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The correct relation between hydrolysis constant `(K_(b))` and degree of hydrolysis `(alpha)` for the following equilibrium isA. `alpha = sqrt((K_(w).C)/(K_(a)))`B. `alpha = sqrt((K_(w))/(K_(a).C))`C. `alpha = sqrt((K_(a).C)/(K_(w))`D. `alpha = sqrt((K_(a))/(K_(w).C))` |
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Answer» Correct Answer - B Salt of weak acid and strong base. `K_(h) = (K_(w))/(K_(a))` `h = sqrt((K_(h))/(c))` `h = sqrt((K_(w))/(K_(a)c))` |
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