If the equilibrium constant of the reaction of weak acid $H A$ with strong base is $10^{9}$ , then pH in $0.1 M$ $N a A$ is:

5

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9

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7

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8

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Solution

The correct option is D $9$
We know
$p H = \frac{1}{2} p K_{w} + \frac{1}{2} (p K_{a} + log C)$
$H A + O H^{-} ⇌ H_{2} O + A^{-}$
Now,
$K_{e q} = \frac{K_{a}}{K_{w}}$
$K_{a} = K_{e q} \times K_{w} = 10^{9} \times 10^{- 14} = 10^{- 5}$

We get $K_{a} = 10^{- 5}$
$p H = 7 + \frac{1}{2} (5 + log 0.1) = 9$

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Similar questions

H A

10^{9}

p H

0.1 M

N a A

K_{b} = 10^{- 9}

K_{b} = 10^{- 9}

10^{x}

1.0 \times 10^{- 5}

(H A)

B O H

10^{11}

p H

0.10 M

B A

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