Question

The pH of a 0.059 M solution of acid HA is found to be 2.36. What is the Ka of the acid? The equation described by the Ka value is

HA(aq)+H2O(l)⇌A−(aq)+H3O+(aq)

Report your answer with two significant figures.

Answer 1

The pH of a 0.059 M solution of acid HA is found to be 2.36, indicating acidity. The equation represents the ionization of the acid in water. The Ka value can be calculated using the concentration of HA and the [H3O+] concentration.

Step-by-step explanation:

The pH of a solution is a measure of its acidity or basicity. The pH scale ranges from 0 to 14, with values less than 7 indicating acidity, values greater than 7 indicating basicity, and a pH of 7 indicating neutrality. In this case, the pH of the 0.059 M solution of acid HA is found to be 2.36, which means it is acidic.

The equation given represents the ionization of the acid HA in water. The acid donates a proton (H+) to water, forming the hydronium ion (H3O+) and the conjugate base (A-). The equilibrium constant for this reaction is called the Ka (acid dissociation constant).

To find the Ka value for the acid, we can use the equation Ka = [H3O+][A-]/[HA]. Given that the pH is 2.36, we can calculate the [H3O+] concentration by taking the antilog of the negative pH value. Then, we can substitute the given concentration of HA and the calculated [H3O+] concentration into the equation to find the Ka value.

Answer 2

To find the Ka of the acid HA with a pH of 2.36 for its 0.059 M solution, we calculate the concentration of H+ from the pH and use it in the expression for Ka. The Ka is found to be 3.24 x 10^-5 M.

Step-by-step explanation:

The question asks to calculate the acid dissociation constant (Ka) for the weak acid HA, given the pH of its 0.059 M solution is 2.36. To find the Ka, we need to determine the concentration of hydrogen ions ([H+]) from the pH and then use the expression for Ka to calculate its value.

First, we convert the pH to the concentration of hydrogen ions using the formula [H+] = 10-pH. For a pH of 2.36, [H+] = 10-2.36 = 4.37 x 10-3 M. In the dissociation of HA, the concentration of H+ will be equal to the concentration of A- in a 1:1 ratio, since one mole of HA produces one mole of H+ and one mole of A-.

Using the ionization equation HA ⇌ H+ + A-, we can write the expression for Ka as follows:

Ka = [H+][A-]/[HA]

We know [H+] and [A-] are both equal to 4.37 x 10-3 M. Assuming x << 0.059 and thus [HA] ≈ 0.059 M, the concentration of HA not significantly changed, the Ka value can be calculated:

Ka = (4.37 x 10-3)2 / 0.059 = 3.24 x 10-5 M

Therefore, the Ka of the acid HA is 3.24 x 10-5 with two significant figures.