calculate determination of molarity ofkmno4 using 0.1M oxalic acid by volumetric analysis

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Solution

Oxalic acid molarity is given as 0.1 M
Normality of oxalic acid = ?
$M = 2 N f o r o x a l i c a c i d \sin c e i t s a d i b a s i c a c i d N o r m a l i t y o f o x a l i c a c i d = \frac{0.1}{2} = 0.05 N V o l u m e o f o x a l i c a c i d = V_{1} m l (p i p e t t e s o l u t i o n) V o l u m e o f p e r m a n g a n a t e = V_{2} m l (b u r e t t e s o l u t i o n) N o r m a i l i t y o f p e r m a n g a n a t e = N_{2} V_{1} N_{1} = V_{2} N_{2} S o e x c e p t N_{2} i s k n o w n c a l c u l a t e t h a t N_{2} = \frac{V_{1} \times N_{1}}{V_{2}} = \frac{V_{1} \times 0.05}{V_{2}} M o l a r i t y o f p e r m a n g a n a t e i s = 5 \times N o r m a l i t y (d u e t o t h e n u m b e r o f e l e c t r o n n e e d e d f o r t h i s r e d o x r e a c t i o n i s 5) M n {O_{4}}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

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