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Salt of Weak Acid and Strong Base

pH of 0.02M ...

Question

pH of 0.02M $(N H_{4})_{2} S O_{4}$ and 0.02M $N H_{4} O H$ buffer solution $(p K_{a} o f N H_{4}^{+} = 9.6)$ is :

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6.5

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9.6

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5.5

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Solution

The correct option is C 9.6
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$

$[s a l t] = [a c i d] = 0.02$

So, $p H = p K_{a} = 9.6$

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