Question

A mixture only contains KCl and NaBr is analysed using Mohr method. A sample of this mixture weighing 0.3172 g is dissolved in 50.00 mL distilled water and titrated to an end-point 36.85 mL of 0.1120 M AgNO3 solution. The blank titration required 0.71 mL to reach the same end-point. Calculate the percentage (w/w) of KCl and NaBr each in the same sample.
[RMM: KCl = 74.6, NaBr = 102.9]

Answer 1

Blank titration required ozim. - To titrate kcl & nabor of sample volume of 07120 m AgNo needed = (36.85 0.71 day = 36 you Dooters of AgNO3 reacted with Kelf NaBr = 0.1120 melhor 36.14 y. - 6:048x10-3 mol Now Ag No react with Kel at lil ratio & similarly AgNoz reart with Noth at 1:1 ratio as AgNO3 (as) + c ) – Agel (6) + KNO3 (2) A No (ar) + Nabrlar) – Ag Br(B) + Nano So, according to problem in the mixlove kaf Nape are present. 4.848x183 and wow let, in solution mixture man. of kel present a gt NaBr Frosent (0.3172 ) g. Now molar man of kat = 74.5513 g/mol in u u u NaBr = 102: 894 g/mo Number of moles of ke was present = x/74.5513 mot & number of moles of Na Br . = 6. 371-27/ 102.894 mol

& ne 74. - According to problem: a 0.31722 – 4.048 x 10-3 102.894 102.8944 & 74.5513 (0.3172-2) = 4.048x11-3 74.5513 8102.894. = 102.894 nt 28.668 - 74.55134 = 74.5513x 102 894 9.048x183 = 28. 3427x 123648 - 31.057 . = 28.3427 = 31.057 - 23 668 = 7.9837 as 74037 28.3427 = 0.26129 won of & c = 0.2612 gf man of NaBr present - (0.3172 - 0.2 oosss In sarple - 2612 7 of cl = 0.25 14 x 10% = 82.346% &%. of Name= (180 - 82 3 46%. = 17.654%