What are the concentrations of HOAc and OAc- in a 0.2 M “acetate” buffer, pH 5.0? The Ka for acetic acid 1.70x10-5 (pKa 4.77)
To find the concentrations of HOAc (acetic acid) and OAc- (acetate ion) in the acetate buffer, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a solution to the concentrations of the acidic and basic components in a buffer solution:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the conjugate base (OAc-) and [HA] represents the concentration of the acid (HOAc).
Given information:
- pH = 5.0
- pKa = 4.77
First, let's calculate the ratio [A-]/[HA] using the pKa and pH:
pH = pKa + log([A-]/[HA])
5.0 = 4.77 + log([A-]/[HA])
Next, let's solve the equation for [A-]/[HA]:
log([A-]/[HA]) = 5.0 - 4.77
log([A-]/[HA]) = 0.23
To find the concentration ratio [A-]/[HA] from the logarithm, we need to take the antilog of both sides of the equation:
[A-]/[HA] = antilog(0.23)
Now, let's calculate the antilog value of 0.23 using the 10^x function on a calculator or software:
[A-]/[HA] = 1.6768
The ratio [A-]/[HA] in the acetate buffer is approximately 1.6768.
Since the initial concentration of the acetate buffer is given as 0.2 M, we can assign the concentration of HOAc (acetic acid) as x M and the concentration of OAc- (acetate ion) as (0.2 - x) M (assuming x is the amount of HOAc that ionizes to form OAc-).
Now, using the concentration ratio, we can assign the values:
[A-]/[HA] = (0.2 - x) / x = 1.6768
Cross-multiplying this equation gives:
0.2 - x = 1.6768x
Simplifying and rearranging:
2.6768x = 0.2
Dividing both sides by 2.6768:
x = 0.2 / 2.6768
Solving this equation gives:
x = 0.0747 M
Therefore, the concentration of HOAc (acetic acid) is approximately 0.0747 M, and the concentration of OAc- (acetate ion) is approximately (0.2 - 0.0747) M, which is approximately 0.1253 M.
So, in a 0.2 M acetate buffer with a pH of 5.0, the concentration of HOAc (acetic acid) is approximately 0.0747 M, and the concentration of OAc- (acetate ion) is approximately 0.1253 M.