Question

When 100 ml of M/10 H2SO4 is mixed with 500ml of M/10 NaOH,then what is the nature of the resulting solution and the normality of the excess reactant left?

Answer 1

Explanation:

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

Answer 2

Answer:

As NaOH is an excess reagent it is not totally consumed during the reaction..... so some (300ml) NaOH is left out.

Therefore, the resulting soln is basic in nature.

Then,

molarity of Na2SO4 = 0.0045/0.6=0.0 075

also molarity× n-factor = normality

so normality of the Na2SO4=0.0075× 2=0.015

Explanation: