Question

. An enzyme-catalyzed reaction was carried out in a 0.2M �Tris� buffer, pH 7.8. As a result of the reaction, 0.03 mole/liter of H+ was produced. (pKa of Tris =8.1) (a) What was the ratio of Tris+ (conjugate acid) / Triso (Conjugate base) at the start of the reaction? (b) What are the concentration of Tris+ and Triso at the start of the reaction? (c) Show the reaction by which the buffer maintained a near constant pH. (d) What were the concentration of Tris+ and Triso at the end of the reaction? (e) What was the pH at the end of the reaction? (f) What would the final pH be if no buffer were present?
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Answer 1

a = acid
b = base

pH = pKa + log (b)/(a)
7.8 = 8.1 + log (b/a)
Solve for b/a. This is equation 1 and is the answer for part(a)

equation 2 is a + b = 0.2

Solve these two equations simultaneously to obtain a and b at the beginning. This gives the answer to part (b).

Can you take it from here.

Answer 2

Good