Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
The acid ionization constant (ka) of an unknown acid if a 0.030M solution of it has a pH of 2.946 is 4.8×10⁻⁵.
What is the relation between pH and [H⁺]?
Relation between the pH and [H⁺] will be shown as:
[H⁺] = [tex]10^{-pH}[/tex]
Given that pH of solution = 2.946
[H⁺] = [tex]10^{-2.946}[/tex] = 0.00113 = 0.0012
Let the ICE table for the unknown acid HA will be:
HA ⇄ H⁺ + A⁻
Initial: 0.03 0 0
Change: -0.0012 0.0012 0.0012
Equilibrium: 0.03-0.0012 0.0012 0.0012
Acid dissociation constant will be calculated as:
Ka = [0.0012][0.0012] / [0.03-0.0012]
Value of 0.0012 is very small as compared to the 0.03, so the equation becomes
Ka = (0.0012)² / 0.03 = 0.000048 = 4.8×10⁻⁵
Hence value of Ka is 4.8×10⁻⁵.
To know more about acid dissociation constant, visit the below link:
https://brainly.com/question/1614600
#SPJ4
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.
