Neutralization?

Question

What volume (mL) of 0.446 M potassium hydroxide solution would just neutralize 111.4 ml of 0.121 M H2SO4 solution?

Answer 1

H2SO4+2NaOH---> Na2SO4 +2H2O

the number of of equivalent acid in 0.121M solution H2SO4 is .242 /L since the acid is bivalent
in 111.4mL = 0.1114 L there are 0.1114*0.242=0.027 eq of H+

to neutralize this you need 0.027eq of NaOH
In 1L of your solution there is 0.446 M or equivalent since the Base as a valence1

so to have 0.027 you need =0.06L = 60 mL NaOH