CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Consider a diprotic acid H2A. The dissociation constants  for H2A and HA are 106 and 1013 respectively.  Then, the pH of 0.25 M aqueous solution of H2A will be :


Solution

Given,  Ka1(H2A)=106Ka2(HA)=1013   [H2A]=0.25 M

Since, the second acid dissociation constant is very small in comparison to the first acid dissociation constant of H2A i.e. Ka2<<Ka1. Thus H+ ions contribution can be taken mainly from the first step i.e. [H+]total=[H+] from H2A dissociation.

                        H2A (aq)HA (aq)+H+ (aq)Initially:       C                       0                0Equilibrium:C(1α)          Cα            CαKa1=[H+][HA][H2A][H+]eq=[HA]eq=Cα1α1[H+]2=Ka1×C[H+]=Ka1×C[H+]=106×0.25[H+]=25×108=5×104 MpH=log[H+]=log(5×104)=(4log5)pH=(40.7)=3.3

Chemistry

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image