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Introduction and Degree of Hydrolysis

The pH of 0...

Question

The pH of $0.01$ $M (N H_{4})_{2} S O_{4}$ and $0.02$ $M N H_{4} O H$ buffer solution $(p K_{a} o f N H_{4}^{+} = 9.26)$ is:

9.26

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4.74

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4.74 + l o g 2

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None of these

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Solution

The correct option is A $9.26$
$p H = p K_{a} + l o g \frac{[s a l t]}{[b a s e]}$

Here, $[s a l t] = [b a s e] = 0.02$

So, $p H = p K_{a} = 9.26 N$

Hence, the pH is $9.26$ .

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