Question

What is the concentration of an naoh solution that requires 15.0 ml of a 0.750 m h2so4 solution to neutralize 17.5 ml of naoh?

Answer 1

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 =
`(15ml * 0.750 mol)/(1000 ml)`

= 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 =
`(1000ml * 0.0225 mol)/(17.5 ml)`

= 1.286 mol

∴ concentration of NaOH is 1.286 mol/L