What is [h] in a 0.270 m solution of acrylic acid, ch, chcooh (ka = 3.16 x 10-5)?

The concentration of [H⁺] is 2.90 x 10⁻³ M

Explanation:

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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.

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