=\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}} \text { Thus } s=\left[\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}}\right]^{\frac{1}{\mathrm{x}+\mathrm{y}}} \end{aligned}$
AMU-2007
Ionic Equilibrium
229401
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in water at $298 \mathrm{~K}$ is $3.2 \times 10^{-11}$. What will be concentration of $\mathrm{CrO}_4^{2-}$ ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$
229437
Solubility of $\mathrm{Ca}(\mathrm{OH})_2$ is $\mathrm{S}$ mol $\mathrm{L}^{-1}$. The solubility product $\left(K_{s p}\right)$ under the same condition is
229400
The molar solubility (s) of the equilibrium, $\mathbf{A}_{\mathbf{x}} \cdot \mathbf{B}_{\mathbf{y}}$ (solid) ?? $\mathbf{x A}^{z^{+}}{ }_{(a q)}+\mathbf{y B}^{z^{-}}{ }_{(\mathrm{aq})}$ in terms of the solubility product $\left(K_{\mathrm{sp}}\right)$ will be
Let us consider the molar solubility of a salt $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ is $s \mathrm{~mol} \mathrm{~L}^{-1}$, then, $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{x}}$ (solid) ??? $\mathrm{xA}_{(\mathrm{aq})}^{z+}+\mathrm{yB}_{(\mathrm{aq})}^{z-}$ $\therefore \mathrm{K}_{\mathrm{sp}}=(\mathrm{xs})^{\mathrm{x}} \cdot(\mathrm{ys})^{\mathrm{y}} \quad \mathrm{xs} \quad \text { ys }$ Hence, molar solubility can be given as : $\begin{aligned} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^z\right]^{\mathrm{x}} \cdot\left[\mathrm{B}^{z-}\right] \\ & \quad=[\mathrm{x} \cdot \mathrm{s}]^{\mathrm{x}}[\mathrm{y} \cdot \mathrm{s}]^{\mathrm{y}} \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{y}} . \\ & \mathrm{K}_{\mathrm{sp}}=\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}} \cdot \mathrm{s}^{\mathrm{x}} \cdot \mathrm{s}^{\mathrm{y}} \cdot \\ & \mathrm{s}^{\mathrm{x}+\mathrm{y}}=\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}} \text { Thus } s=\left[\frac{\mathrm{K}_{\mathrm{sp}}}{\mathrm{x}^{\mathrm{x}} \cdot \mathrm{y}^{\mathrm{y}}}\right]^{\frac{1}{\mathrm{x}+\mathrm{y}}} \end{aligned}$
AMU-2007
Ionic Equilibrium
229401
The solubility product of $\mathrm{Ag}_2 \mathrm{CrO}_4$ in water at $298 \mathrm{~K}$ is $3.2 \times 10^{-11}$. What will be concentration of $\mathrm{CrO}_4^{2-}$ ions in the saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$