Answer:
pH = 4.27. Porcentaje de disociación: 0.03%
Explanation:
El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
Donde la constante de equilibrio, Ka, es
Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]
Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:
[H⁺] = [X⁻]
[HX] es:
20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M
Reemplazando es Ka:
1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]
2.858x10⁻⁹ = [H⁺]²
5.35x10⁻⁵M = [H⁺]
pH = -log[H⁺]
<h3>pH = 4.27</h3>
El porcentaje de disociacion es [X⁻] / [HX] inicial * 100
Reemplazando
5.35x10⁻⁵M / 0.1732M * 100
<h3>0.03%</h3>
Answer:
131 atm
Explanation:
To find the new pressure, you need to use Boyle's Law:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.
P₁ = 3.88 atm P₂ = ? atm
V₁ = 7.74 L V₂ = 0.23 L
P₁V₁ = P₂V₂ <----- Boyle's Law
(3.88 atm)(7.74 L) = P₂(0.23 L) <----- Insert values
30.0312 = P₂(0.23 L) <----- Simplify left side
131 = P₂ <----- Divide both sides by 0.23
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21Explanation:BEACUSE I CAN
