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