HCl+NaOH=NaCl+H2O
取W克盐酸,含氯化氢18.25%W,消耗固体烧碱40*18.25%W/36.5克,生成氯化钠18.25%W*58.5/36.5,物质的量18.25%*W/36.5
溶液总质量=W+40*18.25%W/36.5
溶液总体积=(W+40*18.25%W/36.5)/1.2
氯化钠溶液物质的量浓度=(18.25%W*/36.5)/((W+40*18.25%W/36.5)/1.2)=(18.25%/36.5)/((1+40*18.25%/36.5)/1.2)=0.005摩尔/升