Answer:
58.99mL of 0.0372 M NaOH to reach equivalence point
pH = 8.54
Explanation:
Moles of 42.2 mL of 0.0520 M CH₃COOH are:
0.0422L × (0.0520mol / L) = 2.194x10⁻³ mol CH₃COOH
That react with NaOH, thus:
CH₃COOH + NaOH → CH₃COO⁻ + Na⁺ + H₂O
That means 1 mole of acetic acid reacts per mole of NaOH
Thus, moles of NaOH you need to reach equivalence point are 2.194x10⁻³ mol, in volume:
2.194x10⁻³ mol × (1L / 0.0372mol) = 0.05899L ≡ 58.99mL of 0.0372 M NaOH to reach equivalence point
In the equivalence point, you will have just CH₃COO⁻ that is in equilibrium with water, thus:
CH₃COO⁻(aq) + H₂O(l) ⇄ CH₃COOH(aq) + OH⁻(aq)
And equilibrium is defined by:
Kb = 5.6x10⁻¹⁰ = [CH₃COOH] [OH⁻] / [CH₃COO⁻]
Molarity of CH₃COO⁻ is: 2.194x10⁻³mol / (0.05899L + 0.0422L) = 0.02168M
Thus, concentrations in equilibrium are:
[CH₃COO⁻] = 0.02168M - X
[CH₃COOH] = X
[OH⁻] = X
Replacing in Kb:
5.6x10⁻¹⁰ = [X] [X] / [0.02168M - X]
1.214x10⁻¹¹ - 5.6x10⁻¹⁰X = X²
X² + 5.6x10⁻¹⁰X - 1.214x10⁻¹¹ = 0
Solving for X:
X: -3.48x10⁻⁶ → False answer, there is no negative concentrations
X: 3.484x10⁻⁶
As [OH⁻] = X; [OH⁻] = 3.484x10⁻⁶M
pOH = -log [OH⁻]; pOH = 5.46
As 14 = pOH + pH
pH = 8.54
