\(n_{HCl}=0,01.0,2=0,002\left(mol\right)\)
\(n_{HNO3}=0,03.0,3=0,009\left(mol\right)\)
\(\Rightarrow n_{H^+}=n_{HCl}+n_{HNO3}=0,002+0,009=0,011\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{0,011}{0,2+0,3}=0,022M\Rightarrow pH=-log\left(0,022\right)=1,65\)