A solution of acetic acid (CH3COOH; Ka = 1.74 x 10-5) was titrated to a final pH of 4.76 by using NaOH solution. What % of acetic acid is converted to sodium acetate (CH3COO-Na+)? Show your method
If this were a real organic question I probably could not answer it; however, it looks like a good analytical question to me.
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
pKa = -log(Ka) = 4.76
4.76 = 4.76 + log (base/acid)
Solve for (base/acid). You should get 1 which means the acid was half neutralized.
You can prove this by writing the equation and assuming a number for acetic acid.
CH3COOH + NaOH ==>CH3COONa + H2O
Start with 10 mmoles CH3COOH
Add 5 mmols NaOH.
CH3COOH left = 10-5= 5 mmoles
NaOH left = 0
CH3COONa formed = 5 mmoles
H2O formed = 5 mmoles
You started with 10 mmoles CH3COOH and you have formed 5 mmoles CH3COONa.
%CH3COONa formed = 5/10*100 = 50%
To find the percentage of acetic acid converted to sodium acetate, we need to use the Henderson-Hasselbalch equation. This equation relates the pH of a solution to the pKa of the weak acid and the concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the final pH of the solution
pKa = the negative logarithm of the acid dissociation constant (Ka)
[A-] = the concentration of the conjugate base (sodium acetate, CH3COO-Na+)
[HA] = the concentration of the weak acid (acetic acid, CH3COOH)
In this case, we are given the pH (4.76) and the Ka value (1.74 x 10^-5) for acetic acid. We can rearrange the Henderson-Hasselbalch equation to solve for the ratio [A-]/[HA].
[H+] = 10^-(pH)
[HA]/[A-] = (10^pH)/(10^pKa)
Now, we need to find the concentration of acetic acid ([HA]), which can be calculated using the equation:
[HA] = [HA]0 - [A-]
Where:
[HA]0 = the initial concentration of acetic acid
Since the initial concentration of acetic acid is not given in the question, we cannot proceed further to find the percentage conversion without this information.