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: \[{\rm{0,003 - 0,01x = }}\,0,002 \Rightarrow x = 0,1(M)\]