Question

a) Excess imidazole was removed using a HiPrep 26/10 desalting column (GE Healthcare) that was pre-equilibrated with 25 mM Tris-HCl (pH 7.5), 50 mM KCl, and 10 mM EDTA. Ignoring the KCl, and EDTA, determine the concentrations of the conjugate base and the weak acid in this buffer.

b) Calculate how you would prepare 500 mL of this buffer if you were given the solid chemical form of both the Tris base (conjugate base) and Tris-HCl (weak acid)

c) If 50 mL of 0.1 M NaOH were added to 500 mL of this buffer, what would be the new pH of the solution?

d) Based on the new pH, would Tris still be effective as a buffer? Explain.

Answer 1

a) The pK_a of Tris-HCl is 8.1; the pH of the solution is 7.5.

Use the Henderson-Hasslebach equation to compute the ratio of Tris base and Trish-HCl in the buffer.

pH = pK_a + log [Tris Base]/[Tris-HCl]

=====> 7.5 = 8.1 + log [Tris Base]/[Tris-HCl]

=====> -0.6 = log [Tris Base]/[Tris-HCl]

=====> [Tris Base]/[Tris-HCl] = antilog(-0.6) = 0.25

=====> [Tris Base] = 0.25*[Tris-HCl] ……..(1)

The total concentration of the conjugate base and the weak acid is 25mM; therefore,

[Tris Base] + [Tris-HCl] = 25 mM

=====> 0.25*[Tris-HCl] + [Tris-HCl] = 25 mM

=====> 1.25*[Tris-HCl] = 25 mM

=====> [Tris-HCl] = (25 mM)/(1.25) = 20 mM (ans)

[Tris Base] = 0.25*[Tris-HCl] = 0.25*(20 mM) = 5 mM (ans).

b) The molar masses are listed below.

Tris Base = 121.14 g/mol; Tris-HCl = 157.594 g/mol.

Moles of Tris Base in 500 mL buffer = (500 mL)*(1 L/1000 mL)*(5 mM)*(1 M/1000 mM)*(1 mol.L^-1/1 M) = 0.0025 mole.

Mass of Tris Base required = (0.0025 mole)*(121.14 g/mol) = 0.30285 g ≈ 0.3028 g (ans).

Moles of Tris-HCl in 500 mL buffer = (500 mL)*(1 L/1000 mL)*(20 mM)*(1 M/1000 mM)*(1 mol.L^-1/1 M) = 0.01 mole.

Mass of Tris-HCl required = (0.01 mole)*(157.594 g/mol) = 1.57594 g ≈ 1.5759 g (ans).

c) For NaOH, the molarity and the normality are equivalent; hence, 0.1 N NaOH = 0.1 M NaOH.

Moles of NaOH added to the buffer = (50 mL)*(1 L/1000 mL)*(0.1 M)*(1 mol.L^-1/1 M) = 0.005 mole.

NaOH reacts with Tris-HCl in the buffer to produce Tris Base and water as per the reaction

Tris-HCl (aq) + NaOH (aq) --------> Tris Base (aq) + H₂O (l)

As per the stoichiometric equation,

1 mole Tris-HCl = 1 mole NaOH = 1 mole Tris Base.

Hence, moles of Tris-HCl consumed = moles of Tris Base formed = moles of NaOH used = 0.005 mole.

The buffer solution now contains (0.005 + 0.005) mole = 0.01 mole Tris Base and (0.01 – 0.005) mole = 0.005 mole Tris-HCl.

The volume of the solution stays constant; hence, [Tris Base]/[Tris-HCl] = (0.01 mole)/(0.005 mole) = 2 (volume is constant for both).

Again, use the Henderson-Hasslebach equation for determining the pH.

pH = pK_a + log [Tris Base]/[Tris-HCl] = 8.1 + log (2) = 8.1 + 0.3010 = 8.401 ≈8.4 (ans).

d) A buffer can function effectively when the pH of the buffer satisfies the relation (pK_a – 1) < pH < (pK_a + 1). Since the pH of the buffer (after addition of 50 mL of 0.1 NaOH) satisfies the above relation, the mixture of 500 mL 24 mM Tris Base/Tris-HCl buffer can still function effectively as a buffer (ans).