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The equilibrium constant of the reaction...

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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The correct Answer is:
4
Hydrolysis of a salt is reverse reaction of acid base neutralization reaction.
`therefore K_(b) =K_(w)/K_(a) = 10^(-14)/10^(-7) 10^(-7)`
`therefore [OH^(-)] =h=C xx sqrt(K_(b)/C) = sqrt(C xx K_(b))`
`=sqrt(10^(-8)) = 10^(-4)`
`rArr poH=-log[OH^(-)]`
`=-log[10^(-4)]=4`
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