A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissociation = 0.0134) at 298 K. What is theionization constant of acetic acid ?

A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissocia

1.	A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissociation = 0.0134) at 298 K. What is theionization constant of acetic acid ?
Answer» <html><body><p><br/></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :`{:(,CH_(3)CO OH,hArr,CH_(3)CO O^(-) , +,<a href="https://interviewquestions.tuteehub.com/tag/h-1014193" style="font-weight:bold;" target="_blank" title="Click to know more about H">H</a>^(+),,),("Initial <a href="https://interviewquestions.tuteehub.com/tag/conc-927968" style="font-weight:bold;" target="_blank" title="Click to know more about CONC">CONC</a>.",C "mol" <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>^(-1),,0,,0,,),("At. eqm.",c(1-alpha),, C alpha,,C alpha,,):}` <br/> `<a href="https://interviewquestions.tuteehub.com/tag/k-527196" style="font-weight:bold;" target="_blank" title="Click to know more about K">K</a>=(C alpha.C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha)=(0.01xx(0.0134)^(2))/(1-00.01)=1.8xx10^(-6)`</body></html>