SATHEE: Equilibrium Question 838

Equilibrium Question 838

Question: If the equilibrium constant of the reaction of weak acid HA with strong base is $10^{- 7}$ , then pOH of the aqueous solution of $0.1 M N a A$ is

Options:

A) 8

B) 10

C) 4

D) 5

Show Answer

Answer:

Correct Answer: C

Solution:

Hydrolysis of a salt is reverse reaction of acid-base neutralization reaction.
$∴ K_{h} = \frac{K_{w}}{K_{a}} = \frac{10^{- 14}}{10^{- 7}} = 10^{- 7}$

$[O H^{-}] = c h = c \times \sqrt{\frac{K_{h}}{c}} = \sqrt{c \times K_{h}} = \sqrt{10^{- 8}} = 10^{- 4}$

$\Rightarrow p O H^{-} = - l o g [O H^{-}] = - \log [10^{- 4}] = 4$

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Equilibrium Question 838

Question: If the equilibrium constant of the reaction of weak acid HA with strong base is 10−7 , then pOH of the aqueous solution of 0.1MNaA is

Options:

Answer:

Solution:

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Question: If the equilibrium constant of the reaction of weak acid HA with strong base is $10^{- 7}$ , then pOH of the aqueous solution of $0.1 M N a A$ is