3) A 500.0 mL sample of 0.18 M HCIO4 is titrated with 0.27 M KOH. Determine...

Question

Question

Answer 1

Answer #1

ince HClO4 is a strong acid and strong base so their dissociation will be complete

mmol of HClO4 = concentration * volume = 0.18 * 500= 90 mmol

mmol of KOH = 0.27 * 250 = 67.5 mmol

after mixing mmole of HClO4 left = 90- 67.5 = 22.5 mmol

so, [HClO4] = 22.5 / (500+250) = 0.03 M = [H+]

so, pH = -log[H+]

= -log(0.03) =1.52

amswer os C

Answer 2

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