Which of the following could be used to formulate 100 mls of a 0.10 M acetate bu

Question

Which of the following could be used to formulate 100 mls of a 0.10 M acetate buffer (pK=4.76) at pH 5 if you start with 64 mls of 0.10 M sodium acetate?
34 mls of 0.10 M HCl 3.6 mls of 1 M NaOH 3.6 mls of 1 M HCl 34 mls of 0.10 M NaOH 34 mls of 0.10 M acetic acid
Show Work please
Which of the following could be used to formulate 100 mls of a 0.10 M acetate buffer (pK=4.76) at pH 5 if you start with 64 mls of 0.10 M sodium acetate?
34 mls of 0.10 M HCl 3.6 mls of 1 M NaOH 3.6 mls of 1 M HCl 34 mls of 0.10 M NaOH 34 mls of 0.10 M acetic acid Which of the following could be used to formulate 100 mls of a 0.10 M acetate buffer (pK=4.76) at pH 5 if you start with 64 mls of 0.10 M sodium acetate? 34 mls of 0.10 M HCl 3.6 mls of 1 M NaOH 3.6 mls of 1 M HCl 34 mls of 0.10 M NaOH 34 mls of 0.10 M acetic acid 34 mls of 0.10 M HCl 3.6 mls of 1 M NaOH 3.6 mls of 1 M HCl 34 mls of 0.10 M NaOH 34 mls of 0.10 M acetic acid 34 mls of 0.10 M HCl 3.6 mls of 1 M NaOH 3.6 mls of 1 M HCl 34 mls of 0.10 M NaOH 34 mls of 0.10 M acetic acid 34 mls of 0.10 M HCl

Explanation / Answer

The answer is: 34 mls of 0.10 M acetic acid


Moles of sodium acetate = 64/1000 x 0.10 = 0.0064 mol

Moles of acetic acid = 34/1000 x 0.10 = 0.0034 mol


Henderson-Hasselbalch equation:

pH = pKa + log([sodium acetate]/[acetic acid])

= pKa + log(moles of sodium acetate/moles of acetic acid)

= 4.76 + log(0.0064/0.0034)

= 5.03 (or approximately pH 5)

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

---

---

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