At what initial concentration would a solution of acetic acid (ka=1.8x10^-5) be 2% ionized?
I am stuck. Thanks
-J
Answered above under needs help.
Answered under needs help please.
To determine the initial concentration of a solution of acetic acid where it is 2% ionized, we can use the equilibrium expression for the ionization of acetic acid:
\[ K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} \]
Where:
Ka = acid dissociation constant for acetic acid = 1.8x10^-5
[H+] = concentration of H+ ions formed (in mol/L)
[A-] = concentration of A- ions formed (in mol/L)
[HA] = initial concentration of acetic acid (in mol/L)
Since acetic acid is a weak acid, it does not completely dissociate in water, and the extent of ionization is given by the equation:
\[ \text{% ionization} = \dfrac{[\text{H}^+]}{[\text{HA}]} \times 100 \]
Given that the % ionization is 2%, we can substitute these values into the equation:
\[ 2 = \dfrac{[\text{H}^+]}{[\text{HA}]} \times 100 \]
Now, we need to solve for the initial concentration of acetic acid ([HA]).
Let's assume that x is the concentration of [H+]. Since the acetic acid dissociates into equal amounts of [H+] and [A-], we can also assume that x is the concentration of [A-].
Therefore, the initial concentration of acetic acid ([HA]) will be the sum of [H+] and [A-], which can be represented as 2x.
Substituting these values into the % ionization equation:
\[ 2 = \dfrac{x}{2x} \times 100 \]
Simplifying the equation:
\[ 2 = \dfrac{x}{2x} \times 100 \]
\[ 2 = \dfrac{1}{2}x \times 100 \]
\[ 2 = 50x \]
Solving for x:
\[ x = \dfrac{2}{50} \]
\[ x = 0.04 \, \text{mol/L} \]
Since we assumed that x is the concentration of [H+], the concentration of [HA] (acetic acid) will be twice that value:
\[ [\text{HA}] = 2x \]
\[ [\text{HA}] = 2 \times 0.04 \]
\[ [\text{HA}] = 0.08 \, \text{mol/L} \]
Therefore, the initial concentration of the acetic acid solution would be 0.08 mol/L to have a 2% ionization.