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Q.

pH of 0.01 $M {({NH}_{4})}_{2} {SO}_{4}$ and 0.02M NH₄OH buffer $(p^{Ka} of {NH}_{4}^{+} = 9)$ is

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answer is 9.

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Detailed Solution

$pH = p^{Ka} + \log \frac{[S]}{[A]}$

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