A weak acid of dissociation constant 01^(-5) is being titrated with aqueous NaOHsolution. The pH at the point of one-third neutralization of the acid will be

A weak acid of dissociation constant 01^(-5) is being titrated with aq

1.	A weak acid of dissociation constant 01^(-5) is being titrated with aqueous NaOHsolution. The pH at the point of one-third neutralization of the acid will be
Answer» <html><body><p><a href="https://interviewquestions.tuteehub.com/tag/5log-1900126" style="font-weight:bold;" target="_blank" title="Click to know more about 5LOG">5LOG</a> 2 - log 3<br/>5 - log 2<br/>5 - log 3<br/>5 - log 6</p>Solution :On partial neutralization of the weak acid, <a href="https://interviewquestions.tuteehub.com/tag/salt-1193804" style="font-weight:bold;" target="_blank" title="Click to know more about SALT">SALT</a> is formed. Hence, it becomes a buffer <br/> `pH = pK_(a) + log. (["Salt"])/(["Acid"])` <br/> <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>/3rd <a href="https://interviewquestions.tuteehub.com/tag/neutralisation-577399" style="font-weight:bold;" target="_blank" title="Click to know more about NEUTRALISATION">NEUTRALISATION</a> of the acid meansout of 1 mole of the acid , salt formed = 1/3 mole and acid left = 2/3 mole <br/> Hence, `pH = - log(10^(-5))+log.(1//3)/(2//3)` <br/> `=5+ log .(1)/(2) = 5 - log2`.</body></html>