The concentration of propionic acid (CH3CH2COOH) solution was found to be 5.0 x 10^2 M, at 25 degree Celsius. (Ka for propionic acid is; 1.34 x 10^5, Kw= [H+][OH-]= 1 x 10^14)
Calculate,
A) Hydrogen (H+) ion concentration, ([H3O+] + [H+])
B) Hydroxide (OH-) ion concentration
C) pH of propionic acid solution
I will bet money big time that you omitted the negative sign (-) on three items. I KNOW Ka is 1.34 x 10^-5 and I KNOW Kw = 1.0 x 10^-14. Then I assume the concn of propionic acid is not 500 M but rather that it is 0.05M (that's 5 x 10^-2 M.) So let me call propionic acid (CH3CH2COOH) something much more simple such as HPr.
...................HPr ==> H^+ + Pr^-
I.................0.05M......0.........0
C................-x.............x.........x
E...............0.05-x........x.........x
Substitute the E line into the Ka expression of
Ka = 1.34E-5 = (H^+)(Pr^-)/(HPr) and solve for (H^+).
Then pH = -log(H^+) . Then knowing H^+ and Kw you can solve for OH^-
I'm not sure what you want for H3O^+ + H^+. You may have meant H3O^+ = (H^+)
Post your work if you get stuck or if you want me to check your answers.
To calculate the hydrogen (H+) ion concentration, we can use the relationship between the concentration of the weak acid and its dissociation constant (Ka):
Ka = [H+][A-] / [HA]
In this case, propionic acid (CH3CH2COOH) acts as a weak acid, so we can write the dissociation reaction as follows:
CH3CH2COOH ⇌ CH3CH2COO- + H+
Since the dissociation constant (Ka) is provided as 1.34 x 10^5, we can use the given concentration of the propionic acid solution (5.0 x 10^2 M) to find the concentration of [H+].
Assuming the concentration of [H+] formed due to the dissociation of propionic acid is "x," the concentration of [CH3CH2COO-] will also be "x." However, the concentration of undissociated propionic acid ([HA]) will be (5.0 x 10^2 - x) M.
Using the equation for Ka and substituting the values, we have:
1.34 x 10^5 = x * x / (5.0 x 10^2 - x)
Simplifying the equation, we get:
x^2 - 1.34 x 10^5 x + (1.34 x 10^5)(5.0 x 10^2) = 0
This is a quadratic equation, which can be solved to find the value of x (the concentration of [H+]). However, to simplify the calculation, we can use the assumption that the dissociation of weak acids is generally less than 5%, so "x" will be much smaller than 5.0 x 10^2.
Using this approximation, we can ignore "x" in the denominator, simplifying the equation to:
1.34 x 10^5 = x * x / 5.0 x 10^2
Cross-multiplying, we get:
1.34 x 10^5 * 5.0 x 10^2 = x * x
6.7 x 10^7 = x^2
Taking the square root of both sides, we find:
x = √(6.7 x 10^7)
x ≈ 8196
Therefore, the concentration of the hydrogen (H+) ions, [H+], in the propionic acid solution is approximately 8196 M.
Now, to calculate the hydroxide (OH-) ion concentration, we can use the relationship between the hydrogen ion concentration [H+] and the concentration of hydroxide ions [OH-], which is given by Kw = [H+][OH-].
Given that the value of Kw is 1 x 10^14, we can substitute the concentration of [H+] (8196 M) into the equation:
1 x 10^14 = 8196 * [OH-]
Solving for [OH-], we find:
[OH-] = (1 x 10^14) / 8196
[OH-] ≈ 1.22 x 10^10 M
Thus, the concentration of hydroxide (OH-) ions in the solution is approximately 1.22 x 10^10 M.
Finally, to calculate the pH of the propionic acid solution, we can use the relationship between pH and the hydrogen ion concentration:
pH = -log[H+]
Using the value obtained earlier for the hydrogen ion concentration [H+] (8196 M), we can find the pH:
pH = -log(8196)
pH ≈ 3.91
Therefore, the pH of the propionic acid solution is approximately 3.91.