Em tham khảo nha:
\(\begin{array}{l}
a)\\
{\rm{[}}{H^ + }{\rm{]}} = {C_M}HCl = 0,001M\\
pH = - \log (0,001) = 3\\
b)\\
{\rm{[}}O{H^ - }{\rm{]}} = {C_M}NaOH = 0,001M\\
pOH = - \log (0,001) = 3\\
pH = 14 - 3 = 11\\
c)\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,098}}{{98}} = 0,001\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,001}}{{0,1}} = 0,01M\\
{\rm{[}}{H^ + }{\rm{]}} = 2{C_M}{H_2}S{O_4} = 0,02M\\
pH = - \log (0,02) = 1,7\\
d)\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,245}}{{98}} = 0,0025\,mol\\
{C_M}{H_2}S{O_4} = \dfrac{{0,0025}}{{0,5}} = 0,005M\\
{\rm{[}}{H^ + }{\rm{]}} = 2{C_M}{H_2}S{O_4} = 0,01M\\
pH = - \log (0,01) = 2\\
e)\\
{n_{KOH}} = \dfrac{{0,0224}}{{56}} = 0,0004\,mol\\
{C_M}KOH = \dfrac{{0,0004}}{{0,4}} = 0,001M\\
{\rm{[O}}{{\rm{H}}^ - }{\rm{]}} = {C_M}KOH = 0,001M\\
pOH = - \log (0,001) = 3\\
pH = 14 - 3 = 11
\end{array}\)
