A certain weak acid has K_(a)=1.0xx10^(-4). Calculate the equilibrium constant for its reaction with a strong base.

A certain weak acid has K_(a)=1.0xx10^(-4). Calculate the equilibrium

1.	A certain weak acid has K_(a)=1.0xx10^(-4). Calculate the equilibrium constant for its reaction with a strong base.
Answer» Solution :`{:(HA,+,BOH,hArr,BA,+,H_(2)O),("weak",,"strong",,,,,):}`:}` or HA+B^(+)+OH^(-) hArr B^(+)+A^(-)+H_(2)O or HA + OH^(-) hArr A^(-)+ H_(2)O` `K=([A^(-)])/([HA][OH^(-)])` ...(i) Further, for the weak acid, `HA hArr H^(+)+A^(-), K_(a) = ([H^(+)][A^(-)])/([HA])` ...(ii) Also `K_(w)=[H^(+)][OH^(-)]`...(iii) From eqns. (i), (ii) and (iii), `K=(K_(a))/(K_(w))=(10^(-4))/(10^(-14))=10^(10)`