What was the pH of the resulting buffer?

13.45 mL of 15.2 M potassium hydroxide (KOH) was added to 364 mL of 0.885 M

hypochlorous acid (HClO).

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  • 4 months ago

    pKa for HClO = 7.53

    moles HClO initially present = 0.346 L X 0.885 mol/L = 0.306 mol HClO

    moles KOH added = 0.01345 L X 15.2 mol/L = 0.204 mol KOH

    The added KOH neutralizes 0.204 mol of the acid forming 0.204 mol ClO- and leaving 0.102 mol HClO

    pH = pKa + log [ClO-]/[HClO]

    pH = 7.53 + log (0.204 / 0.102) = 7.83

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