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2 NaOH + H2SO4 → Na2SO4 +2H2O
Thus, first we would have to neutralize the NaOH before H2SO4 can accumulate in the system.
For that, we can use N1V1=N2V2
Normality of H2SO4=2*molarity=0.2N
For NaOH=0.1N
Thus, for neutralization,
V=0.1*50/0.2=25ml
Thus at this point, solution volume=75ml and molarity of H2SO4 is 0M
Now whatever we add will be free H2SO4
Molarity=0.1M of solution added
Thus if V volume more is added the new molarity of the solution will be,
(0.1*V)/(75+V)
We have to find V at which this is 0.05
Thus solving,
(0.1*V)/(75+V)=0.05
Implies V=3.75/0.05=75ml
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