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%
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
1 answer
1 answer