Consider a 0.5 L buffer that contains 0.12 M HOCl and 0.080 M KOCl. What is the pH of this buffer after adding 2.5 mL of 10M KOH to the solution

pH = 7.73

Explanation:

The pKa of the HOCl/KOCl buffer is 7.46. To determine the pH of this buffer we can use H-H equation:

pH = pKa + log [A⁻] / [HA]

pH = 7.46 + log [KOCl] / [HOCl]

Where [] is molarity -or moles- of each compound

Initial moles of HClO and NaClO:

HOCl: 0.500L * (0.12mol / L) = 0.06 moles HOCl

KOCl: 0.500L * (0.080mol / L) = 0.04 moles KOCl

Now, HOCl reacts with KOH producing KOCl:

HOCl + KOH → KOCl + H₂O

The moles of KOH in the reaction are:

2.5x10⁻³L * (10mol /L) = 0.025 moles KOH

That means after the reaction, 0.025 moles of HOCl are consumed and 0.025 moles of KOCl are produced.

And after the reaction, moles are:

Final moles:

HOCl: 0.06mol - 0.025mol = 0.035 mol

KOCl: 0.04mol + 0.025mol = 0.065 mol

Replaing in H-H equation:

pH = 7.46 + log [0.065mol] / [0.035mol]

the answer is 303.15 or 303

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