Tricine (N-[Tris(hydroxymethyl)methyl]glycine) is commonly used as a buffer. Bas

Question

Tricine (N-[Tris(hydroxymethyl)methyl]glycine) is commonly used as a buffer. Based on its structure, over approximately what pH range would a 50 mM Tricine solution be expected to be an effective buffer? Explain. Draw a graph to illustrate your reasoning. Which solution will provide the most buffering against added KOH: 0.1 M acetic acid (pK_a = 4.8) at pH 5.5; 0.1 M acetic acid at pH 4.8; 0.1 M acetic acid at pH 4.0; or 0.1 M methylamine (pK_a = 10.7) at pH 4.5? Explain your reasoning in words and/or diagrams. How would your answer change if HCl were added instead of KOH?

Explanation / Answer

Tricine is a zwitterionic amino acid that has a pKa1 value of 2.3 at 25 °C, while its pKa2 at 20 °C is 8.15.

SO the best buffering range will be

pH = pKa + log base/acid

pH = 8.15 + log base/acid

Log base/acid is 0.5 is the best buffer activity

pH = 8.15 + 0.5 =8.65

pH = 8.15 - 0.5 = 7.65

SO the best buffer range is 7.65 - 8.65 this is irrespective of the concentration

Since you are adding KOH which is a base the best buffer will be the one which is acidic. Buffer action means it will not let pH change by more than +/-1 unit in pH

pKa of acetic acid is 4.76

pH = pKa + log base/acid

5.5 = 4.76 + log base/acid

base/acid = 5.5

Since total base+ acid is 0.1

base is 0.085 and acid is 0.015

Since you are adding base new pH max should be 6.5

6.5 = 4.76 + log base/acid

base/acid = 55

base = 0.098 and acid is 0.002

SO amount of base that can be added is 0.013 M

4.0 = 4.76 + log base/acid

base/acid = 0.18

Since total base+ acid is 0.1

base is 0.015 and acid is 0.085

Since you are adding base new pH max should be 5.0

5.0 = 4.76 + log base/acid

base/acid = 1.74

base = 0.064 and acid is 0.036

SO amount of base that can be added is 0.049 M

For methyl amine

4.5 = 10.7 + log base/acid

base/acid = 6.31 x 10-7

Since total base+ acid is 0.1

base is 0.0000001 and acid is 0.0999999

Since you are adding base new pH max should be 5.5

5.5 = 10.7 + log base/acid

base/acid = 6.31 x 10-6

base = 0.000001 and acid is 0.099999

SO amount of base that can be added is very small

SO the best buffer is 0.1M acetic acid at pH 4.0

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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