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).
I've answered this before. You need two equations. eqn 1 is the HH equation
pH = pKa + log (base)/(acid)
Substitute the number and solve for the ratio of base/acid.
The second equation you need is
(acid) + (base) = 0.2 M
where (acid ) = (HOAc)
(base) = (OAc^-)
Post your work if you get stuck.
Is it correct like that?
Thanks in advance
5=4.77+log(b/a)
b/a=1.698
b+a=0.2
(1.698*a)+a=0.2
A=0.074M
B=0.1259M
That looks OK to me. I obtained b/a = 1.7 but that doesn't change anything with the remainder of the problem. Thanks for showing your work. I suspect I have my calculator set to round to 2 places and your calculator is different. Your work is good.
To determine the concentrations of HOAc (acetic acid) and OAc- (acetate ion) in the acetic acid buffer, we need to use the Henderson-Hasselbalch equation, which is expressed as:
pH = pKa + log([OAc-]/[HOAc])
First, let's determine the ratio of [OAc-]/[HOAc] using the Henderson-Hasselbalch equation. Rearranging the equation, we have:
[OAc-]/[HOAc] = 10^(pH - pKa)
Substituting the given values, we get:
[OAc-]/[HOAc] = 10^(5.0 - 4.77)
Calculating this:
[OAc-]/[HOAc] = 10^0.23
[OAc-]/[HOAc] ≈ 1.708
This means that the ratio of OAc- to HOAc in the buffer is approximately 1.708.
Next, we know that the total concentration of the acetate buffer is 0.2 M. Since the ratio of OAc- to HOAc is 1.708, we can set up the following equation:
0.2 M = [OAc-] + [HOAc]
From the ratio above, we can substitute [OAc-] as 1.708x and [HOAc] as x:
0.2 M = 1.708x + x
Simplifying:
0.2 M = 2.708x
Dividing by 2.708:
x = 0.2 M / 2.708
Calculating this:
x ≈ 0.0738 M
Thus, the concentration of HOAc is approximately 0.0738 M, and the concentration of OAc- is approximately 1.708 times that, which is:
0.0738 M * 1.708 = 0.1259 M
Therefore, in a 0.2 M acetate buffer with a pH of 5.0, the concentration of HOAc is approximately 0.0738 M, and the concentration of OAc- is approximately 0.1259 M.