100 ml each of 0.5 n naoh, n / 5 hcl and n / 10 h2so4 are mixed together. The resulting solution will be
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Gram equivalent of h2so4=normality *volume(in litre)=1/5×100/1000=0.02
Gram eq. Of Naoh=1/10×500/1000=0.05
So nature of solution will be basic as gram eq of Naoh are more..
So after use of 0.02 eq H2so4, eq left=0.05–0.02=0.03 (600 ml/0.6 lit)
So normality of excess reactant left is=gram Eq/vol(lit.)=0.03/0.6=0.05N
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rupasamadrita:
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the resulting solution will be alkaline ...
so answer is Alkaline
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