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A certain weak acid has a dissociation constant `1.0xx10^(-4)`. The equilibrium constant for its reaction with a strong base is :A. `1xx10^(-4)`B. `1xx10^(-10)`C. `oo`D. `1xx10^(10)` |
Answer» Correct Answer - D `HA hArr H^(+)+A^(-)` or `K_(a)=([H^(+)][A^(-)])/([HA])=10^(-4)" "....(i)` The reaction with strong base BOH can be expressed as `HA+OH^(-) hArr H_(2)O+A^(-)` `K=([A^(-)][H_(2)O])/([OH^(-)][HA])" "....(ii)` Also `K_(w)=[H^(+)][OH^(-)]=10^(-14)` From eqns. (i) , (ii) and (iii) `K=K_(a)//K_(w)=(10^(-4))/(10^(-14))=1010` |
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