The acid dissociation constant of trimethyl acetic acid is . the pH of a solution of trimethyl acetic acid is 2.22.
the trimethyl acetic acid dissociates as :
HC(CH₃)₃CO₂ ---> C(CH₃)₃CO₂⁻ + H⁺
the concentration of acid Ca = 3.8 M
the acid dissociation constant expression is given as :
[ H⁺ ] = √Ca × Ka
= √3.8 × 9.33 × 10⁻⁶
= 5.9 × 10⁻³ M
pH = -log [ H⁺ ]
= - log (5.9 × 10⁻³)
= -log 5.9 - log 10⁻³
= - 0.77 + 3
= 2.22
The question is in complete the complete question is :
The acid dissociation constant Ka of trimethylacetic acid is 9.33 × 10⁻⁶ . calculate the pH of a 3.8 M solution of trimethylacetic acid. round your answer to decimal place.
To learn more about acid dissociation constant here
https://brainly.com/question/28187564
#SPJ4
