The correct option is D 1×10−4 M
Given, [H2A]=0.02, pH=4,Ka1=5×10−7 M
pH=4−log[H+]=4[H+]=10−4 M
The dissociation reaction for finding Ka1 of H2A is :
H2A (aq)⇌H+ (aq)+HA− (aq)Ka1=[HA−][H+][H2A]
Now by rearranging, the value of [HA−] can be obtained.
[HA−]=Ka1×[H2A][H+][HA−]=5×10−7×0.0210−4[HA−]=10−4 M