Answer:
3,48 mL of 8,00wt% H₂SO₄
Explanation:
The equilibrium in water is:
H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)
The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = [tex]10^{-7,2}[/tex]
Thus, you need to add:
[H⁺] = [tex]10^{-7,2} - 10^{-8}[/tex] = 5,31x10⁻⁸ M
The total volume of the pool is:
5,00 m × 15,0 m ×1,50 m = 112,5 m³ ≡ 112500 L
Thus, moles of H⁺ you need to add are:
5,31x10⁻⁸ M × 112500 L = 5,97x10⁻³ moles of H⁺
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
5,97x10⁻³ moles of H⁺ × [tex]\frac{1 H_{2}SO_{4} moles}{2H^+ mole}[/tex] = 2,99x10⁻³ moles of H₂SO₄
These moles comes from:
2,99x10⁻³ moles of H₂SO₄ × [tex]\frac{98,1g}{1mol}[/tex] × [tex]\frac{100 gsolution}{8g H_{2}SO_{4} }[/tex] × [tex]\frac{1mL}{1,052 g}[/tex] = 3,48 mL of 8,00wt% H₂SO₄
I hope it helps!
