Find the ph of the equivalence point and the volume (ml) of 0.0372 m naoh needed to reach the equivalence point in the titration of 42.2 ml of 0.0520 m ch3cooh.

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

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