At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
The equilibrium constant, ka, for the monoprotic acid is 7.08 × 10⁻¹⁶.
How do we calculate Ka?
Acid dissociation constant (Ka) for monoprotic acid will be calculated according to the given equation:
Ka = [H₃O⁺][A⁻] / [HA]
Relation between H₃O⁺ and pH will be represented as:
[H₃O⁺] = [tex]10^{-pH}[/tex]
pH = 6.44 (given)
[H₃O⁺] = [tex]10^{-6.44}[/tex] = 3.63×10⁻⁷
Given chemical reaction with ICE table is:
HA(aq) + H₂O(l) ⇄ H₃O⁺(aq) + A⁻
Initial: 0.0186 0 0
Change: -3.63×10⁻⁷ 3.63×10⁻⁷ 3.63×10⁻⁷
Equilibrium: 0.0186-3.63×10⁻⁷ 3.63×10⁻⁷ 3.63×10⁻⁷
Now Ka for this will be:
Ka = (3.63×10⁻⁷)² / (0.0186-3.63×10⁻⁷)
Value of 3.63×10⁻⁷ is negligible as compared to the 0.0186 value and equation becomes,
Ka = (3.63×10⁻⁷)² / 0.0186 = 13.18 × 10⁻¹⁴ / 0.0186
Ka = 7.08 × 10⁻¹⁶
Hence value of equilibrium constant is 7.08 × 10⁻¹⁶.
To know more about equilibrium constant, visit the below link:
https://brainly.com/question/12858312
#SPJ4
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
