A 100.0 mL sample of 0.18 M HClO₄ is titrated with 0.27 M LiOH.Determine the pH of the solution after the addition of 100.0 mL of LiOH. A)13.13 B)12.65 C)1.35 D)0.87 E)12.95

Question

Quizplus · Accepted Answer

1. Write the balanced chemical equation for the reaction between HClO and LiOH:

HClO + LiOH -> LiClO + H2O

2. Calculate the initial number of moles of HClO:

0.18 M HClO x 0.100 L = 0.018 mol HClO

3. Calculate the number of moles of LiOH added:

0.27 M LiOH x 0.100 L = 0.027 mol LiOH

4. Determine the limiting reagent:

HClO:LiOH = 0.018 mol:0.027 mol = 0.67:1

Since LiOH is in excess, it is the limiting reagent.

5. Calculate the number of moles of LiClO produced:

0.027 mol LiOH x 1 mol LiClO/1 mol LiOH = 0.027 mol LiClO

6. Calculate the concentration of LiClO:

0.027 mol LiClO/0.200 L = 0.135 M LiClO

7. Determine the pH of the solution:

To determine the pH, we need to calculate the concentration of H+ ions in the solution. Since LiOH is a strong base, it will react completely with HClO to form LiClO and H2O.

HClO + LiOH -> LiClO + H2O

Since HClO is a weak acid, the reaction can be approximated by:

HClO + OH- -> ClO- + H2O

Kb = Kw/Ka = 1.0 x 10^-14/3.0 x 10^-8 = 3.3 x 10^-7

Since [OH-] = [LiOH] = 0.027 mol/0.200 L = 0.135 M, we can use the Kb expression to calculate the concentration of H+:

Kb = [ClO-][H+]/[OH-] = 3.3 x 10^-7

[H+] = Kb[OH-]/[ClO-] = 3.3 x 10^-7 x 0.135 M/0.018 M = 2.5 x 10^-6 M

pH = -log[H+] = 12.65

A 1000 ML Sample of 0