Đáp án:
\(\begin{array}{l}
a)\\
{\rm{[}}{H^ + }{\rm{]}} = {\rm{[N}}{{\rm{a}}^ + }{\rm{]}} = 0,25M\\
{\rm{[}}C{l^ - }{\rm{]}} = 0,5M\\
pH = 0,6\\
b)\\
{V_{{\rm{dd}}NaOH}} = 50ml
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
HCl + NaOH \to NaCl + {H_2}O\\
{n_{HCl}} = 0,1 \times 1 = 0,1\,mol\\
{n_{NaOH}} = 0,1 \times 0,5 = 0,05\,mol\\
\dfrac{{0,1}}{1} > \dfrac{{0,05}}{1} \Rightarrow \text{ HCl dư}\\
{n_{NaCl}} = {n_{NaOH}} = 0,05\,mol\\
{n_{HCl}} \text{ dư}= 0,1 - 0,05 = 0,05\,mol\\
{n_{{H^ + }}} = {n_{HCl}}\text{ dư} = 0,05\,mol\\
{n_{N{a^ + }}} = {n_{NaCl}} = 0,05\,mol\\
{n_{C{l^ - }}} = 0,05 + 0,05 = 0,1\,mol\\
{V_{{\rm{dd}}}} = 0,1 + 0,1 = 0,2\,l\\
{\rm{[}}{H^ + }{\rm{]}} = {\rm{[N}}{{\rm{a}}^ + }{\rm{]}} = \dfrac{{0,05}}{{0,2}} = 0,25M\\
{\rm{[}}C{l^ - }{\rm{]}} = \dfrac{{0,1}}{{0,2}} = 0,5M\\
pH = - \log (0,25) = 0,6\\
b)\\
NaOH + HCl \to NaCl + {H_2}O\\
{n_{NaOH}} = {n_{HCl}}\text{ dư} = 0,05mol\\
{V_{{\rm{dd}}NaOH}} = \dfrac{{0,05}}{1} = 0,05l = 50ml
\end{array}\)
