1. Tardigrade
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  3. Chemistry
  4. The equilibrium constant of the reaction of weak acid HA with strong base is 10-7. Find the pOH of the aqueous solution of 0.1M NaA.

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Q. The equilibrium constant of the reaction of weak acid $HA$ with strong base is $10^{-7}$. Find the $pOH$ of the aqueous solution of $0.1M \,NaA$.

Equilibrium

Solution:

Hydrolysis of a salt is reverse reaction of acid base neutralization reaction.
$\therefore K_h = \frac{K_w}{K_a} = \frac{10^{-14}}{10^{-7}} = 10 ^{-7}$
$[OH^-] = h = C \times \sqrt{\frac{K_h}{C}} $
$= \sqrt{C \times K_h}$
$= \sqrt{10^{-8}} = 10^{-4}$
$\Rightarrow pOH^- = -log [OH^-]$
$ = -log [10^{-4}] = 4$