A50.00-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

A) pH = 8.11

B) pH = 7.96

Explanation:

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

pKa = -logKa = -log(3.0 ✕ 10⁻⁸)

pKa = 7.5

[salt] = 0.667

[Acid] = 0.214

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

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

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