Answer:
C) 0.24 M
Explanation:
The given chemical reaction is presented as follows;
H₂SO₄(aq) + 2 KOH(aq) → K₂SO₄(aq) + 2 H₂O(l)
The titration experiment results are;
Volume of H₂SO₄(aq) used = 12.0 mL
Volume of KOH (aq) used = 36.0 mL
Concentration of KOH (aq) = 0.16 M
The number of moles of KOH present, n = 0.16 M × 36/1000 = 0.00576 moles
From the given reaction, 1 mole of H₂SO₄ reacts with 2 moles of KOH to give 1 mole of K₂SO₄ and 2 moles of H₂O
Therefore, 0.00576 moles of KOH reacts with (1/2) × 0.00576 moles = 0.00288 moles of H₂SO₄
Therefore, for the reaction;
The number of moles of H₂SO₄ in 12.0 mL of H₂SO₄ = 0.00288 moles
The concentration of the H₂SO₄ = 0.00288 M/(12.0 mL) = 0.24 M
The concentration of H₂SO₄ in the reaction = 0.24 M.
