Answer :
So the first thing we must do is write a balanced equation for the reaction and we know the equation is balnced when all the species on the RHS is equal to the species on the LHS
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
So now it's time to identify what you know from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (volume of NaOH)
If 1000 ml of H₂SO₄ contain 0.100 mol [0.100 M is the amount of moles in
1 L (1000 ml)]
then let 20 ml of H₂SO₄ contain x mol [20 ml is the amount of the acid that took place in the reaction]
⇒ x = [tex] \frac{20 ml * 0.1 mol}{1000 mol} [/tex]
= 0.002 mol
Mole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
2NaOH + 1H₂SO₄ → Na₂SO₄ + 2H₂O
thus mole ratio of NaOH to H₂SO₄ is 2 : 1
∴ if mole of of H₂SO₄ = 0.002 mol
then moles of NaOH = (0.002 mol) × 2
= 0.004 mol
Now molarity = [tex] \frac{MOLES}{MOLARITY} [/tex]
∴ by transposition
Volume = [tex] \frac{moles}{molarity} [/tex]
= [tex] \frac{0.004 mol}{0.1 mol/L} [/tex]
= 0.04 L
= 40 mL
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
So now it's time to identify what you know from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (volume of NaOH)
If 1000 ml of H₂SO₄ contain 0.100 mol [0.100 M is the amount of moles in
1 L (1000 ml)]
then let 20 ml of H₂SO₄ contain x mol [20 ml is the amount of the acid that took place in the reaction]
⇒ x = [tex] \frac{20 ml * 0.1 mol}{1000 mol} [/tex]
= 0.002 mol
Mole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
2NaOH + 1H₂SO₄ → Na₂SO₄ + 2H₂O
thus mole ratio of NaOH to H₂SO₄ is 2 : 1
∴ if mole of of H₂SO₄ = 0.002 mol
then moles of NaOH = (0.002 mol) × 2
= 0.004 mol
Now molarity = [tex] \frac{MOLES}{MOLARITY} [/tex]
∴ by transposition
Volume = [tex] \frac{moles}{molarity} [/tex]
= [tex] \frac{0.004 mol}{0.1 mol/L} [/tex]
= 0.04 L
= 40 mL