Let CH3COOH = HAc
.......HAc + NaOH = NaAc + H2O
At the equivalence point you have NaAc, sodium acetate. The acetate ion (C2H3O2^- or Ac^-) is hydrolyzed as follows and I assume that is solid NaOH placed in the acetic acid, there is no volume change, and (Ac^-) = 1M:
.........Ac^- + HOH ==> HAc + OH^-
I........1.0M............0.....0
C........-x..............x.....x
E........1.0-x...........x......x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(1.0-x)
Solve for x = (OH^-) and convert to pH. I'll let you finish.
The reason is #2.
A 1.0 M acetic acid solution (CH3COOH, pKa = 4.7) is neutralized by dissolving NaOH(s) (a strong base) in the solution. Estimate the pH at the equivalence point of the neutralization process:
4.7
7.0
9.3
Which one of the following statements best explains your prediction?
1. Since a weak acid does not react easily, higher basicity values are needed to fully neutralize it.
2. At the equivalence point, the most abundant specie in solution is the acetate ion, which acts as a base.
3. At equal concentrations of acid and base, the strong base is more dominant than the weak acid.
4. Neutralization means a neutral (pH = 7) solution.
Being a weak acid, the acetic acid doesn't produce as much H3O+(aq) as NaOH produces OH−(aq).
1 answer
2 answers