Based on my molarities found in questions 1 and 2, how is question 3 found? 1. A

Question

Based on my molarities found in questions 1 and 2, how is question 3 found? 1. A student titrated a buffer with o HCI. The observed curve your is shown titration below. is the concentration base in the buffer? document use of the weak not a dilution (do not including mole conversion. (Hint: this is equi point and this is a titration) int 20.0 mL valence Buffer pH with Added HCI Hal molar mol 126 M O.020 L mo les 2. S3 x 10 0.002 moles mol base O. loog 2. The student then titrated 25.0 mL of the same buffer with 0.115 M NaoH. The observed fo5 IOLM 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. Buffer pH With Added NaOH equivalence point 10.5 ml SL les 700 aixlo me les X Me Acid, Volume NaoH Added 0.025 L 3. What is the buffer strength (i.e., the total of HA] and [AT? 4. the pH of the buffer (before is titrated) is 5.70, what is the pKa of the weak acid? Hint solve the Henderson-Hasselbalch equation for pKa. ka pKa lor l pka to

Explanation / Answer

1. Titration of buffer with HCl

Volume of HCl to reach equivalence point = 20 ml

moles of HCl used = 0.126 M x 20 ml

                              = 2.52 mmol

moles of base reacted = 2.52 mmol

Volume of buffer = 25 ml

concentration of base in buffer = 2.52 mmol/25 ml

                                                 = 0.1008 M

2. 25 ml of buffer titrated with 0.115 M NaOH

Volume of NaOH to reach equivalence point = 10.5 ml

moles of NaOH used = 0.115 M x 10.5 ml

                                   = 1.2075 mmol

moles of acid present in buffer = 1.2075 mmol

concentration of acid in buffer = 1.2075 mmol/25 ml

                                                = 0.0483 M

3. Buffer strength = [HA] + [A-]

                             = 0.0483 + 0.1008

                             = 0.1491 M

4. If the pH of buffer before titration is 5.70

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

with,

pKa is for the weak acid

[base] = 0.1008 M

[acid] = 0.0483 M

we get,

pKa = 5.70 - log(0.1008/0.0483)

       = 5.38

So,

pKa of weak acid would be 5.38

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.