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If the pH of 0.26 M HNO(2) is 2.5, what ...

If the `pH` of `0.26 M HNO_(2)` is `2.5`, what will be its dissociation constant.

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The correct Answer is:
A, C
`C` is fairly high, then `(1-alpha) ~~1`
So,
`pH_(W_(A)) = (1)/(2) (pK_(a) - logC)`
`2.5 xx2 = pK_(a) - log 0.26`
`5 = pK_(a) + 0.585`
`pK_(a) = 4.415`
`K_(a) = "Antilog" (-4.415)`
`= "Antilog" (-4-0.415 + 1-1)`
`= "Antilog" (bar(5).585)`
`= 3.8 xx 10^(-5)`
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