Grade 12th passPhysical Chemistry

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

sahil kumar sahu

7 Years agoGrade 12th pass

3 Answers

Arun

7 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

7 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: N_R= N_bVb -NaVa /Vb+Va

= 0.1x500-0.1x100/ 500+100

=50-10/600

=40/600 = 0.667N

hope this helps, give a thumbsup!

Rishi Sharma

5 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