Correct Answer - B
`pH=3 implies [H_(3)O^(+)]_(1)=10^(-3)`
`pH=4 implies [H_(3)O^(+)]_(2)=10^(-4)`
As equal volumes are mixed so after mixing
`[H_(3)O^(+)]_(1)=(10^(-3))/(2)`
`[H_(3)O^(+)]_(2)=(10^(-4))/(2)`
`[H^(+)]_(mixture)=[0.5xx10^(-3)+0.5+10^(-4)]lt`
`pH=3.26`