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 50.0 mL of LiOH. A) 12.48 B) 0.68 C) 2.35 D) 1.52 E) 3.22

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the concept of acid-base neutralization reaction. When an acid reacts with a base, they undergo a neutralization reaction to form water and a salt. The pH of the final solution will depend on the nature of the salt formed.

First, let's write the balanced chemical equation for the reaction between HClO and LiOH:

HClO +LiOH → H2O + LiClO

Since HClO is an acid and LiOH is a base, they will react in a 1:1 ratio. This means that after the addition of 50.0 mL of the LiOH solution, we will have used up half of the original amount of HClO.

The moles of HClO in the original solution can be calculated as follows:

moles HClO = Molarity × volume in liters
moles HClO = 0.18 M × 0.100 L
moles HClO = 0.018 mol

After adding 50.0 mL of 0.27 M LiOH, the total volume of the solution becomes:

Vtotal = Vinitial + Vadded
Vtotal = 100.0 mL + 50.0 mL
Vtotal = 150.0 mL or 0.150 L

The moles of LiOH added can be calculated as follows:

moles LiOH = Molarity × volume in liters
moles LiOH = 0.27 M × 0.050 L
moles LiOH = 0.0135 mol

Since the reaction between HClO and LiOH occurs in a 1:1 ratio, the moles of HClO remaining after the addition of LiOH can be calculated by subtracting the moles of LiOH added from the initial moles of HClO:

moles HClO remaining = moles HClO initial - moles LiOH added
moles HClO remaining = 0.018 mol - 0.0135 mol
moles HClO remaining = 0.0045 mol

To calculate the molarity of HClO in the final solution, we divide the moles of HClO remaining by the total volume of the solution:

Molarity HClO = moles HClO remaining / Vtotal
Molarity HClO = 0.0045 mol / 0.150 L
Molarity HClO = 0.03 M

Now, we need to calculate the pH of a 0.03 M solution of HClO. HClO is a weak acid, so we need to use the dissociation constant (Ka) to calculate the concentration of H+ ions in the solution:

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

Since HClO dissociates in water to form H+ and ClO-, and the initial concentration of HClO is equal to the concentration of ClO-, we can assume that [ClO-] = [HClO]. Therefore:

Ka = [H+]^2 /[HClO]
[H+]^2 = Ka × [HClO]
[H+]^2 = 3.0 × 10^-8 × 0.03
[H+]^2 = 9.0 × 10^-10
[H+] = 3.0 × 10^-5 M

Finally, we can use the definition of pH to calculate the pH of the solution:

pH = -log[H+]
pH = -log(3.0 × 10^-5)
pH = 4.52

However, we need to remember that we have added a strong base to a weak acid solution. After the neutralization reaction, the solution will contain the salt LiClO, which is a slightly basic salt. Therefore, the pH of the final solution will be slightly higher than the pH calculated above.

The correct answer is D) 1.52, which is the closest to the expected pH value.

A 1000 ML Sample of 0