Question

Given a balanced chemical equation between H2SO4(aq) and KOH(aq) H2SO4(aq) + 2 KOH(aq) → K2SO4(aq) + 2 H2O(l) What volume (in mL) of 0.33 M H2SO4(aq) solution is necessary to completely react with 110 mL of 0.76 M KOH(aq)?

Note: (1) The unit of volume of H2SO4(aq) is in mL (2) Insert only the numerical value of your answer (do not include the units or chemical in your answer).

Answer 1

The balanced chemical equation for the reaction is given as

H₂SO₄ (aq) + 2 KOH (aq) ----------> K₂SO₄ (aq) + 2 H₂O (l)

As per the balanced stoichiometric equation,

1 mole H₂SO₄ = 2 moles KOH.

Millimol(s) KOH added = (110 mL)*(0.76 M) = 83.6 mmol.

Millimol(s) H₂SO₄ required to titrate 83.6 mmol KOH = (83.6 mmol KOH)*(1 mole H₂SO₄)/(2 moles KOH) = 41.8 mmol H₂SO₄.

Volume of 0.33 M H₂SO₄ corresponding to 41.8 mmol H₂SO₄ = (41.8 mmol H₂SO₄)/(0.33 M)

= 126.667 mL

≈ 126.67 mL (ans, correct to 2 decimal places).

The number of mL of 0.33 M H₂SO₄ required = 126.67