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If the equilibrium constant of the reaction of weak acid HA with strong base is 109, then pH of 0.1 M NaA is:
 

Option: 1

2


Option: 2

5


Option: 3

9


Option: 4

11


Answers (1)

best_answer

Given, equilibrium constant for the reaction is equal to \mathrm{10^9}

\mathrm{HA + OH^- \rightleftharpoons A^- + H_2O, K_{eq}=10^9}

Now, hydrolysis reaction of \mathrm{A^-} is given by 

\mathrm{ A^- + H_2O \rightleftharpoons HA + OH^-, K_h= \frac{1}{Keq}=10^{-9}}

and the hydrolysis constant is given by

\mathrm{ K_h= \frac{1}{Keq}=10^{-9}}

\therefore \mathrm{ [OH^{-}]= \sqrt{C \times K_h}=\sqrt{0.1 \times 10^{-9}}=10^{-5}}

\mathrm{Thus,\ pOH = 5 \Rightarrow pH = 9}

Therefore, option(3) is correct

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Shailly goel

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