Answer:
7
Explanation:
Assume we have 1 L of each solution.
Solution 1
Solution 2
pH = 12
pOH = 14.00 - pOH = 14.00 - 12 = 2.0
3. pH after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 10⁻² 10⁻²
C/mol: -10⁻² -10⁻²
E/mol: 0 0
The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.
pH = 7
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⁺] = , thus,
Thus, you need to add:
[H⁺] = = 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⁺ × = 8,96x10⁻³ moles of H₂SO₄
These moles comes from:
8,96x10⁻³ moles of H₂SO₄ × ×
×
=
6,78 mL of 12,0wt% H₂SO₄
I hope it helps!
