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» <html><body><p></p>Solution :`{:(HA,+,BOH,hArr,<a href="https://interviewquestions.tuteehub.com/tag/ba-389206" style="font-weight:bold;" target="_blank" title="Click to know more about BA">BA</a>,+,H_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)O),("weak",,"strong",,,,,):}`:}` <br/> or HA+<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a>^(+)+OH^(-) hArr B^(+)+A^(-)+H_(2)O or HA + OH^(-) hArr A^(-)+ H_(2)O` <br/> `K=([A^(-)])/([HA][OH^(-)])` ...(i) <br/> Further, for the weak acid, `HA hArr H^(+)+A^(-), K_(a) = ([H^(+)][A^(-)])/([HA])` ...(ii) <br/> Also `K_(w)=[H^(+)][OH^(-)]`...(iii) <br/> From eqns. (i), (ii) and (iii), `K=(K_(a))/(K_(w))=(<a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a>^(-4))/(10^(-<a href="https://interviewquestions.tuteehub.com/tag/14-272882" style="font-weight:bold;" target="_blank" title="Click to know more about 14">14</a>))=10^(10)`</body></html>