a) \(n_{H^+}=2n_{H_2SO_4}=\dfrac{2.9,8}{98}=0,2\) mol
\(\Rightarrow\left[H^+\right]=\dfrac{0,2}{10}=0,02\) mol/lít
\(\Rightarrow pH=-lg\left[H^+\right]=-lg0,02=1,7\)
b) \(n_{OH^-}=n_{NaOH}=\dfrac{4}{40}=0,1\) mol
\(\left[OH^-\right]=\dfrac{0,1}{1}=0,1\) mol/lít
\(\Rightarrow pOH=1\Rightarrow pH=14-pOH=13\)
c) Ta có \(n_{OH^-}=2n_{H_2}=\dfrac{2.0,672}{22,4}=0,06\) mol
\(\left[OH^-\right]=\dfrac{0,06}{0,5}=0,12\) mol/lít
\(pOH=-lg\left[OH^-\right]=0,92\)
\(\Rightarrow pH=14-pOH=13,08\)