Answer:
the pH of the pond is 4.0963
Explanation:
Given the data in the question;
4FeS₂(s) + 11O₂(g) ------> 2Fe₂O₃(s) + 8SO₂(g)
and oxidation of sulfur dioxide to trioxide
SO₃(g) + H₂O(l) --------> H₂SO₄(aq)
given that; coal = 61.7 metric tons = 61700 kg
1.20 % by weight S, therefore mass of sulfur in coal will be;
⇒ 61700 × 1.20% = 740.4 kg
moles of sulfur = (740.4 × 1000 g) / 32 g/mole = 23,137.5 moles
moles of so₂ produced
= same moles of so₃ and same moles of H₂SO₄ = 23,137.5 moles
volume of pond = 397 m x 258 m x 5.64 m = 577,682.64 m³ = 577682640 L
Molarity = moles / volume = 23,137.5 / 577682640 = 4.0052 × 10⁻⁵ M
now each mole of H₂SO₄ will give 2 moles of H⁺
[H⁺] = 2 × 4.0052 × 10⁻⁵ = 8.0104 × 10⁻⁵ M
pH = -log[H⁺]
pH = -log [ 8.0104 × 10⁻⁵ M ]
pH = 4.0963
Therefore, the pH of the pond is 4.0963