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» 5LOG 2 - log 3 5 - log 2 5 - log 3 5 - log 6 Solution :On partial neutralization of the weak acid, SALT is formed. Hence, it becomes a buffer `pH = pK_(a) + log. (["Salt"])/(["Acid"])` 1/3rd NEUTRALISATION of the acid meansout of 1 mole of the acid , salt formed = 1/3 mole and acid left = 2/3 mole Hence, `pH = - log(10^(-5))+log.(1//3)/(2//3)` `=5+ log .(1)/(2) = 5 - log2`.