What is the [H3O+] concentration of a pH 8.2 solution produced by a buffer made from acetic acid (pKa= 4.8) and sodium acetate?
To determine the [H3O+] concentration of a pH 8.2 solution produced by a buffer made from acetic acid and sodium acetate, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
First, let's determine the concentration of acetic acid ([HA]) and sodium acetate ([A-]). In a buffer solution, the concentration of the acid and its conjugate base are usually in a 1:1 ratio. Therefore, let's assume we have equal concentrations of acetic acid and sodium acetate, denoted as [HA] and [A-], respectively.
Now, we know that acetic acid (CH3COOH) dissociates in water as follows: CH3COOH ⇄ H3O+ + CH3COO-
The sodium acetate (CH3COONa) dissociates as follows: CH3COONa ⇄ Na+ + CH3COO-
The dissociation of acetic acid and sodium acetate helps maintain the buffer solution at a specific pH.
Using the given pKa value of acetic acid (4.8), we can substitute the values into the Henderson-Hasselbalch equation:
pH = 4.8 + log([A-]/[HA])
Since we assumed equal concentrations of [A-] and [HA], we can use the variable x to represent both concentrations:
pH = 4.8 + log(x/x)
Simplifying further, we have:
pH = 4.8 + log(1)
log(1) is equal to zero, so the equation becomes:
pH = 4.8 + 0
Therefore, the pH of the buffer solution is equal to the pKa of the acid, which is 4.8.
However, in the question, it states that the pH of the solution is 8.2. This is higher than the pKa, indicating that the solution is basic rather than acidic. This suggests that the buffer capacity of the acetic acid-sodium acetate system is exceeded, and the pH is no longer under the control of the buffer.
To calculate the [H3O+] concentration for a given pH, we can use the equation:
[H3O+] = 10^(-pH)
Plugging in the pH value of 8.2, we get:
[H3O+] = 10^(-8.2)
Using a calculator, the [H3O+] concentration is approximately 6.31 x 10^(-9) M.
pH=-log[H3O+]
10^-(pH)=[H3O+]
All other information given in the problem is irrelevant.