g A 25.00 mL sample of 0.0988 M benzoic acid (HC7H5O2, Ka = 6.5 · 10-5) is a monoprotic acid that is titrated with 0.115 M sodium hydroxide. Calculate the pH at the equivalence point.

pH at equivalence point is 8.47

Explanation:

Benzoic acid react with NaOH, thus:

HC₇H₅O₂ + NaOH → C₇H₅O₂⁻ + H₂O + Na⁺

You reach equivalence point when moles of the acid = moles of NaOH.

Moles of benzoic acid are:

0.025L ₓ (0.0988mol / L) = 0.00247 moles

To have 0.00247 moles of NaOH in solution and reach equivalence point you need to add:

0.00247 moles NaOH ₓ (1L / 0.115mol) = 0.0215L of NaOH solution.

Total volume is 0.0465L.

There are produced 0.00247 moles of C₇H₅O₂⁻ and its molarity will be:

0.00247 mol C₇H₅O₂⁻ / 0.0465L = 0.0531M C₇H₅O₂⁻

C₇H₅O₂⁻ is in equilibrium with water, thus:

C₇H₅O₂⁻(aq) + H₂O ⇄ HC₇H₅O₂(aq) + OH⁻(aq)

Where Kb = Kw / Ka = 1x10⁻¹⁴ / 6.5x10⁻⁵ = 1.54x10⁻¹⁰ is:

Kb = 1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

The concentrations in equilibrium of the species are:

[HC₇H₅O₂] = X

[OH⁻] = X

[C₇H₅O₂⁻] = 0.0531M - X

Where X represents how much C₇H₅O₂⁻ react, X is reaction coordinate

Replacing in Kb expression:

1.54x10⁻¹⁰ = [HC₇H₅O₂] [OH⁻] / [C₇H₅O₂⁻]

1.54x10⁻¹⁰ = [X] [X] / [0.0531 - X]

8.169x10⁻¹² - 1.54x10⁻¹⁰X = X²

8.169x10⁻¹² - 1.54x10⁻¹⁰X - X² = 0

Solving for X:

X = -2.858x10⁻⁶M → False solution, there is no negative concentrations

X = 2.858x10⁻⁶M → Right solution

As [OH⁻] = X

[OH⁻] = 2.858x10⁻⁶M

pOH is -log [OH⁻]

pOH = 5.54

pH = 14 - pOH

pH = 8.46

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