Answer:
6,78 mL of 12,0 wt% 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^{-7,20}[/tex], thus,
Thus, you need to add:
[H⁺] = [tex]10^{-7,2} -10^{-8,0}[/tex] = 5,31x10⁻⁸ M
The total volume of the pool is:
9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L
Thus, moles of H⁺ you need to add are:
5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
1,792x10⁻² moles of H⁺ × [tex]\frac{1H_{2}SO_4 mol}{2H^{+} mol}[/tex] = 8,96x10⁻³ moles of H₂SO₄
These moles comes from:
8,96x10⁻³ moles of H₂SO₄ × [tex]\frac{98,1 g}{1 mol}[/tex] × [tex]\frac{100 gSolution}{12 gH_{2}SO_4 }[/tex] × [tex]\frac{1 mL}{1,080 g}[/tex] =
6,78 mL of 12,0wt% H₂SO₄
I hope it helps!
