CaCO3试样0.2500g,溶解于25.00ml,0.2006mol*l的HCL溶液中,过量盐酸用15.50ml的NaOH溶液进行返滴定,.
求此试样中CaCO3的质量分数(详)
C NaOH=0.2050mol*l
数学人气:885 ℃
优质解答
NaOH溶液的浓度请给出来再帮你分析o 实在不好意思我没打上去......C NaOH=0.2050mol*ln(NaOH)=0.2050*0.01550=0.0031775mol NaOH---------HCl1mol1mol 0.0031775 0.0031775n(HCl)=0.025*0.2006=0.005015与CaCO3反应的HCl的物质的量为:0.005015-0.0031775=0.0018375CaCO3---------------------------2HCl100 73x= 0.00251712 0.0018375CaCO3的质量分数:0.00251712/0.25=1.006848%