A sample of gas at 1.0 atm pressure has a volume of 20.0 L. If the
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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
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<u>Answer:</u> The concentration of at equilibrium is 0.00608 M
<u>Explanation:</u>
As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.
In the second dissociation, the ions will remain in equilibrium.
We are given:
Concentration of sulfuric acid = 0.025 M
Equation for the first dissociation of sulfuric acid:
0.025 0.025 0.025
Equation for the second dissociation of sulfuric acid:
<u>Initial:</u> 0.025 0.025
<u>At eqllm:</u> 0.025-x 0.025+x x
The expression of second equilibrium constant equation follows:
We know that:
Putting values in above equation, we get:
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of sulfate ion = x = 0.00608 M
Hence, the concentration of at equilibrium is 0.00608 M