An nzyme-catalyzed reaction is carried out in a 50 mL solution containing 0.100

Question

An nzyme-catalyzed reaction is carried out in a 50 mL solution containing 0.100 M Tris (tris(hydroxymethyl)aminomethane) buffer (pKa=8.3 at 20 degrees C). The pH of the reaction mixture was initially 8.0. As a result of the reaction, 2.0 X 10^-4 moles of H3O+ were produced.
A. What is the ratio of Tris base to Tris acid at the start of the experiment? B. What is the final pH? C. Did the Tris buffer have sufficient buffering capacity to maintain the solution at a constant pH 8.0 throughout the experiment? D. If not, explain how you would alter the reaction conditions. An nzyme-catalyzed reaction is carried out in a 50 mL solution containing 0.100 M Tris (tris(hydroxymethyl)aminomethane) buffer (pKa=8.3 at 20 degrees C). The pH of the reaction mixture was initially 8.0. As a result of the reaction, 2.0 X 10^-4 moles of H3O+ were produced.
A. What is the ratio of Tris base to Tris acid at the start of the experiment? B. What is the final pH? C. Did the Tris buffer have sufficient buffering capacity to maintain the solution at a constant pH 8.0 throughout the experiment? D. If not, explain how you would alter the reaction conditions.
A. What is the ratio of Tris base to Tris acid at the start of the experiment? B. What is the final pH? C. Did the Tris buffer have sufficient buffering capacity to maintain the solution at a constant pH 8.0 throughout the experiment? D. If not, explain how you would alter the reaction conditions.

Explanation / Answer

a]

Use the Henderson-Hasselbalch equation.
pH = pKa + log ([Tris] / [Tris HCl]). Here the pH is 8 and the pKa of the Tris is 8.3, so you can solve for the ratio ([Tris] / [Tris HCl]).

[Tris] / [Tris HCl] = 0.5

b]

Moles of Tris + Moles of Tris HCl = M*V in L = 50*0.1 = 0.005 moles

Moles of Tris = 0.5 *Moles of Tris HCl

From the two equtions

Moles of Tris HCl = 3.33*10^-3

Moles of Tris = 1.66*10^-3

Moles of H+ = 0.0002

Tris base + H+ -------> Tris HCl

1.66*10^-3 0.0002 3.33*10^-3

1.66*10^-3 -0.0002 3.33*10^-3 + 0.0002

pH = pKa + log [Tris] / [Tris HCl]

pH = 7.91

C]

Yes it has sufficient buffering capacity

Because if Base is added .....Tris HCl converts to Tris base

and vice-versa

SO it has buffering capacity to maintain at pH = 8

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