n\(_{H_2SO_4}\)= \(\dfrac{100\cdot1}{1000}\)=0,1(mol)
H\(_2\)SO\(_4\) + 2NaOH → Na\(_2\)SO\(_4\) + 2H\(_2\)O
(mol) 0,1 → 0,2
⇒ V\(_{NaOH}\) = \(\dfrac{n_{NaOH}}{C_{M_{NaOH}}}\) = \(\dfrac{0,2}{1}\) = 0,2(lít)
2NaOH + H2SO4 → Na2SO4 + 2H2O
\(n_{H_2SO_4}=0,1\times1=0,1\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow V_{ddNaOH}=\dfrac{0,2}{1}=0,2\left(l\right)\)
Vậy muốn trung hòa 100 ml dung dịch H2SO4 1M cần 0,2 lít dung dịch NaOH 1M