The dissociation constant of a substituted benzoic acid at 250C is 1.0×10–4. Find the pH of a 0.01M solution of its sodium salt.

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The dissociation constant of a substituted benzoic acid at $25^{0} C$ is $1.0 \times 10^{- 4}$ . Find the pH of a 0.01M solution of its sodium salt.

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answer is 8.

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It is given to find out the pH of a 0.01M solution

pH = $\frac{1}{2}$ ( $p K_{w} + p K_{a}$ + log c)

Sodium salt of benzoic acid is a strong base-weak acid salt.

pH = $\frac{1}{2}$ ( $p K_{w} + p K_{a}$ + log c)

pH = $\frac{1}{2}$ [ $(- l o g K_{w}) + (- l o g K_{a})$ + log 0.01]

= $\frac{1}{2}$ [14 + 4 -2]

= $\frac{1}{2}$ [16] = 8.

Hence, the answer is 8.

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