A 20.0-mL sample of 0.30 M HClO was titrated with 0.30 M NaOH. The following data were collected during the titration.
What is the K_a for HClO?

Question

A 20.0-mL sample of 0.30 M HClO was titrated with 0.30 M NaOH. The following data were collected during the titration. What is the K_a for HClO?

Quizplus · Accepted Answer

The balanced chemical equation for the reaction between HClO and NaOH is:

HClO + NaOH -> NaClO + H2O

Since the initial concentration of HClO is 0.30 M and the volume of the solution is 20.0 mL (0.0200 L), the initial moles of HClO can be calculated as:

moles of HClO = (0.30 M) x (0.0200 L) = 0.00600 moles

During the titration, NaOH is added until the equivalence point is reached, which occurs when all of the HClO has reacted with NaOH. At the equivalence point, the moles of NaOH added are equal to the moles of HClO present in the initial solution. Therefore, the moles of NaOH added can be calculated as:

moles of NaOH = (0.30 M) x (0.0230 L) = 0.00690 moles

The volume of NaOH required to reach the equivalence point is 23.0 mL (0.0230 L).

From the balanced chemical equation, it can be seen that one mole of NaOH reacts with one mole of HClO. Therefore, the moles of HClO at the equivalence point can be calculated as:

moles of HClO = 0.00690 moles

The volume of the solution at the equivalence point can be calculated as:

volume of solution at equivalence point = 20.0 mL + 23.0 mL = 43.0 mL = 0.0430 L

The concentration of HClO at the equivalence point can be calculated as:

[HClO] = moles of HClO / volume of solution at equivalence point
[HClO] = 0.00690 moles / 0.0430 L
[HClO] = 0.161 M

The Ka expression for the ionization of HClO is:

Ka = [H3O+][ClO-] / [HClO]

At the equivalence point, [HClO] = 0.161 M, and [OH-] = [H3O+] since the solution is neutral. The concentration of [OH-] can be calculated from the amount of NaOH added:

moles of NaOH = (0.30 M) x (0.0230 L) = 0.00690 moles
moles of H3O+ = moles of OH- = 0.00690 moles
[OH-] = moles of OH- / volume of solution at equivalence point
[OH-] = 0.00690 moles / 0.0430 L
[OH-] = 0.161 M

Substituting the values into the Ka expression gives:

Ka = (0.161 M)^2 / 0.161 M = 0.161 M

Therefore, the pKa is:

pKa = -log(Ka) = -log(0.161) = 0.791

Using the relationship between pKa and pKb for conjugate acid-base pairs:

pKa + pKb = 14.00

the pKb can be calculated as:

pKb = 14.00 - pKa = 14.00 - 0.791 = 13.209

Finally, the Kb can be calculated using:

Kb = 10^(-pKb) = 3.5 x 10^(-14)

The correct answer is B: 3.5 × 10^-8.

A 200-ML Sample of 0