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A certain weak acid has a dissocation constant of `1.0 xx 10^(-4)`. The equilibrium constant for its reaction with a strong base isA. `1.0 xx 10^(-4)`B. `1.0 xx 10^(-10)`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(14)` |
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Answer» Correct Answer - C `HA: K_(a) = 10^(-4)` `HA +NaOH hArr NaA +H_(2)O` Clearly, the reverse reaction is the hydrolysis reaction. `rArr K_("Required") = (1)/(K_(h)) = (K_(a))/(K_(w)) = (10^(-10))/(10^(-14)) = 10^(10)` |
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