Question

A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Kaof hypochlorous acid is 3.0x10^-8.

a) 10.20
b) 7.00
c) 6.48
d) 7.52
e) 14.52

Answer 1

The pH at the equivalence point of the titration between hypochlorous acid and NaOH is 7.52, as it's equivalent to the pKa of hypochlorous acid, calculated from the given Ka. So the correcct option is d.

Step-by-step explanation:

The student is asked to determine the pH at the equivalence point of a titration of 25.0 mL of 0.150 M hypochlorous acid with a 0.150 M NaOH solution. To solve this, first understand that at the equivalence point, all the hypochlorous acid has been neutralized to form its conjugate base, hypochlorite ion (ClO-).

Use the formula pH = pKa + log([A-]/[HA]), but since we are at the equivalence point, the ratio of [A-]/[HA] is 1, so pH = pKa. The pKa can be calculated from the given Ka value using the formula pKa = -log(Ka), which gives pKa = -log(3.0 x 10-8) = 7.52. Therefore, the pH at the equivalence point is 7.52.

Answer 2

pH = 10.20

Step-by-step explanation:

The HClO reacts with NaOH as follows:

HClO + NaOH → H2O + NaClO

Where HClO and NaOH react in a 1:1 reaction.

As the concentration of both reactions is the same and the reaction is 1:1, to reach equivalence point are required the same 25.0mL.

And the NaClO produced decreases its concentration in 2 because the volume is doubled.

The concentration of NaClO is: 0.150M / 2 = 0.075M

The equilibrium of NaClO is:

NaClO(aq) + H2O(l) ⇄ HClO(aq) + OH-(aq)

Where Kb of reaction is 1.0x10⁻¹⁴ / Ka =

1.0x10⁻¹⁴ / 3.0x10⁻⁸ = 3.33x10⁻⁷ = [HClO] [OH-] / [NaClO]

[NaClO] = 0.075M

As both HClO and OH- comes from the same equilibrium,

[HClO] = [OH-] = X

Where X is the reactoin coordinate

Replacing:

3.33x10⁻⁷ = [X] [X] / [0.075M]

2.5x10⁻⁸ = X²

X = 1.58x10⁻⁴M = [OH-]

pOH = -log [OH-]

pOH = 3.80

pH = 14 - pOH

pH = 10.20