1毫升等于1毫克氯离子的硝酸银标准溶液如何配

2025-03-15 05:32:53
推荐回答(1个)
回答1:

1)计算硝酸银标准溶液的浓度:
氯的原子量是35.45
T=C*35.45
C=1/35.45=0.02821mol/L
2)计算称取硝酸银的质量:
硝酸银的摩尔质量是169.88
0.02821*1=m/169.88
m=4.7923g
即配制1升0.02821mol/L的硝酸银标准溶液需称取硝酸银4.7923克。
由于硝酸银没有基准试剂,可用分析纯硝酸银配制,称取比理论量多一点的硝酸银(如4.9克),配置成1升浓度高一点标准溶液,准确标定其浓度C1,按公式C1*V1=0.02821*(V1+V)计算出加水量V,摇匀后再进行标定,其中,V1是加水前的溶液体积,要将洗滴定管及滴定消耗的部分体积全部扣除。

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