Answer:
pH = 8.75
Explanation:
100.0mL of a 0.8108M of a butanoic acid (HC₃H₇CO₂, pKa 4.82) solution is titrated with 0.0520M KOH.
The reaction is:
HC₃H₇CO₂ + KOH → H₂O + KC₃H₇CO
Moles of butanoic acid are:
0.1000L × (0.8108mol / L) = 0.08108 moles of butanoic
For a complete reaction, volume of KOH must be added is the volume in which 0.08108 moles of KOH are added, that is:
0.08108 mol × (L / 0.0520mol) = 1.56L of KOH.
Total volume in equilibrium is 1.56L + 0.10L = 1.66L
That means concentration of butanoic acid is:
0.08108 mol / 1.66L = 0.04884M HC₃H₇CO₂
At equivalence point, there is just C₃H₇CO⁻ in solution
Kb of butanoic acid is:
C₃H₇CO⁻ + H₂O ⇄ HC₃H₇CO₂ + OH⁻
Kb = Kw / Ka
Ka = 10^-pKa
Ka = 1.51x10⁻⁵
Kb = 1x10⁻¹⁴ / 1.51x10⁻⁵ = 6.61x10⁻¹⁰
The equilibrium of Kb is:
Kb = 6.61x10⁻¹⁰ = [HC₃H₇CO₂] [OH⁻] / [C₃H₇CO⁻]
As at equivalence point there is just C₃H₇CO⁻, the equilibrium concentrations are:
[C₃H₇CO⁻] = 0.04884M - X
[HC₃H₇CO₂] = X
[OH⁻] = X
Replacing in Kb:
6.61x10⁻¹⁰ = X² / [0.04884M - X]
0 = X² + 6.61x10⁻¹⁰X - 3.23x10⁻¹¹
Solving for X:
X = -5.68x10⁻⁶ → False solution. There is no negative concentrations
X = 5.683x10⁻⁶ → Right solution.
As [OH⁻] = X, [OH⁻] = 5.683x10⁻⁶.
pOH = - log [OH⁻]
pOH = 5.245
pH = 14 - pOH
