Question

HOW TO FIND SECOND equivalent point Ph!!!Anyone can give some idea how to determine second equivalent point PH. For first half-equivalent PH=Pka1 first equivalent PH=(Pka1+pka2)/2 , second half equivalent point Ph= Pka2, so how to caculate or any equation to figure out second equivalent point Ph.

half evinalept pint pH- pka, PH PlaitPa -7 eqvivalent point p 2 ceond half eqhidont pint plt pka, sond eqiralent pat ??|

Answer 1

This is for titration of Dibasic acid with strong base (as per your details ) :

At second equivalence point all dibasic acid has been completely neutralized by base , and it is in form of dianion (A^2-) : pH at 2nd equivalence point is decided by A^2- ; we have following equilibrium at this point :

(hydrolysis of A^2-)

A^2-(aq) + H₂O (l) $\rightleftharpoons$ HA^-(aq) + OH^-(aq) Kh = Kw/Ka2 = ((HA^-) ( OH^-)/ (A^2-))

(HA^-) = ( OH^-) = Kw/Ka2 * (A^2-)  

or ( OH^-) = ( Kw/Ka2 * (A^2-) ) ^1/2

pH = pKw - pOH

or pH = 1/2(pKw) + 1/2(pKa2) + 1/2 log (A^2-)