229501
The solubility of $\mathrm{Sb}_2 \mathrm{~S}_3$ in water is $10 \times 10^{-5}$ $\mathrm{mol} / \mathrm{L}$ at $298 \mathrm{~K}$. What will be its solubility product?
1 $108 \times 10^{-25}$
2 $10 \times 10^{-25}$
3 $144 \times 10^{-25}$
4 $126 \times 10^{-24}$
Explanation:
Key Idea Write the equation for dissociation of $\mathrm{Sb}_2 \mathrm{~S}_3$ and find relationship between $\mathrm{K}_{\text {pp }}$ and solubility. From this relationship, calculate $\mathrm{K}_{\mathrm{sp}}$. Let solubility of $\mathrm{Sb}_2 \mathrm{~S}_3=\mathrm{x} \mathrm{mol} / \mathrm{L}$ $\begin{array}{rl} \qquad \mathrm{Sb}_2 \mathrm{~S}_3 \text { ? } & \mathrm{BH} / 2 \mathrm{Sb}^{3+}+3 \mathrm{~S}^{2-} \\ \therefore \text { solubility } \mathrm{x} \mathrm{mol} / \mathrm{L} & 2 \mathrm{x} \quad 3 \mathrm{x} \\ \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Sb}^{3+}\right]^2\left[\mathrm{~S}^2\right]^3 \\ & =(2 \mathrm{x})^2(3 \mathrm{x})^3 \\ \mathrm{~K}_{\mathrm{sp}} & =108 \mathrm{x}^5 \\ \text { Given, } \mathrm{x} & =10 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \\ \mathrm{K}_{\mathrm{sp}} & =108 \times\left(10 \times 10^{-5}\right)^5 \\ & =108 \times 10^{-25} \end{array}$
UP CPMT-2004
Ionic Equilibrium
229503
The solubility product of calcium fluoride is 3.2 $\times 10^{-11} \mathrm{M}^3$. Its solubility in saturated solution is
229505
The solubility in water of a sparingly soluble salt $\mathrm{AB}_2$ is $1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility Product number will be
229501
The solubility of $\mathrm{Sb}_2 \mathrm{~S}_3$ in water is $10 \times 10^{-5}$ $\mathrm{mol} / \mathrm{L}$ at $298 \mathrm{~K}$. What will be its solubility product?
1 $108 \times 10^{-25}$
2 $10 \times 10^{-25}$
3 $144 \times 10^{-25}$
4 $126 \times 10^{-24}$
Explanation:
Key Idea Write the equation for dissociation of $\mathrm{Sb}_2 \mathrm{~S}_3$ and find relationship between $\mathrm{K}_{\text {pp }}$ and solubility. From this relationship, calculate $\mathrm{K}_{\mathrm{sp}}$. Let solubility of $\mathrm{Sb}_2 \mathrm{~S}_3=\mathrm{x} \mathrm{mol} / \mathrm{L}$ $\begin{array}{rl} \qquad \mathrm{Sb}_2 \mathrm{~S}_3 \text { ? } & \mathrm{BH} / 2 \mathrm{Sb}^{3+}+3 \mathrm{~S}^{2-} \\ \therefore \text { solubility } \mathrm{x} \mathrm{mol} / \mathrm{L} & 2 \mathrm{x} \quad 3 \mathrm{x} \\ \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Sb}^{3+}\right]^2\left[\mathrm{~S}^2\right]^3 \\ & =(2 \mathrm{x})^2(3 \mathrm{x})^3 \\ \mathrm{~K}_{\mathrm{sp}} & =108 \mathrm{x}^5 \\ \text { Given, } \mathrm{x} & =10 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \\ \mathrm{K}_{\mathrm{sp}} & =108 \times\left(10 \times 10^{-5}\right)^5 \\ & =108 \times 10^{-25} \end{array}$
UP CPMT-2004
Ionic Equilibrium
229503
The solubility product of calcium fluoride is 3.2 $\times 10^{-11} \mathrm{M}^3$. Its solubility in saturated solution is
229505
The solubility in water of a sparingly soluble salt $\mathrm{AB}_2$ is $1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility Product number will be
229501
The solubility of $\mathrm{Sb}_2 \mathrm{~S}_3$ in water is $10 \times 10^{-5}$ $\mathrm{mol} / \mathrm{L}$ at $298 \mathrm{~K}$. What will be its solubility product?
