Question

H3AsO4 reacts with water as follows:
H3aso4+h20<=>h2aso4- + h30+
The conjugate base of h3aso4 in this reaction is

Answer 1

You need to do this yourself and it's really simple.
H3AsO4 + H2O ==> H3O^+ + H2AsO4^-
In the Bronsted-Lowry theory of acids and bases, the acid is the proton donor (the one giving away H's) and the base is the proton acceptor (the one that now has an extra H). You have only two reactants; i.e., H3AsO4 and H2O. You started with H3AsO4 and it ended up as H2AsO4^- in the product so which reactant donated a proton, Was it H3AsO4 or was it H2O. Now for the base. Where did the proton go? Did it go to the H2AsO4^- or did it go to the H2O to form H3O^- ?
So now you have identified the acid as H3AsO4 and the base as H2O.
So when H3AsO4, the acid, becomes H2AsO4^- and that is called the conjugate base. Every acid becomes a conjugate base while every base (H2O), a base, forms a conjugate acid, H3O^+.
Common acid/conjugate base pairs are
HCl/Cl^-
HNO3/NO3^-
H3O^+/H2O
NH4^+/NH3