Em tham khảo nha :
\(\begin{array}{l}
a)\\
{n_{HCl}} = \dfrac{{0,224}}{{22,4}} = 0,01mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,01}}{{0,5}} = 0,02M\\
{\rm{[}}{H^ + }{\rm{]}} = {C_{{M_{HCl}}}} = 0,02M\\
pH = - \log (0,02) = 1,699\\
b)\\
{n_{HBr}} = 0,01 \times 1 = 0,01mol\\
{C_{{M_{HBr}}}} = \dfrac{{0,01}}{{0,1}} = 0,1M\\
{\rm{[}}{H^ + }{\rm{]}} = {C_{{M_{HBr}}}} = 0,1M\\
pH = - \log (0,1) = 1
\end{array}\)