Đáp án:
Giải thích các bước giải:
a)PTHH:
\[\begin{array}{l}
{\rm{HCl + NaOH }} \to {\rm{NaCl + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\
{\rm{NaOH + HN}}{{\rm{O}}_3} \to NaN{O_3} + {H_2}O
\end{array}\]
b) Gọi nồng độ mol của dung dịch HCl ban đầu là x
\[\begin{array}{l}
\,\,\,\,\,\,\,\,\,\,\,{\rm{HCl}}\,\,\,\,{\rm{ + }}\,\,\,{\rm{NaOH }} \to {\rm{NaCl + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\\
{\rm{bd: }}\,\,\,\,{\rm{0,01x}}\,\,\,\,\,\,\,\,{\rm{0,003}}\\
{\rm{pu}}\,\,\,\,\,\,\,{\rm{0,01x}} \to {\rm{0,01x}}\\
{\rm{du}}\,\,\,\,\,\,\,\,\,{\rm{0}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0,003 - 0,01x}}\\
\,\,\,\,\,{\rm{NaOH}}\,\,\,\,\,\,{\rm{ + }}\,\,\,\,\,\,\,{\rm{HN}}{{\rm{O}}_3} \to NaN{O_3} + {H_2}O\\
{\rm{0,003 - 0,01x }}\,\,\,\,\,\,\,\,\,\,\,\,0,002
\end{array}\]
Trung hòa NaOH dư bằng HNO3 vừa đủ nên:
\[{\rm{0,003 - 0,01x = }}\,0,002 \Rightarrow x = 0,1(M)\]
