Terephthalic( C8H6O4) is a diprotic acid with pka's 3.51 and 4.82.
a. What is the pH of a 0.15M solution of this acid?
b. What is the concentration of C8H4O4-2 in a 0.15M solution f this acid?
Let's call terephthalic acid H2T to make it simpler to write. You will have (H^+) from both ka1 and ka2. Take them one at a time.
............H2T ==> H^+ HT^-
I............0.15........0......0
C............-x............x......x
E,,,,,,,,0.15-x..........x......x
Ka1 = (H^+)(HT^-)/(H2T)
Plug in the E line, solve for x = (H^+) and convert to pH. I've estimated x = 0.007 M but that's just an estimate. You need to calculate it yourself and make sure you can use the assumption that x is about zero if you choose to go that route. Then for Ka2.
............HT^- ==> H^+ T^2-
I...........0.007......0.007...0
C...........-x.............x..........x
E........0.007-x....0.007+x....x
plug the E line into Ka2 expression and solve for x = H^+ from Ka2.
Total H^+ = H from ka1 + H from ka2.
For part B that is x from Ka2 calculation.
Post your work if you get stuck.
To find the pH of a 0.15M solution of Terephthalic acid (C8H6O4), we need to consider the dissociation of the acid and the ionization of water.
a. First, let's determine the major acid dissociations that occur in water. Terephthalic acid is a diprotic acid, meaning it can donate two protons (H+ ions). Its two dissociation reactions are as follows:
C8H6O4 + H2O ⇌ C8H5O4- + H3O+ (pKa = 3.51)
C8H5O4- + H2O ⇌ C8H4O4-2 + H3O+ (pKa = 4.82)
We can see that the pKa value of 3.51 corresponds to the reaction where one proton (H+) is released, while the pKa value of 4.82 corresponds to the reaction where the second proton (H+) is released.
Since the acid concentration is 0.15M, and we're assuming that both protons will be released, we need to consider the acid's concentration at equilibrium as the concentration of C8H4O4-2.
To calculate the pH, we need to determine the concentrations of H3O+ and C8H4O4-2 at equilibrium. At equilibrium, the concentration of H3O+ will be equal to the concentration of C8H4O4-2.
Let's assign 'x' to the concentration of C8H4O4-2.
Therefore,
[H3O+] = [C8H4O4-2] = x
Using the first dissociation equation and the equation for the ionization of water (Kw = [H3O+][OH-] = 1.0 x 10^-14), we can set up an equation:
Ka1 = [H3O+][C8H5O4-] / [C8H6O4] = 10^(-3.51)
Substituting the values we know:
(10^(-3.51)) = (x)(x) / (0.15 - x)
Since x is assumed to be much smaller than 0.15, we can approximate 0.15 - x as 0.15. This allows us to simplify the equation to:
(10^(-3.51)) = (x)(x) / 0.15
By solving this equation, we find x ≈ 0.0123. This gives us the concentration of C8H4O4-2 ions at equilibrium, which is also equal to [H3O+].
Using the equation for pH: pH = -log[H3O+], we can calculate the pH value:
pH = -log(0.0123) ≈ 1.91
Therefore, the pH of a 0.15M solution of Terephthalic acid is approximately 1.91.
b. To find the concentration of C8H4O4-2 (Terephthalate ions) in a 0.15M solution of the acid, we determined in part a that the concentration of C8H4O4-2 at equilibrium is approximately 0.0123M.
Therefore, the concentration of C8H4O4-2 in a 0.15M solution of Terephthalic acid is approximately 0.0123M.