40 ml of N10 HCl and 60 ml of N20 KOH are mixed together then the normality of salt formed is ?

m. eq. of HCl= N.V (ml) 0.1×40=4m. eq. of KOH = N.V (ml) =0.05×60=3HCl+ KOH→KCl+H2O1 m. eq 1 meq → 3 meq4 meq 3 meq→04−3=1 3−3=0excessreagent Limitingreagentm.eq of HCl left =1Total volume of Soln=60+40=100mlm.eq of salt formed =N×Vml=3 ⇒N×100=3 ⇒N=3100=0.03

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40 ml of $(\frac{N}{10})$ HCl and 60 ml of $(\frac{N}{20})$ KOH are mixed together then the normality of salt formed is ?

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answer is 0.03.

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Detailed Solution

m. eq. of HCl= N.V (ml) $0.1 \times 40 = 4$

m. eq. of KOH = N.V (ml) = $0.05 \times 60 = 3$

$\begin{array}{l} H C l + K O H \to K C l + H_{2} O \\ 1 m . e q 1 m e q \to 3 m e q \\ 4 m e q 3 m e q \to 0 \\ (4 - 3) = 1 (3 - 3) = 0 \\ (\begin{array}{l} e x c e s s \\ r e a g e n t \end{array}) (\begin{array}{l} L i m i t i n g \\ r e a g e n t \end{array}) \\ m . e q o f H C l l e f t = 1 \\ T o t a l v o l u m e o f S o l^{n} = 60 + 40 = 100 m l \end{array}$

$\begin{array}{l} m . e q o f s a l t f o r m e d = N \times V (m l) = 3 \\ \Rightarrow N \times 100 = 3 \\ \Rightarrow N = \frac{3}{100} = 0.03 \end{array}$