What is 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)
You solve two equation simultaneously. Eqn 1 is this.
pH = pKa + log (base)/(acid). Base is OAc and acid is HOAc. You know pH and pKa and you solve for the ratio of base/acid. Eqn 2 is this.
base + acid = 0.2
Simplified:
eqn 1..........base/acid = ? some number
eqn 2 ........ base + acid = 0.2
Post your work if you get stuck.
Let's take acetic acid, HAc. It ionizes as HAc ==> H^+ + Ac^-. When you say 0.2 M you are measuring that out as HAc. However, as soon as water is added it ionizes and form some H^+ and some Ac^-. So the total is still 0.2 M which means if you add the acid (in this case HAc) and the base (in this case Ac-) it will add up to 0.2 M. so (acid) + ( base) = total molarity.
could you please say why base+acide = 0.2 ??
Describe the preparation of 5 liters of a 0.3 M acetate buffer, pH 4.47, starting from a 2 M solution of acetic acid and a 2.5 M solution of KOH (pKa 4.77).
so the OAC= 0.9999 M
and the HOAC= 1x10^-5 that right ?
To determine the concentrations of HOAc (acetic acid) and OAc- (acetate ion) in the given acetate buffer, we need to apply the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the ratio of the concentrations of its conjugate acid and conjugate base:
pH = pKa + log([A-]/[HA])
Where:
pH = the given pH of the buffer (5.0 in this case)
pKa = the negative logarithm of the acid dissociation constant (1.70x10^-5 in this case)
Since we are given the pH and pKa values, we can rearrange the equation to solve for [A-]/[HA].
log([A-]/[HA]) = pH - pKa
Now, let's substitute the values:
log([A-]/[HA]) = 5.0 - 4.77
By subtracting, we find:
log([A-]/[HA]) = 0.23
To find the [A-]/[HA] ratio, we need to convert the logarithmic form into the exponential form. We will use the antilog function (10^x).
[A-]/[HA] = 10^(0.23)
Using a calculator, we find:
[A-]/[HA] ≈ 1.778
Now, we can determine the concentrations of HOAc and OAc-.
Since the buffer is 0.2 M, we know that [HOAc] + [OAc-] = 0.2 M.
Let's denote [HOAc] as x and [OAc-] as 0.2 - x.
Using the ratio we derived earlier:
[A-]/[HA] = ([OAc-]/[HOAc]) = 1.778
Substituting the values:
1.778 = (0.2 - x)/x
Now, solve for x:
1.778x = 0.2 - x
2.778x = 0.2
x = 0.2 / 2.778
x ≈ 0.072 M
Hence, the concentration of HOAc (acetic acid) is approximately 0.072 M, and the concentration of OAc- (acetate ion) is approximately 0.128 M in the 0.2 M acetate buffer at pH 5.0.