Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
<h3>pKa = 5.01</h3>
1. 100.67
2. 168.55
3. 2.747
So I’m not sure if 2 or 3 are right sorry
Answer is: pH of aniline is 9.13.<span>
Chemical reaction: C</span>₆H₅NH₂(aq)+
H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq).
pKb(C₆H₅NH₂) = 9.40.
Kb(C₆H₅NH₂) = 10∧(-9.4) = 4·10⁻¹⁰.
c₀(C₆H₅NH₂) = 0.45 M.
c(C₆H₅NH₃⁺) = c(OH⁻) = x.
c(C₆H₅NH₂) = 0.45 M - x.
Kb = c(C₆H₅NH₃⁺) · c(OH⁻) / c(C₆H₅NH₂).
4·10⁻¹⁰ = x² / (0.45 M - x).
Solve quadratic equation: x = c(OH⁻) = 0.0000134 M.
pOH = -log(0.0000134 M.) = 4.87.
pH = 14 - 4.87 = 9.13.
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:
B 144.0 s is the best answer of this question