If the pH of a 0.714 M HC₃H5O₃ solution is 2.00,what is the ionization constant for lactic acid?

Question

Quizplus · Accepted Answer

The ionization constant for lactic acid (HC₃H5O₃) is given by the expression:
Ki = [H3O+][C3H5O3-]/[HC3H5O3]
where [H3O+] is the concentration of hydronium ions, [C3H5O3-] is the concentration of lactate ions, and [HC3H5O3] is the concentration of undissociated lactic acid.

Since the pH of the solution is 2.00, we can calculate the concentration of hydronium ions as follows:

pH = -log[H3O+]
2.00 = -log[H3O+]
[H3O+] = 10^-2.00 = 0.01 M

Now, we can use the dilution equation to calculate the concentration of lactate ions and undissociated lactic acid:

0.714 M = [C3H5O3-] + [HC3H5O3]

Assuming that the degree of ionization (α) is small, we can make the approximation that the concentration of undissociated lactic acid is approximately equal to the initial concentration:

[HC3H5O3] ≈ 0.714 M

Now, we can substitute these values into the expression for Ki and solve for the ionization constant:

Ki = [H3O+][C3H5O3-]/[HC3H5O3]
Ki = (0.01 M)(α)/0.714 M
Ki ≈ 1.4 x 10^-4

Therefore, the answer is C) Ki = 1.4 x 10^-4.

If the PH of a 0