SATHEE: Equilibrium Question 789

Equilibrium Question 789

Question: The solubility of $P b I_{2}$ at $25^{\circ} C$ is $0.7 g L^{- 1}$ . The solubility product of $P b I_{2}$ at this temperature is (molar mass of $P b I_{2} = 461.2 g m o l^{- 1}$ )

Options:

A) $1.40 \times 10^{- 9}$

B) $0.14 \times 10^{- 9}$

C) $140 \times 10^{- 9}$

D) $14.0 \times 10^{- 9}$

Show Answer

Answer:

Correct Answer: D

Solution:

$P b I_{2} \to \underset{s}{P b^{+ +}}, + \underset{2 s}{2 I^{-}},$

$K_{s p} = s \times {(2 s)}^{2} = 4 s^{3}$

$= 4 \times {(\frac{0.7}{461.2})}^{3} = 14.0 \times 10^{- 9}$

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Equilibrium Question 789

Question: The solubility of PbI2 at 25∘C is 0.7gL−1 . The solubility product of PbI2 at this temperature is (molar mass of PbI2=461.2gmol−1 )

Options:

Answer:

Solution:

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Question: The solubility of $P b I_{2}$ at $25^{\circ} C$ is $0.7 g L^{- 1}$ . The solubility product of $P b I_{2}$ at this temperature is (molar mass of $P b I_{2} = 461.2 g m o l^{- 1}$ )