100ml of dibasic acid is completely neutralised with 10g NaOH .calculate the normality of acidic solution

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Solution

$A s p e r t h e v o l u m e t r i c p r i n c i p l e V_{1} M_{1} = V_{2} M_{2} L e t u s t a k e T h e v o l u m e o f t h e a c i d V_{1} = 100 m l M o l a r i t y o f t h e d i b a s i c a c i d = M_{1} T h e v o l u m e o f t h e b a s e V_{2} = 1000 m l M o l a r i t y o f t h e N a O H M_{2} = \frac{N u m b e r o f m o l e s o f s o l u t e}{V o l u m e o f t h e s o l u t i o n i n L} = \frac{10 / 40}{1} = 0.25 M S u b s t i t u t e a l l t h e s e 100 \times M_{1} = 1000 \times 0.25 M_{1} = \frac{1000 \times 0.25}{100} = 2.5 M N o r m a l i t y = M o l a r i t y \times n (n f a c t o r i s 2 a s i t s a d i b a s i c a c i d) = 2.5 \times 2 = 5 N$

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