Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{0,098}{98}=0,001\left(mol\right)\\n_{HNO_3}=0,2.0,04=0,008\left(mol\right)\end{matrix}\right.\)
=> \(\sum n_{H^+}=2n_{H_2SO_4}+n_{HNO_3}=0,01\left(mol\right)\)
=> \(\left[H^+\right]=\dfrac{0,01}{0,5+0,2}=\dfrac{1}{70}M\)
=> \(pH=-\log\left(\dfrac{1}{70}\right)=1,845\)