What is the pH at the equivalence point in the titration of 10.0 mL of 0.39 M HZ with 0.200 M NaOH? Ka = 2.4 × 10−6 for HZ.

pH = 9.37

Explanation:

The reaction of HZ with NaOH occurs as follows:

HZ + NaOH → H₂O + Na⁺ + Z⁻

When the reaction is in the equivalence point, the moles added of HZ = Moles of NaOH.

At equivalence point, you will have in solution just NaZ that is in equilibrium with water as follows:

Z⁻(aq) + H₂O(l) ⇄ OH⁻(aq) + HZ(aq)

Where K of equilibrium is Kw / Ka = 1x10⁻¹⁴ / 2.4x10⁻⁶ = 4.17x10⁻⁹

And K is defined as:

K = 4.17x10⁻⁹ = [OH⁻] [HZ] / [Z⁻]

To solve this question we must find  [Z⁻] and as [OH⁻] and [HZ] comes from the same equilibrium, both are equal and we can find [OH⁻] in order to solve pH:

Moles NaOH added = Moles HZ:

0.010L * (0.39mol / L) = 3.9x10⁻³ moles HZ = Moles NaOH added

The volume of 0.200M NaOH must be:

3.9x10⁻³ moles * (1L / 0.200moles) = 0.0195L.

The moles of HZ = Moles Z⁻ after the reaction and the volume is 0.010L + 0.0195L = 0.0295L. Thus, [Z⁻] is:

[Z⁻] = 3.9x10⁻³ moles / 0.0295L

[Z⁻] = 0.132M

Replacing in K expression:

4.17x10⁻⁹ = [OH⁻] [HZ] / [Z⁻]

4.17x10⁻⁹ = [X] [X] / [0.132]

5.51x10⁻¹⁰ = X²

X = 2.35x10⁻⁵M = [OH⁻]

As pOH = -log [OH⁻]

pOH = 4.63

And pH is

pH = 14 -pOH