Write a balanced equation for the reaction between NaOH and sulfuric acid. A conical flask contains 25cm^3 of 0.124M NaOH, a burette contains 0.0625M sulfuric acid. Find the minimum amount of acid required to completely react with the NaOH in the flask.

first, consider balanced equation: consider reaction occuring: NaOH + H₂SO₄ --> Na₂SO₄ + H₂O (neutralisation) ions present in each reactant: NaOH: Na⁺, OH^-  H₂SO₄: 2H⁺, SO₄^2- - SO₄^2- requires 2 Na⁺ ions to form a neutral compound - the 2 H⁺ ions each require 1 OH^- to form water answer: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O now consider calculation: how many moles of NaOH are there in the flask? recall: number of moles = volume (in dm³) x concentration (in moldm^-3) first, convert 25 cm³ into dm³, by dividing by 1000: 25 cm³ = 25 / 1000 dm³ = 0.025 dm³ now, plug numbers into equation: number of moles of NaOH = 0.025 dm³ x 0.124 moldm^-3 = 0.0031 mol what is the minimum number of moles of H₂SO₄ required to react completely with the NaOH? from balanced equation (see above), NaOH : H₂SO₄ = 2 : 1 (stoichiometry of reaction) - this means that 2 molecules of NaOH are required to react with 1 molecule of H₂SO₄ so, number of moles of H₂SO₄ required = 0.0031 mol / 2 = 0.00155 mol what is the minimum volume of H₂SO₄ required? recall: number of moles = volume (in dm³) x concentration (in moldm^-3) rearrange equation to give: volume (in dm³) = number of moles / concentration (in moldm^-3) so, minimum volume of H₂SO₄ required = 0.00155 mol / 6.25 x 10^-2 moldm^-3 = 0.0248 dm³ = 24.8 cm³ answer: 24.8 cm³