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Balanced equation will be as follows
2NaOH + H2SO4=Na2SO4 + 2H2O
from MaVa/MbVb=Na/Nb where Ma is molarity of acid=?
Va is volume of acid=25.0 mL
Mb is molarity of base= 0.05 M
Vb is volume of base=18.3 mL
Nb number of moles of base= 2
Na is number of moles of acid =1
Ma*25 mL / 18.3mL*0.05M = 1/2
Ma=18.3*0.05M / 50 = 0.0183M
The molarity of sulphuric acid = 0.0183M
5.50M NaOH must be added to 680.0 mL of a buffer, The amount of NaOH is mathematically given as
V=2.107mL
<h3>What is the volume of 5.50M NaOH?</h3>
Generally, the equation for Chemical Reaction is mathematically given as
CH3COOH+OH^{-} ⇄ CHCOO^{-} + H2O
Therefore
pka=-log(1.8*10{-5})
pKa=5-0.25527
pKa=4.3447
Hence

x=0.0115898
Volume of NaOH
V=0.0021072L
V=2.107mL
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