Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq).?

What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?

What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?

Calculate the pH at the halfway point.

Update:

How did you get 3 mml for C6H5COOH? when calculating the halfway point

Update 2:

How did you get 3 mmol for C6H5COOH? when calculating the halfway point**

1 Answer

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  • 9 years ago
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    start mmol HOBz = 30.0 mL x 0.20 mmol/mL = 6.0 mmol

    mmol base added = 15 mL x 0.30 mmol/mL = 4.5 mmol

    mmol HObz remaining = 6.0 - 4.5 = 1.5 mmol

    pH = pKa + log(OBz-/HOBz) = 4.20 + log(4.5/1.5) = 4.68

    ? mL KOH = 3.0 mmol HOBz x 1 mmol KOH / mmol HOBz x 1 mL / 0.30 mmol KOH

    = 10. mL KOH

    If Cs=Ca, then log 1 = 0 and pH = pKa = 4.20

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