What is the ph at the equivalence point if it requires 32.5 mL of 0.15 M KOH to completely titrate 20 mL of a solution of propanoic acid, HC3H5O2

pH = 8.92

Explanation:

To solve this question we must know that the reaction of KOH with HC3H5O2 is:

KOH + HC3H5O2 → H2O + KC3H5O2

At equivalence point, all propanoic acid reacts to produce KC3H5O2.

This KC3H5O2 = C3H5O2⁻ is in equilibrium in water as follows:

C3H5O2⁻(aq) + H₂O(l) → OH⁻(aq) + HC3H5O2(aq)

Where Kb = Kw / ka = 1x10⁻¹⁴/ 1.34x10⁻⁵ = 7.46x10⁻¹⁰

is defined as:

Kb = 7.46x10⁻¹⁰ = [OH⁻] [HC3H5O2] / [C3H5O2⁻]

As both [OH⁻] [HC3H5O2] ions comes from the same equilibrium,

[OH⁻] = [HC3H5O2] = X

[C3H5O2⁻] is:

Moles KOH = Moles C3H5O2⁻:

0.0325L * (0.15mol / L) = 0.004875 moles

In 32.5 + 20mL = 52.5mL = 0.0525L:

0.004875 moles / 0.0525L = 0.09286M.

Replacing:

7.46x10⁻¹⁰ = [X] [X] / [0.09286M]

6.927x10⁻¹¹ = X²

X = 8.323x10⁻⁶M = [OH-]

As pOH = -log [OH-]

pOH = 5.08

pH = 14 -pOH

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