$$\begin{gathered} a, \left[H^{+}\right]=2.C_{M H_{2}S{O}_4}=2.0,01=0,02 M \\ =>pH=-\log \left[H^{+}\right]=1,7 \\ Vậy pH=1,7 \\ b, Ca{\left(OH\right)}_2->C{a}^{2+}+2O{H}^{-} \\ 0,005 ----->0,01 \left(M\right) \\ =>\left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}={10}^{-12}M \\ =>pH=-\log \left[H^{+}\right]=12 \\ Vậy pH=12 \\ c, n_{O{H}^{-}}=n_{KOH}=\frac{0,56}{56}=0,01 mol \\ =>\left[O{H}^{-}\right]=\frac{0,01}{0,1}=0,1M \\ =>pOH=-\log \left[O{H}^{-}\right]=1 \\ =>pH=14-pOH=13 \\ Vậy pH=13 \end{gathered}$$

$$\begin{gathered} d, n_{O{H}^{-}}=n_{NaOH}+n_{KOH}=0,1.0,1+0,1.0,1=0,02 mol \\ Vsau=0,2 lít \\ =>\left[O{H}^{-}\right]=\frac{0,02}{0,2}=0,1M \\ =>pOH=-\log \left[O{H}^{-}\right]=1 \\ =>pH=14-pOH=13 \\ Vậy pH=13 \\ e, n_{H^{+}}=2n_{H_{2}S{O}_4}=2.0,1.0,1=0,02 mol \\ {n}_{O{H}^{-}}=n_{NaoH}=0,1.0,1=0,01 mol \\ {H}^{+}+O{H}^{-}->H_{2}O \\ 0,01 0,01 \left(mol\right) \\ =>n_{H^{+}} dư=0,01 mol \\ Vsau=0,2 lít \\ =>\left[H^{+}\right]=\frac{0,01}{0,2}=0,05 M \\ =>pH=-\log \left[H^{+}\right]=1,3 \\ Vậy pH=1,3 \end{gathered}$$.