Question

The next 11 questions are related to the titration of 25.00 mL of a 0.0750 M acetic acid solution with 0.0750 M KOH. Assume that the temperature is 25 oC.

What is the initial pH of the analyte solution?

2.94

You are correct.

Your receipt no. is 150-8783 Help: Receipt Previous Tries

What volume of KOH is required to reach the equivalence point of the titration?

25.00 mL

You are correct.

Your receipt no. is 150-8154 Help: Receipt Previous Tries

How many mmol of acetate are present at the equivalence point? Report the analytical amount, not the equilibrium amount of acetate at the equivalence point.

1.875

You are correct.

Your receipt no. is 150-3167 Help: Receipt Previous Tries

What is the total volume of the solution at the equivalence point?

50.00 mL

You are correct.

Your receipt no. is 150-6918 Help: Receipt Previous Tries

What is the analytical concentration of acetate ions at the equivalence point?

3.750×10-2 M

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Your receipt no. is 150-6387 Help: Receipt Previous Tries

What is the pOH at the equivalence point?

5.33

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Your receipt no. is 150-1376 Help: Receipt Previous Tries

What is the pH at the equivalence point?

8.67

You are correct.

Your receipt no. is 150-1908 Help: Receipt Previous Tries

What volume of KOH has been added at the half-equivalence point?

12.50 mL

You are correct.

Your receipt no. is 150-4896 Help: Receipt Previous Tries

What is the pH at the half-equivalence point?

4.76

You are correct.

Your receipt no. is 150-537 Help: Receipt Previous Tries

What is the pH when 5.00 mL of the KOH solution have been added?

Tries 0/10

What is the pH when 30.00 mL of the KOH solution have been added?

Tries 0/10

Answer 1

What is the pH when 5.00 mL of the KOH solution have been added? pH = 4.16

What is the pH when 30.00 mL of the KOH solution have been added? pH = 11.83

Explanation

(a) concentration acetic acid = 0.0750 M

volume of acetic acid = 25.00 mL

moles of acetic acid = (concentration acetic acid) * (volume of acetic acid)

moles of acetic acid = (0.0750 M) * (25.00 mL)

moles of acetic acid = 1.875 mmol

moles KOH added = (concentration KOH) * (volume KOH)

moles KOH added = (0.0750 M) * (5.00 mL)

moles KOH added = 0.375 mmol

KOH is a very strong acid which will consume acetic acid in neutralization reaction

moles acetic acid left = (initial moles acetic acid) - (moles KOH added)

moles acetic acid left = (1.875 mmol) - (0.375 mmol)

moles acetic acid left = 1.500 mmol

moles conjugate base formed = moles KOH added

moles conjugate base formed = 0.375 mmol

According to Henderson - Hasselbalch equation,

pH = pK_a + log([conjugate base] / [weak acid])

pH = pK_a + log(moles conjugate base formed / moles acetic acid left)

pH = 4.76 + log(0.375 mmol / 1.500 mmol)

pH = 4.16