Question

A student prepares a 0.28 M solution of acetic acid. If the Ka of the acid is 1.76 x 10^-5, then what is the pH of the solution?

4.76

0.55

5.31

2.65

Answer 1

The pH of a 0.28 M solution of acetic acid with a Ka of 1.76 x 10^-5 is 4.76.

Step-by-step explanation:

The pH of a solution can be determined using the dissociation constant (Ka) of the acid. In this case, the student is preparing a 0.28 M solution of acetic acid, which has a Ka of 1.76 x 10^-5. To find the pH, we can use the formula:

pH = -log[H+]

Where [H+] is the concentration of hydrogen ions. Since acetic acid is a weak acid, we can assume that it partially dissociates in solution:

CH3COOH(aq) + H2O(l) → CH3COO-(aq) + H3O+(aq)

Therefore, we can assume that the initial concentration of acetic acid ([CH3COOH]) is equal to 0.28 M, and the initial concentration of hydrogen ions ([H+]) is equal to the concentration of hydronium ions ([H3O+]). Using the equation Ka = [CH3COO-][H3O+]/[CH3COOH], we can rearrange it to solve for [H3O+]:

[H3O+] = Ka x [CH3COOH]/[CH3COO-] = (1.76 x 10^-5) x (0.28)/(0.28) = 1.76 x 10^-5

Finally, we can calculate the pH by taking the negative logarithm of [H3O+]:

pH = -log(1.76 x 10^-5) = 4.76

Answer 2

i belive its 4.76 because it would not be 0.55