Question

Be sure to answer all parts.

Tris [tris(hydroxymethyl)aminomethane] is a common buffer for studying biological systems.

(K_a = 5.01 ×10⁻⁹ and pK_a = 8.3)

(a) Calculate the pH of the Tris buffer after mixing 10.0 mL of 0.20 M HCl solution with 25.0 mL of 0.10 M Tris.


(b) This buffer was used to study an enzyme-catalyzed reaction. As a result of the reaction, 0.00015 mol of H⁺ was produced. What is the pH of the buffer at the end of the reaction?


(c) What would be the final pH if no buffer were present?

Answer 1

a)

mmol of TRIS = MV = 25*0.1 = 2.5

mmol of acid = MV = 0.2*10.0 = 2.0

after reaction

TRIS-H+ = 2.0

TRIS left = 2.5-2.0 = 0.5

pH = pKa + log(TRIS-H+ /TRIS)

pH = 8.3+log(0.5/2.0) =

pH = 7.6979

b)

addition of = 0.00015 mol of H+ = 0.15 mmol

mmol fo TRIS base left --> 2.5-2.0 = 0.5

mmol of TRIS acid formed = 2.0 + 0.15 = 2.15

pH = pKa + log(TRIS/TRIS acid)

pH = 8.3 + log(0.5/2.15)

ph = 7.666

c)

final pH if no buffer... V = 25 mL

[H+] = mol/V = (0.15)/25) = 0.006

pH = -log(0.006) = 2.22

pH will drop drastically