\(n_{H^+\left(HCl\right)}=0,09.0,3=27.10^{-3}\left(mol\right)\)
\(n_{H^+\left(H_2SO_4\right)}=2.0,06.0,001.V=12V.10^{-5}\left(mol\right)\)
\(\Rightarrow n_{H^+}=27.10^{-3}+12V.10^{-5}\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{\left(27+0,12V\right).10^{-3}}{0,001.V+0,3}=10^{-1,0963}\Rightarrow V=595\left(ml\right)\)