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.

Question

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. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Hydrolysis of a salt is reverse reaction of acid base neutralization reaction.

∴ K_{h} = \frac{K _{w}}{K _{a}} = \frac{1 0 ^{- 14}}{1 0 ^{- 7}} = 1 0^{- 7}

[O H^{-}] = h = C \times \frac{K _{h}}{C}

= C \times K_{h}

= 1 0^{- 8} = 1 0^{- 4}

\Rightarrow pO H^{-} = - l o g [O H^{-}]

= - l o g [1 0^{- 4}] = 4

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 0.1MNaA.

Hydrolysis of a salt is reverse reaction of acid base neutralization reaction. ∴Kh​=Ka​Kw​​=10−710−14​=10−7 [OH−]=h=C×CKh​​​ =C×Kh​​ =10−8​=10−4 ⇒pOH−=−log[OH−] =−log[10−4]=4

Q. The equilibrium constant of the reaction of weak acid $H A$ with strong base is $1 0^{- 7}$ . Find the $pO H$ of the aqueous solution of $0.1 M N a A$ .