Đáp án:

 \(\begin{array}{l}
a)\\
{\rm{[}}N{a^ + }{\rm{]}} = 0,009M\\
{\rm{[}}C{l^ - }{\rm{]}} = 0,008M\\
{\rm{[}}O{H^ - }{\rm{] = }}0,001M\\
b)\\
pH = 11\\
c)\\
m = 0,0508g
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
a)\\
{n_{NaOH}} = 0,05 \times 0,018 = 0,0009\,mol\\
{n_{HCl}} = 0,05 \times 0,016 = 0,0008\,mol\\
NaOH + HCl \to NaCl + {H_2}O\\
\dfrac{{0,0009}}{1} > \dfrac{{0,0008}}{1} \Rightarrow\text{NaOH dư} \\
{n_{NaOH}} \text{ dư}= 0,0009 - 0,0008 = 0,0001\,mol\\
{n_{NaCl}} = {n_{HCl}} = 0,0008\,mol\\
{n_{N{a^ + }}} = 0,0008 + 0,0001 = 0,0009\,mol\\
{n_{O{H^ - }}} = {n_{NaOH}} \text{ dư} = 0,0001\,mol\\
{n_{C{l^ - }}} = {n_{NaCl}} = 0,0008\,mol\\
{\rm{[}}N{a^ + }{\rm{]}} = \dfrac{{0,0009}}{{0,1}} = 0,009M\\
{\rm{[}}C{l^ - }{\rm{]}} = \dfrac{{0,0008}}{{0,1}} = 0,008M\\
{\rm{[}}O{H^ - }{\rm{]}} = \dfrac{{0,0001}}{{0,1}} = 0,001M\\
b)\\
pH = 14 - {\rm{[}} - \log (0,001){\rm{]}} = 11\\
c)\\
{m_{NaCl}} = 0,0008 \times 58,5 = 0,0468g\\
{m_{NaOH}} \text{ dư}= 0,0001 \times 40 = 0,004g\\
m = 0,0468 + 0,004 = 0,0508g
\end{array}\)