At what initial concentration would a solution of acetic acid (ka=1.8x10^-5) be 2% ionized?
Thanks
Let's call acetic acid HAc. Then the ionization is
HAc ==> H^+ + Ac^-
We will call the initial concn of HAc x. If it is 2% ionized, then (H^+) = 0.02*x
(Ac^-) = 0.02*x
and (HAc) = x-(0.02*x) = 0.98*x
Now write the Ka expression and plug in those concns, then solve for x. Post your work if you get stuck. The answer is approximately 0.05 M
what is the difference between hydration and hydrogenation reactions?
How did you get 0.98? Why were you allowed to subtract 0.02 to get that? I'm still confused as to how to plug this in and solve, just because of the 0.98. Thanks
To find the initial concentration at which a solution of acetic acid is 2% ionized, we can use the concept of percent ionization and the Ka value of acetic acid.
Percent ionization is defined as the ratio of the concentration of ionized acid to the initial concentration of the acid, multiplied by 100%.
Percent ionization = (concentration of ionized acid / initial concentration of acid) x 100%
In this case, we want the percent ionization to be 2%, so we can rewrite the equation as:
2 = (concentration of ionized acid / initial concentration of acid) x 100%
Now, let's consider the ionization reaction of acetic acid:
CH3COOH ⇌ CH3COO- + H+
The equation for Ka is:
Ka = ([CH3COO-] * [H+]) / [CH3COOH]
Since the initial concentration of acetic acid is equal to its concentration after ionization, we can write:
2 = ([CH3COO-] / [CH3COOH]) x 100%
Now, let's substitute [CH3COOH] with 'x' to represent the initial concentration of acetic acid. Since the acid ionizes in a 1:1 ratio, the concentration of CH3COO- and H+ will be equal.
2 = (x / x) x 100%
2 = 1 x 100%
2 = 100%
Since 2 does not equal 100%, we made an error. This means that acetic acid cannot be 2% ionized.
Therefore, there is no initial concentration of acetic acid that could result in a 2% ionization.