\(\begin{array}{l}
a)\\
{\rm{[}}{H^ + }{\rm{]}} = {C_M}HCl = 0,1\,M\\
pH = - \log (0,1) = 1\\
b)\\
N{a_2}O + {H_2}O \to 2NaOH\\
nN{a_2}O = \dfrac{{6,2}}{{62}} = 0,1\,mol\\
nNaOH = 2nN{a_2}O = 0,2\,mol\\
nO{H^ - } = nNaOH = 0,2\,mol\\
\Rightarrow {\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,2}}{{0,1}} = 2\\
pH = 14 - ( - \log (2)) = 14,3\\
c)\\
nNaOH = 0,2 \times 0,1 = 0,02\,mol\\
nHCl = 0,2 \times 0,05 = 0,01\,mol\\
nNaOH\,spu = 0,02 - 0,01 = 0,01\,mol\\
V{\rm{dd}}spu = 0,2 + 0,2 = 0,4l\\
{\rm{[O}}{{\rm{H}}^ - }{\rm{]}} = \dfrac{{0,01}}{{0,4}} = 0,025\,M\\
pH = 14 - ( - \log (0,025)) = 12,4
\end{array}\)
