when 1 L of 0.1 M sulphuric acid solution is allowed to react with 1L of 0.1M of sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution is obtained is

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$H_{2} S O_{4} + 2 N a O H \to N a_{2} S O_{4} + 2 H_{2} O I t s g i v e n t h a t 1 L o f 0.1 M H_{2} S O_{4} = 0.1 m o l H_{2} S O_{4} 1 L o f 0.1 M N a O H = 0.1 m o l N a O H A s p e r t h e g i v e n e q u a t i o n, 1 m o l e o f s u l p h u r i c a c i d r e a c t s w i t h 2 m o l e s o f N a O H 0.1 m o l o f s u l p h u r i c a c i d r e a c t s w i t h = \frac{2}{1} \times 0.1 = 0.2 m o l e I t s g i v e n t h a t 0.1 m o l e o f N a O H o n l y . S o N a O H a c t s a s a l i m i t i n g r e a g e n t . T h e a c i d l e f t o u t = 0.05 m o l e 2 m o l e s o f N a O H g i v e s 1 m o l e o f N a_{2} S O_{4} 0.1 m o l e o f N a O H g i v e s = \frac{1}{2} \times 0.1 = 0.05 m o l e M o l e = \frac{G i v e n m a s s}{M o l a r m a s s o f N a_{2} S O_{4}} G i v e n m a s s = 0.05 \times M o l a r m a s s o f N a_{2} S O_{4} = 0.05 \times 142 = 7.10 g A s p e r t h e v o l u m e t r i c p r i n c i p l e V_{1} M_{1} + V_{2} M_{2} = V_{3} M_{3} S i n c e t h e a c i d a n d b a s e a r e g i v e n i t u n d e r g o e s n e u t r a l i s a t i o n . S o t h e l e f t o u t a c i d o r b a s e i s c a l c u l a t e d . L e f t o u t a c i d = 0.05 m o l e s M o l a r i t y = \frac{N o o f m o l e s}{V o l u m e o f t h e s o l u t i o n} = \frac{0.05}{2} = 0.025 m o l / L$

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