Question

a 50.00-mL sample of bleach solution contains 0.214 M HClO and 0.667 M NaClO. The Ka of hypochlorous acid is 3.0 ✕ 10−8. Find the pH of the solution. The solution is then divided in half. A) To one half of the original solution, 10.00 mL of 0.100 M NaOH is added. What is the final pH of this solution? B) To the other half of the original solution, 1.00 mL of 0.100 M HCl is added. What is the final pH of this solution

Answer 1

Answer:

A) pH = 8.11

B) pH = 7.96

Explanation:

The pH of buffer is calculated using Henderson Hassalbalch's equatio, which is

$pH=pKa+log\frac{[salt]}{[acid]}$

pKa = -logKa = -log(3.0 ✕ 10⁻⁸)

pKa = 7.5

[salt] = 0.667

[Acid] = 0.214

$pH = 7.5 + log\frac{0.667}{0.214}=7.99$

A) If the solution is halved it means the concentration of both salt and acid will be the same.

The volume will change

Number of moles of acid = molarity X volume = 0.214 X 25 = 5.35 mmol

Number of moles of salt = molarity X volume = 0.667 X 25 = 16.68 mmol

moles of NaOH added = molarity X volume = 0.100 X 10 = 1 mmol

These moles of NaOH will react with acid to give same amount of salt and thus the moles of acid will decrease

New moles of acid = 5.35 - 1 = 4.35

New moles of salt = 16.68 + 1 = 17.68

the new pH will be

$pH = 7.5 + log\frac{17.68}{4.35}=8.11$

B)

moles of HCl added = molarity X volume = 0.100 X 10 = 1 mmol

These moles of HCl will react with salt to give same amount of acid and thus the moles of salt will decrease

New moles of acid = 5.35  + 1 = 6.35

New moles of salt = 16.68 - 1 = 15.68

the new pH will be

$pH = 7.5 + log\frac{15.68}{5.35}=7.97$