Ta có: \(pH=2\) \(\Rightarrow\left[H^+\right]=10^{-2}\left(M\right)\) \(\Rightarrow n_{H^+}=10^{-2}\cdot0,1=10^{-3}\left(mol\right)\)
\(\Rightarrow\left[H^+\right]_{\left(sau\right)}=\dfrac{10^{-3}}{1}=10^{-3}\left(M\right)\) \(\Rightarrow pH=-log\left(10^{-3}\right)=3\)