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WHEN 100 ml OF M/10 H2SO4 IS MIXED WITH 500 ml OF M/10 NAOH THEN NATURE OF RESULTING SOLUTION AND NORMALITY OF EXCESS OF REACTANT LEFT IS

WHEN 100 ml OF M/10 H2SO4 IS MIXED WITH 500 ml OF M/10 NAOH THEN NATURE OF RESULTING SOLUTION
AND NORMALITY OF EXCESS OF REACTANT LEFT IS
 

Grade:12th pass

3 Answers

Arun
25750 Points
5 years ago

This mixture would lead to a neutralization reaction as follows:

H2SO4 + 2 NaOH → Na2SO4 + 2 H2O

Thus for every mole of H2SO4 two moles of NaOH are used.

If u see, H2SO4 is limiting reagent.

Thus, all H2SO4 will be used up while 0.03 (0.05–0.02) moles of NaOH are left.

Thus solution would be basic.

Normality of NaOH is=Equivalents of NaOH left(=moles for NaOH)/ Volume(L)

=0.03/0.6 (Volume = 500ml + 100ml = 600ml = 0.6L)

Normality = 1/20 N = 0.05 N

Apurva Sharma
220 Points
5 years ago
n equivalent (n eq) = molarity x volume
n eq of H2SO4 = 0.1 x 100 =10
n eq of NaOH = 0.1 x 500 =50 
now n eq of NaOH > H2SO4 i.e 50>40
thus the resultant solution will be BASIC
 
now Normality: NR= NbVb -NaVa /Vb+Va 
                              = 0.1x500-0.1x100/ 500+100
                             =50-10/600
                             =40/600 = 0.667N 
 
hope this helps, give a thumbsup! 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

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

Thanks and Regards

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