Read the background material from both parts I and II. 1. A student titrated 25.

Question

Read the background material from both parts I and II. 1. A student titrated 25.0 mL of a buffer with 0.126 M HCI. The observed titration curve is shown below. What is the concentration of the weak base in the buffer? Thoroughly document your logic including equivalence point and mole conversion. (Hint this is not a dilution (do not use M,V M2V2) this is a titration) Buffer pH with Added HO 2 The student then ttrated 25.0 mL of the same buffer with 0.115 M NaOH. The observed titration curve is shown below. What is the concentration of the weak acid in the buffer? Thoroughly document your logic including equivalence point and mole conversion. (Hint this is not a dilution (do not use M,V M2V2) this is a titration Bufter pH With Added NaOH 5.00 20 elume NaOH Added( 3. What is the buffer strength (ie, the total of [HA] and [AT)? 4. If the pH of the buffer (before it is titrated) is 5.70, what is the pKa of the weak acid? Hint solve the Henderson-Hasselbaich equation for pKa 15

Explanation / Answer

Determine pKa of weak acid

1. Titration of buffer with 0.126 M HCl

weak base component of buffer would react with HCl added

Volume of HCl used to reach equivalence point = 20 ml

moles of HCl added = moles of weak base present in buffer

moles of weak base = 0.126 M x 20 ml = 2.52 mmol

concentration of weak base in the buffer = 2.52 mmol/25 ml = 0.1008 M

2. Titration of buffer with 0.115 M NaOH

weak acid component of buffer would react with NaOH added

Volume of NaOH used to reach equivalence point = 11 ml

moles of NaOH added = moles of weak acid present in buffer

moles of weak acid = 0.115 M x 11 ml = 1.265 mmol

concentration of weak acid in the buffer = 1.265 mmol/25 ml = 0.0506 M

3. Buffer strength = total weak acid + total weak base in buffer

                             = 0.1008 M + 0.0506 M

                             = 0.1514 M

4. pH of buffer = 5.70

Using Hendersen-Hasselbalck relation,

pH = pKa + log(weak base/weak acid)

5.70 = pKa + log(0.1008/0.0506)

Thus,

pKa of weak acid = 5.40

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