For a sparingly soluble salt , the equilibrium concentrations of ions and ions are M and M, respectively. The solubility product of is:

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Q.
For a sparingly soluble salt ${AB}_{2}$ , the equilibrium concentrations of $A^{2 +}$ ions and $B^{-}$ ions are $1.2 \times 10^{- 4}$ M and $0.24 \times 10^{- 3}$ M, respectively. The solubility product of $A B_{2}$ is: [2024]

1 $0.276 \times 10^{- 12}$

2 $27.65 \times 10^{- 12}$

3 $0.069 \times 10^{- 12}$

4 $6.91 \times 10^{- 12}$

Ans.
(4)

${AB}_{2} ⇌ A^{2 +} (a q) + 2 B^{-} (aq)$

$K_{s p} = [A^{2 +}] {[B^{-}]}^{2} = 1.2 \times 10^{- 4} \times {(0.24 \times 10^{- 3})}^{2} = 6.912 \times 10^{- 12}$

Solution

Q.
For a sparingly soluble salt ${AB}_{2}$ , the equilibrium concentrations of $A^{2 +}$ ions and $B^{-}$ ions are $1.2 \times 10^{- 4}$ M and $0.24 \times 10^{- 3}$ M, respectively. The solubility product of $A B_{2}$ is: [2024]

1 $0.276 \times 10^{- 12}$

2 $27.65 \times 10^{- 12}$

3 $0.069 \times 10^{- 12}$

4 $6.91 \times 10^{- 12}$

Ans.
(4)

${AB}_{2} ⇌ A^{2 +} (a q) + 2 B^{-} (aq)$

$K_{s p} = [A^{2 +}] {[B^{-}]}^{2} = 1.2 \times 10^{- 4} \times {(0.24 \times 10^{- 3})}^{2} = 6.912 \times 10^{- 12}$

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Solution

Q. For a sparingly soluble salt AB2, the equilibrium concentrations of A2+ ions and B- ions are 1.2×10-4 M and 0.24×10-3 M, respectively. The solubility product of AB2 is: [2024]

1 0.276×10-12

2 27.65×10-12

3 0.069×10-12

4 6.91×10-12

Ans. (4) AB2⇌A2+(aq)+2B-(aq) Ksp=[A2+][B-]2=1.2×10-4×(0.24×10-3)2=6.912×10-12

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Q.
For a sparingly soluble salt ${AB}_{2}$ , the equilibrium concentrations of $A^{2 +}$ ions and $B^{-}$ ions are $1.2 \times 10^{- 4}$ M and $0.24 \times 10^{- 3}$ M, respectively. The solubility product of $A B_{2}$ is: [2024]

1 $0.276 \times 10^{- 12}$

2 $27.65 \times 10^{- 12}$

3 $0.069 \times 10^{- 12}$

4 $6.91 \times 10^{- 12}$

Ans.
(4)

${AB}_{2} ⇌ A^{2 +} (a q) + 2 B^{-} (aq)$

$K_{s p} = [A^{2 +}] {[B^{-}]}^{2} = 1.2 \times 10^{- 4} \times {(0.24 \times 10^{- 3})}^{2} = 6.912 \times 10^{- 12}$