Here diacidic base is present which give two end point one with phenalphthein and second with methy orange
For phenolphthalein end point-
"\\frac{1}{2}\\ Meq\\ of\\ Diacidic\\ base = Meq.\\ of\\ of \\ 8ml \\ \\frac{N}{50}\\ HCl"
="\\frac{1\\times 8}{50\\times1000}=\\frac{1}{6225}\\ Meq\\\\"
"Meq.\\ of\\ Diacidic\\ base=2\\times Meq.\\ of\\ HCl=\\frac{2}{6225}\\ Meq."
For methyl orange end point-
"Meq.\\ of\\ monoacidic\\ base\\ formed=Meq.\\ of\\ 1ml of\\ extra\\ HCl\\ used"
"=\\frac{1\\times1}{50\\times 1000}=\\frac{1}{50000\\ Meq.}"
Normality of diacidic alkalinity present in 100 ml water = "\\frac{2\\times 1000}{6225\\times 100}=\\frac{20}{6225}=\\frac{4}{1245}\\ N"
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