EP 中含量检验方法

15107377166

各位大神，EP药典中某原料药含量的方法是这样的：Dissolve 0.250 g in a mixture of 20 mL of dilute hydrochloric acid R and 50 mL of water R. Cool the solution in iced water. Carry out the determination of primary aromatic aminonitrogen (2.5.8), determining the end-point electrometrically.1 mL of 0.1 M sodium nitrite is equivalent to 27.83 mg of C12H14N4O2S.
其中2.5.8：Dissolve the prescribed quantity of the substance to be examined in 50 mL of dilute hydrochloric acid R or in another prescribed solvent and add 3 g of potassium bromide R. Cool in ice-water and titrate by slowly adding 0.1 M sodium nitrite with constant stirring.
检验方法就是将0.250 g溶解在20 mL稀盐酸R和50 mL水R的混合物中。将溶液在冰水中冷却。用电测法测定终点吗，2.5.8里的内容是什么意思？

xyy0250119

2.5.8就是这个the determination of primary aromatic aminonitrogen

xyy0250119

2.5.7. Unsaponifiable matter.................................................... 169
2.5.8. Determination of primary aromatic aminonitrogen......................................................................169
2.5.9. Determination of nitrogen by sulfuric acid digestion.

xyy0250119

2.5.8. DETERMINATION OF PRIMARY
AROMATIC AMINO-NITROGEN
Dissolve the prescribed quantity of the substance to be
examined in 50 mL of dilute hydrochloric acid R or in another
prescribed solvent and add 3 g of potassium bromide R. Cool
in ice-water and titrate by slowly adding 0.1 M sodium nitrite
with constant stirring.
Determine the end-point electrometrically or by the use of
the prescribed indicator.EP通则2.5.8里面的内容就这些

hn_xj

操作方法：将0.250 g溶于20ml稀盐酸R和50ml水R的混合物中，用冰水冷却，加入3g溴化钾R，不断搅拌，缓慢加入0.1 M亚硝酸钠滴定，用电位滴定法测定终点。1 mL 0.1 M的亚硝酸钠相当于27.83 mg的C12H14N4O2S。

15107377166

xyy0250119 发表于 2023-4-19 15:15
2.5.8. DETERMINATION OF PRIMARY
AROMATIC AMINO-NITROGEN
Dissolve the prescribed quantity of the su ...

懂了，谢谢大神指点

15107377166

hn_xj 发表于 2023-4-19 15:49
操作方法：将0.250 g溶于20ml稀盐酸R和50ml水R的混合物中，用冰水冷却，加入3g溴化钾R，不断搅拌，缓慢加入 ...

感谢