SATHEE: Equilibrium Question 907

Equilibrium Question 907

Question: The solubility of $A g C l$ in water at $25^{\circ} C$ is $1.79 \times 10^{- 3} g / L$ . The $K_{s p}$ of $A g C l$ at $25^{\circ} C$ is

Options:

A) $1.68 \times 10^{- 12}$

B) $1.55 \times 10^{- 10}$

C) $12.4 \times 10^{- 8}$

D) $1.73 \times 10^{- 14}$

Show Answer

Answer:

Correct Answer: B

Solution:

Solubility of $A g C l$ in water $= 1.79 \times 10^{- 3} g / L$

$= \frac{1.79 \times 10^{- 3}}{143.5} m o l / L$

$= 1.247 \times 10^{- 5} m o l / L$

For $A g C l$

$K_{s p} = S \times S = S^{2}$

$= {(1.247 \times 10^{- 5})}^{2}$

$= 1.55 \times 10^{- 10} m o l^{2} L^{- 2}$

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Equilibrium Question 907

Question: The solubility of AgCl in water at 25∘C is 1.79×10−3g/L . The Ksp of AgCl at 25∘C is

Options:

Answer:

Solution:

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Question: The solubility of $A g C l$ in water at $25^{\circ} C$ is $1.79 \times 10^{- 3} g / L$ . The $K_{s p}$ of $A g C l$ at $25^{\circ} C$ is