A 0.1 M solution of weak acid HA is 1% dissociated at 25^(@)C To this solution NaA is added till [NaA] = 0.2 M, if the new degree of dissociation of HA = yxx 10^(-5) then what is 'y'?

A 0.1 M solution of weak acid HA is 1% dissociated at 25^(@)C To this

1.	A 0.1 M solution of weak acid HA is 1% dissociated at 25^(@)C To this solution NaA is added till [NaA] = 0.2 M, if the new degree of dissociation of HA = yxx 10^(-5) then what is 'y'?
Answer» <html><body><p><br/></p>Solution :` alpha = 0.01 , C =0.1` <br/> ` K_a = <a href="https://interviewquestions.tuteehub.com/tag/calpha-2020052" style="font-weight:bold;" target="_blank" title="Click to know more about CALPHA">CALPHA</a> ^(2)= 0.1 xx ( 0.01 )^(2)= <a href="https://interviewquestions.tuteehub.com/tag/10-261113" style="font-weight:bold;" target="_blank" title="Click to know more about 10">10</a> ^(5) ` <br/> ` K_a = ([<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+) ] [A^(-) ]) /( [<a href="https://interviewquestions.tuteehub.com/tag/ha-479237" style="font-weight:bold;" target="_blank" title="Click to know more about HA">HA</a>]) =([H^(+) ]xx 0.2 )/(0.1 )` <br/> ` <a href="https://interviewquestions.tuteehub.com/tag/rarr-1175461" style="font-weight:bold;" target="_blank" title="Click to know more about RARR">RARR</a> [H^(+) ] = calpha =10 ^(-5)xx ( 0.1)/( 0.2 ) rArr alpha = 5 xx 10 ^(-5)`</body></html>