1 $108 \times 10^{-25}$
2 $10 \times 10^{-25}$
3 $144 \times 10^{-25}$
4 $126 \times 10^{-24}$
Explanation:
Key Idea Write the equation for dissociation of $\mathrm{Sb}_2 \mathrm{~S}_3$ and find relationship between $\mathrm{K}_{\text {pp }}$ and solubility. From this relationship, calculate $\mathrm{K}_{\mathrm{sp}}$. Let solubility of $\mathrm{Sb}_2 \mathrm{~S}_3=\mathrm{x} \mathrm{mol} / \mathrm{L}$ $\begin{array}{rl} \qquad \mathrm{Sb}_2 \mathrm{~S}_3 \text { ? } & \mathrm{BH} / 2 \mathrm{Sb}^{3+}+3 \mathrm{~S}^{2-} \\ \therefore \text { solubility } \mathrm{x} \mathrm{mol} / \mathrm{L} & 2 \mathrm{x} \quad 3 \mathrm{x} \\ \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Sb}^{3+}\right]^2\left[\mathrm{~S}^2\right]^3 \\ & =(2 \mathrm{x})^2(3 \mathrm{x})^3 \\ \mathrm{~K}_{\mathrm{sp}} & =108 \mathrm{x}^5 \\ \text { Given, } \mathrm{x} & =10 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \\ \mathrm{K}_{\mathrm{sp}} & =108 \times\left(10 \times 10^{-5}\right)^5 \\ & =108 \times 10^{-25} \end{array}$
UP CPMT-2004
Ionic Equilibrium
229503
The solubility product of calcium fluoride is 3.2 $\times 10^{-11} \mathrm{M}^3$. Its solubility in saturated solution is
229505
The solubility in water of a sparingly soluble salt $\mathrm{AB}_2$ is $1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility Product number will be
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
229501
The solubility of $\mathrm{Sb}_2 \mathrm{~S}_3$ in water is $10 \times 10^{-5}$ $\mathrm{mol} / \mathrm{L}$ at $298 \mathrm{~K}$. What will be its solubility product?
1 $108 \times 10^{-25}$
2 $10 \times 10^{-25}$
3 $144 \times 10^{-25}$
4 $126 \times 10^{-24}$
Explanation:
Key Idea Write the equation for dissociation of $\mathrm{Sb}_2 \mathrm{~S}_3$ and find relationship between $\mathrm{K}_{\text {pp }}$ and solubility. From this relationship, calculate $\mathrm{K}_{\mathrm{sp}}$. Let solubility of $\mathrm{Sb}_2 \mathrm{~S}_3=\mathrm{x} \mathrm{mol} / \mathrm{L}$ $\begin{array}{rl} \qquad \mathrm{Sb}_2 \mathrm{~S}_3 \text { ? } & \mathrm{BH} / 2 \mathrm{Sb}^{3+}+3 \mathrm{~S}^{2-} \\ \therefore \text { solubility } \mathrm{x} \mathrm{mol} / \mathrm{L} & 2 \mathrm{x} \quad 3 \mathrm{x} \\ \mathrm{K}_{\mathrm{sp}} & =\left[\mathrm{Sb}^{3+}\right]^2\left[\mathrm{~S}^2\right]^3 \\ & =(2 \mathrm{x})^2(3 \mathrm{x})^3 \\ \mathrm{~K}_{\mathrm{sp}} & =108 \mathrm{x}^5 \\ \text { Given, } \mathrm{x} & =10 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \\ \mathrm{K}_{\mathrm{sp}} & =108 \times\left(10 \times 10^{-5}\right)^5 \\ & =108 \times 10^{-25} \end{array}$
UP CPMT-2004
Ionic Equilibrium
229503
The solubility product of calcium fluoride is 3.2 $\times 10^{-11} \mathrm{M}^3$. Its solubility in saturated solution is
229505
The solubility in water of a sparingly soluble salt $\mathrm{AB}_2$ is $1.0 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1}$. Its solubility Product number will be
