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 reactant you know the most for from the question (volume & conc. of H₂SO₄) and use that info to find the unknown (conc. of NaOH)
If 1000 ml of H₂SO₄ contain 0.750 mol [0.750 M is the amount of moles in
1 L (1000 ml)]
then let 15 ml of H₂SO₄ contain x mol [15 ml is the amount of the acid that took part in the reaction]
⇒ x = [tex] \frac{15ml * 0.750 mol}{1000 ml} [/tex]
= 0.01125 mol
Mole ratio of NaOH to H₂SO₄ can be obtained from the balanced equation
0 2NaOH + 1H₂SO₄ → Na₂SO₄ + 2H₂O
mole ratio of NaOH to H₂SO₄ is 2 : 1
∴ if mole of of H₂SO₄ = 0.01125 mol
then moles of NaOH = (0.01125 mol) × 2
= 0.0225 mol
If 17.5 ml of NaOH contain 0.0225 mol [this was given in the question]
then let 1000 ml of NaOH contain x
⇒ x = [tex] \frac{1000ml * 0.0225 mol}{17.5 ml} [/tex]
= 1.286 mol
∴ concentration of NaOH is 1.286 mol/L
