Question

An enzyme-catalyzed reaction is carried out in a solution containing 0.2 M TRIS buffer. The pH of the reaction mixture at the start was 7.8. As a result of the reaction, 0.03 mole/liter of OH- were produced. What is the ratio of TRIS^o / Tris^+ at the end of the reaction ? b] what was the final ph of the reaction mixture , C] what would the final ph be if no buffer were present ? e] write the chemical equation showing how the tris buffer maintained a near constant ph during the reaction pka of tris = 8.1 ?

Answer 1

a) The initial acid and salt molar ratio is calculated:

[Tris] / [Tris +] = 10 ^ (pH - pKa) = 10 ^ (7.8 - 8.1) = 0.5

It has:

1) [Tris] - 0.5 [Tris +] = 0

2) [Tris] + [Tris +] = 0.2

System of equations is applied and you have:

[Tris] = 0.07 M

[Tris +] = 0.13 M

The OH- reacts with Tris + (decreasing it) and forms Tris (increasing it), the new molar ratio is calculated:

[Tris] / [Tris +] = (0.07 + 0.03) / (0.13 - 0.03) = 1

b) The final pH is calcified:

pH = pKa + log [Tris] / [Tris +] = 8.1 + log 1 = 8.1

c) The pOH and pH from the production of OH- are calculated:

pOH = - log [OH-] = - log 0.03 = 1.5

pH = 14 - 1.5 = 12.5

e) The reaction that occurs is:

Tris + + OH- = Tris + H2O

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