Answer:
3,78 mL of 12,0wt% 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:
7,00 m × 18,0 m ×1,50 m = 189 m³ ≡ 189000 L
Thus, moles of H⁺ you need to add are:
5,31x10⁻⁸ M × 189000 L = 1,00x10⁻² moles of H⁺
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
1,00x10⁻² moles of H⁺ × [tex]\frac{1 H_{2}SO_{4} moles}{2H^+ mole}[/tex] = 5,00x10⁻³ moles of H₂SO₄
These moles comes from:
5,00x10⁻³ moles of H₂SO₄ × [tex]\frac{98,1g}{1mol}[/tex] × [tex]\frac{100 gsolution}{12 g H_{2}SO_{4} }[/tex] × [tex]\frac{1mL}{1,080 g}[/tex] = 3,78 mL of 12,0wt% H₂SO₄
I hope it helps!
