what is the equilibrium concentration of acetate ion if the equilibrium concentration of acetic acid is .40 M, the pH is 3.85, and Ka is 1.8E-05?
To find the equilibrium concentration of acetate ion, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = 3.85 (given)
pKa = -log(Ka) = -log(1.8E-05) ≈ 4.74 (from the given Ka value)
[HA] = concentration of acetic acid = 0.40 M (given)
[A-] = concentration of acetate ion (unknown)
Now, we can substitute the values into the Henderson-Hasselbalch equation:
3.85 = 4.74 + log([A-]/0.40)
Rearranging the equation:
log([A-]/0.40) = 3.85 - 4.74
log([A-]/0.40) = -0.89
Now, let's take the antilog of both sides of the equation to eliminate the logarithm:
[A-]/0.40 = antilog(-0.89)
[A-] = 0.40 × 10^(-0.89)
Calculating this value:
[A-] ≈ 0.40 × 0.126
[A-] ≈ 0.0504 M
Therefore, the equilibrium concentration of acetate ion ([A-]) is approximately 0.0504 M.
To find the equilibrium concentration of acetate ion, we need to consider the dissociation of acetic acid, which can be represented by the following equation:
CH3COOH ⇌ CH3COO- + H+
The equilibrium constant expression for this dissociation reaction is given by:
Ka = [CH3COO-][H+] / [CH3COOH]
Given that the equilibrium concentration of acetic acid ([CH3COOH]) is 0.40 M and the pH is 3.85, we can determine the concentration of hydrogen ions ([H+]) using the relationship:
[H+] = 10^(-pH)
[H+] = 10^(-3.85)
Solving this calculation, we find that the concentration of hydrogen ions is approximately 1.77 x 10^(-4) M.
Since the concentration of acetate ion ([CH3COO-]) is the same as the concentration of hydrogen ions in this reaction (1:1 ratio), the equilibrium concentration of acetate ion is also approximately 1.77 x 10^(-4) M.
CH3COOH ==> H^+ + CH3COO^-
Ka = (H^+)(CH3COO^-)/CH3COOH)
It looks to me as if you have everything in the equation except for the one unknown. Solve for that.