Question

How many millilitres of 0.200M of NaOH are required to neutralize 37.0 mL of 0.125 M sulphuric acid

Answer 1

$MaVa=MbVb$

$Ma=0.200M$

$Va=?$

$Mb=0.125M$

$Vb=37.0ml$

$0.200×Va=0.125×37$

$Va=0.125×37/0.200$

$Va=23.125ml$