How many grams of lithium nitrate will be needed to make 250 grams of lithium sulfate assuming that you have an adequate amount
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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⁺] =
Thus, you need to add:
[H⁺] = = <em>5,31x10⁻⁸ M</em>
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 = 1,00<em>x10⁻² moles of H⁺</em>
These moles comes from
H₂SO₄ → 2H⁺ +SO₄²⁻
Thus:
1,00x10⁻² moles of H⁺ × = <em>5,00x10⁻³ moles of H₂SO₄</em>
These moles comes from:
5,00x10⁻³ moles of H₂SO₄ × ×
×
= <em>3,78 mL of 12,0wt% H₂SO₄</em>
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I hope it helps!