Respuesta :
The value of Ka for butanoic acid is 1.53 X 10⁻⁵
Explanation:
Given-
Initial concentration of butanoic acid = 0.1M
Ionization percentage = 1.23%
Dissociation of butanoic acid:
HC₄H₇O₂ ⇒ C₄H₇O₂⁻ + H⁺
Initial 0.1 0 0
Change -x x x
Equilibrium 0.1 - x x x
The value of x can be calculated from the percent ionization.
percent ionization = |H⁺| X 100 / |HC₄H₇O₂|
1.23% = |H⁺| X 100% / 0.1M
|H⁺| = 0.1M X 1.23% / 100%
|H⁺| = 0.00123M
|H⁺| = |C₄H₇O₂⁻|
|C₄H₇O₂⁻| , x= 0.00123 M
Solving for the equilibrium concentration of HC₄H₇O₂
|HC₄H₇O₂| = 0.1 - x
|HC₄H₇O₂| = 0.1 - 0.00123 M
|HC₄H₇O₂| = 0.09877 M
Ka can be calculated as:
Ka = |C₄H₇O₂⁻| |H⁺| / |HC₄H₇O₂|
Ka = (0.00123) X (0.00123) / (0.09877)
Ka = 1.53 X 10⁻⁵
The value of Ka for butanoic acid is 1.53 X 10⁻⁵
The value of Ka for butanoic acid is 1.53 * 10⁻⁵
Dissociation of butanoic acid:
HC₄H₇O₂ ⇒ C₄H₇O₂⁻ + H⁺
Given:
Initial concentration of butanoic acid = 0.1M
Ionization percentage = 1.23%
Initial 0.1 0 0
Change -x x x
Equilibrium 0.1 - x x x
The value of x can be calculated from the percent ionization.
Percent ionization = |H⁺| * 100 / |HC₄H₇O₂|
1.23% = |H⁺| * 100% / 0.1M
|H⁺| = 0.1M * 1.23% / 100%
|H⁺| = 0.00123M
|H⁺| = |C₄H₇O₂⁻|
|C₄H₇O₂⁻| , x= 0.00123 M
Solving for the equilibrium concentration of HC₄H₇O₂
|HC₄H₇O₂| = 0.1 - x
|HC₄H₇O₂| = 0.1 - 0.00123 M
|HC₄H₇O₂| = 0.09877 M
Calculation for Ka:
Ka = |C₄H₇O₂⁻| |H⁺| / |HC₄H₇O₂|
Ka = (0.00123) * (0.00123) / (0.09877)
Ka = 1.53 X 10⁻⁵
Thus, the value of Ka for butanoic acid is 1.53 * 10⁻⁵
Find more information about Dissociation constant here:
brainly.com/question/3006391
