The buffer bis-Tris is a commonly used in biochemistry labs. It has a Ka = 3.16

Question

The buffer bis-Tris is a commonly used in biochemistry labs. It has a Ka = 3.16 x 10-7 and a molar mass of 209.2 g/mol. Show all of your calculations clearly and with correct units and reasonable significant figures. The small x approximation may be used throughout.

a) How many grams of bis-Tris are needed to make 1-L of a 0.100 M solution?

b) What is the pKa of bis-Tris?

c) What is the pH of the 0.100M solution of bis-Tris from part (a)?

d) What stoichiometric ratio of the conjugate base to weak acid would be necessary to create a bisTris buffer with a pH of 7.00?

e) How much 0.100M NaOH would you need to add to the solution in part (a) to obtain the ratio in part (d)? Hint: calculate the number of moles of the conjugate base in part (d).

Explanation / Answer

a)

M = mol/V

mol = MV = 0.1M*1L = 0.1 mol of bis.tris

mass = mol*MW = 0.1 * 209.2 = 20.92 g of bis-tris required for a 0.1 M solution

b)

pKa = -log(Ka)

so

pKa = -log(3.16*10^-7) = 6.5003

c)

find pH for a 0.1 M solution

so

Tris + H2O <--> H+ + Tris

Ka = [H+][Tris]/[bistris]

3.16*10^-7 = x*x/(0.1-x)

assume x is too small so

3.16*10^-7 = x*x/(0.1)

x = sqrt((3.16*10^-7)(0.1))

x =1.77763*10^-4

pH = -log(x) = -log(1.77763*10^-4)

pH = 3.7501

d)

for a buffer:

pH = pKa + log(ratio)

ratio = 10^(pH-pKa) = 10^-(7-6.5003)

ratio = 0.3164

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