Answer:
The concentration of [H⁺] is 2.90 x 10⁻³ M
Explanation:
Hi there!
Acrylic acid is a weak acid, then, it does not dissociate completely:
CHCHCOOH ⇄ CHCHCOO⁻ + H⁺
In the equilibrium, the concentrations of H⁺ and CHCHCOO⁻ are equal but unknown. Let´s call these concentrations as "x"
The concentration of CHCHCOOH in the equilibrium is the initial concentration minus the concentration of H⁺or CHCHCOO⁻. Then, in the equilibrium, the concentration of CHCHCOOH will be: 0.270 M - x
Concentrations in equilibrium:
CHCHCOOH ⇄ CHCHCOO⁻ + H⁺
0.270 - x x x
The constant Ka is calculated as:
ka = product of the concentrations of products / product of the concentrations of reactans
For a generic reaction:
HA ⇄ A⁻ + H⁺
ka = [A⁻][H⁺] / [HA]
where:
[A⁻], [H⁺] and [HA] are molar concentrations
In our problem:
ka = [ CHCHCOO⁻][H⁺] / [CHCHCOOH]
ka = x * x / 0.270 - x
3.16 x 10⁻⁵ = x² / 0.270 - x
3.16 x 10⁻⁵ * ( 0.270 - x) = x²
8.53 x 10⁻⁶ - 3.16 x 10⁻⁵ * x = x²
8.53 x 10⁻⁶ - 3.16 x 10⁻⁵ * x - x² = 0
Solving the quadratic equation (a = -1, b = -3.16 x 10⁻⁵ and c = 8.53 x 10⁻⁶) :
x = 2.90 x 10⁻³ M ( the other vlaue of x is negative and therefore discarded)
Then, the concentration of [H⁺] is 2.90 x 10⁻³ M.
The equilibrium has been the condition in which the concentration of reactant and product equal. The concentration of hydrogen ion in the solution has been [tex]\rm 2.90\;\times\;10^{-3}[/tex] M.
The Ka has been the equilibrium concentration of the reaction. The Ka can be given as:
[tex]\rm Ka=\dfrac{Product}{Reactant}[/tex]
The equilibrium reaction can be given as:
[tex]\rm CHCHCOOH\;\leftrightharpoons\;CHCHCOO^-\;+\;H^+[/tex]
The equilibrium concentration can be given in the ICE table attached.
Substituting the values for the Ka:
[tex]\rm Ka=\dfrac{[CHCHCOO^-]\;[H^+]}{[CHCHCOOH]}\\\\ 3.16\;\times\;10^-^5=\dfrac{x\;\times\;x}{0,27-x}\\\\ x=2.90\;\times\;10^{-3\;}M[/tex]
The concentration of hydrogen ion in the solution has been [tex]\rm 2.90\;\times\;10^{-3}[/tex] M.
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