given pH= 2.59
-log [H+] =2.59
[H+] =10-2.59 =0.00257
for mono protic acid HA ---> H+ +A-
Ka= [H+] [A-]/[HA] Ka =equilirium constant
[H+] =[A-] = 0.00257 at equilibrium and [HA] at equulibrium= 0.0165-0.00257=0.01393
Ka= 0.00257*0.00257 0.01393=0.000474
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