A 0.15 M aqueous solution of the weak acid HA at 25.0 °C has a pH of 5.35. The value of K_afor HA is .

Question

Quizplus · Accepted Answer

Using the given information, we can apply the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base of HA.

We know that the pH = 5.35 and [HA] = 0.15 M. To find [A-], we need to use the equation [HA] + [A-] = [total] = 0.15 M (since HA is a weak acid and only partially dissociates).

Rearranging the Henderson-Hasselbalch equation, we get:

log([A-]/[HA]) = pH - pKa

Plugging in the given values and solving for pKa, we get:

pKa = pH - log([A-]/[HA])

[HA] = 0.15 M and [A-] = x (letting x = the concentration of the conjugate base)

0.15 + x = 0.15 M

x = 0.15 - 0.15 = 0

So, [A-] = 0 M. This means that all of the weak acid remains in the HA form and none of it has dissociated to form its conjugate base.

Therefore:

log(0/0.15) = undefined

So, pKa = pH in this case.

pKa = 5.35

So, Ka (the acid dissociation constant) can be found from:

pKa = -log(Ka)

Ka = 1.4 × 10^-6

Since Ka is the equilibrium constant for the dissociation of the acid, the answer is choice (B).

A 015 M Aqueous Solution of the Weak Acid HA at at 25.0