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20/10/2021 10:42 am
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If the equilibrium constant of the reaction of weak acid HA with strong base is 10-7, then pOH of the aqueous solution of 0.1M NaA is
(a) 8
(b) 10
(c) 4
(d) 5
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20/10/2021 10:48 am
Correct answer: (c) 4
Explanation:
Hydrolysis of a salt is reverse reaction of acid-base neutralization reaction.
∴ Kh = \(\frac{K_w}{K_a}\) = \(\frac{10^{-14}}{10^{-7}}\)
= 10-7
[OH-] = ch = c x \(\sqrt{\frac{K_h}{c}}\)
= \(\sqrt{c \times K_h}\)
= \(\sqrt{10^{-8}}\) = 10-4
⇒ pOH- = -log[OH-]
= -log[10-4] = 4
This post was modified 3 years ago by Mayank1444