Question

(show calculations) make a 40mL, 1 M pH 8 Tris buffer from the following reagent:

a) What is the pKa of Tris? b)The amount of powder you will need to weigh out. c)The amount of water you will use. d) the volume of 1 N HCl or NaOH needed to make the result pH 8. e) Plot the pH as a function of the number of moles of HCl of NaOH.

Answer 1

a) Tris, C₄H₁₁NO₃ forms an acid/base buffer with the conjugate acid, Tis-H⁺. The pK_a of Tris-H⁺ is 8.07 and the acid ionization reaction is given as

Tris-H⁺ (aq) --------> Tris (aq) + H⁺ (aq); pK_a = 8.07

b) Tris is available as a white crystalline powder. The pH of the buffer is given by the Henderson-Hasslebach equation as

pH = pK_a + log [Tris]/[Tris-H⁺]

=====> 8.00 = 8.07 + log [Tris]/[Tris-H⁺]

=====> log [Tris]/[Tris-H⁺] = -0.7

=====> [Tris]/[Tris-H⁺] = antilog (-0.7) = 0.1995 ≈ 0.2000

=====> [Tris] = 0.2000*[Tris-H⁺] ……..(1)

The total concentration of Tris is given as 1 M; therefore,

[Tris] + [TriS-H⁺] = 1 M

=====> [Tris] + 0.2000*[Tris] = 1 M

=====> 1.2000*[Tris] = 1 M

=====> [Tris] = (1 M)/(1.2000) = 0.8333 M

[Tris-H⁺] = 0.2000*[Tris] = 0.2000*(0.8333 M) = 0.1667 M.

Mole(s) of Tris-H⁺ in 40 mL buffer solution = (40 mL)*(1 L/1000 mL)*(0.1667 M)*(1 mol/L/1 M) = 0.006668 mole.

Tris-H⁺ is produced by the action of HCl on Tris as below.

Tris (aq) + HCl (aq) ------> Tris-H⁺ (aq) + Cl^- (aq)

As per the stoichiometry of the reaction,

1 mole Tris = 1 mole HCl = 1 mole Tris-H⁺.

Therefore, 0.006668 mole Tris-H⁺ = 0.006668 mole Tris.

Moles of Tris required to prepare the buffer = (40 mL)*(1 L/1000 mL)*(0.8333 M)*(1 mol/L/1 M) + 0.006668 mole = 0.0400 mole.

Molar mass of Tris = (4*12.01 + 11*1.008 + 1*14.0067 + 3*15.9994) g/mol = 121.1329 g/mol.

Mass of solid Tris powder required = (0.0400 mole)*(121.1329 g/mol) = 4.845316 g ≈ 4.8453 g (ans).

c) Mole(s) of HCl required = 0.006668 mole (as per the stoichiometry of the reaction).

Volume of 1 N HCl required = (0.006668 mole)/(1 mol/L) = 0.006668 L = (0.006668 L)*(1000 mL/1 L) = 6.668 mL ≈ 6.67 mL.

Weigh out 4.8453 g solid Tris powder, add 6.67 mL of 1 N HCl and (40.00 – 6.67) mL = 33.33 mL water.

d) The volume of 1 N HCl required = 6.67 mL (ans).