What is the pH of a 0.500 M solution of acetylsalicylic acid, pK 3.52?

Question

Serenity Allen · Accepted Answer

#sf(1.91)# For an acid that is weak: #sf(pH=1/2(pK_a-loga))# Where #sf(a)# is the concentration of the acid. #:.##sf(pH=1/2[3.52-log(0.5)]# #sf(pH=1/2[3.52-(-0.301)]=1.91)#

Olivia Anton · Answer

undefined To find the pH of a solution of acetylsalicylic acid with a concentration of 0.500 M and a pKa of 3.52, you can use the Henderson-Hasselbalch equation:

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

In this equation:
- pKa is the negative logarithm of the acid dissociation constant (Ka), which is given as 3.52.
- [A^-] is the concentration of the conjugate base of the acid.
- [HA] is the concentration of the acid.

Since acetylsalicylic acid (ASA) is a weak acid, it partially dissociates into its conjugate base (aspirin ion, A^-) and hydrogen ions (H^+).

First, calculate [A^-] and [HA]:
- [HA] is the initial concentration of acetylsalicylic acid, which is 0.500 M.
- [A^-] can be calculated using the equation [A^-] = [H^+] = √(Ka * [HA]).

Given that Ka = 10^(-pKa), we have Ka = 10^(-3.52).

Now, calculate [A^-]:
[A^-] = [H^+] = √(Ka * [HA]) = √((10^(-3.52)) * 0.500) ≈ √(2.63 * 10^(-4)) ≈ 5.13 * 10^(-3) M

Now, substitute the values into the Henderson-Hasselbalch equation:
pH = 3.52 + log(5.13 * 10^(-3) / 0.500)

pH ≈ 3.52 + log(1.026 * 10^(-5))

pH ≈ 3.52 - 4.99

pH ≈ -1.47

Since pH cannot be negative, the pH of the solution is approximately 1.47.