Question

1) Using the Henderson-Hasselbach equation, what will the pH become when buffers P1 and P2 are mixed during plasmid purification?____________ . Ignore the EDTA and the cells in your calculation and assume Tris-HCl has a pKa of 8.0.

Information: Buffer P1: 50mM Tris-HCl, pH8, 10mM EDTA, .1mg/ml RNase A

Buffer P2: Sodium hydroxide (200mM) and SDS (1%)

Teachers tip: "the pKa of protonated Tris as 8.0. Since the NaOH concentration is larger than the Tris concentration the pH is based on the left over NaOH. Remember pOH+pH = 14."

Using the information above and the Henderson-Hasselbach equation, what will the pH become when P1, P2, and N3 are mixed during plasmid purification? ___________ . Assume the pKa of acetic acid is 4.8 and remember to take into account the different volumes used.

Information: Buffer N3: Acetate buffer (.9M, pH 4.8) to neutralize the sodium hydroxide, and guanidine hydrochloride (4.2M), which denatures protein. N3 provides a high salt concentration, so that the DNA sticks to the silica gel membrane in the small column.

Volumes being used:

P1: 250uL(microliter)

P2: 250uL

N3: 350uL

There is a partial answer previously posted, but without any explanations to the procedure. Please explain the solution along with the work. Thanks!

Answer 1

1) The question is not very clearly presented; however, from what I understood, you mixed 250 µL of buffer P1 with 250 µL of sodium hydroxide (NaOH), which is not a buffer, but a very strong base. The buffer P1 consists of Tris-HCl and the conjugate base. The reaction taking place between Tris-HCl and NaOH is shown below.

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

NaCl and H₂O do not participate in the pH calculations; hence leave them out.

As per the balanced stoichiometric equation,

1 mole Tris-HCl = 1 mole NaOH

Mole Tris-HCl added = (250 µL)*(1 L/10⁶ L)*(50 mM)*(1 M/1000 mM)*(1 mol/L/1 M) = 1.25*10^-5 mole.

Mole NaOH added = (250 µL)*(1 L/10⁶ L)*(200 mM)*(1 M/1000 mM)*(1 mol/L/1 M) = 5.00*10^-5 mole.

Mole NaOH unreacted = (mole NaOH added) – (mole Tris-HCl added) = (5.00*10^-5 mole) – (1.25*10^-5 mole) = 3.75*10^-5 mole.

Total volume of the solution = (250 + 250) µL = 500 µL.

Molar concentration of excess NaOH in the solution = (mole NaOH)/(total volume) = (3.75*10^-5 mole)/[(500 µL)*(1 L/10⁶ µL)] = 0.075 mol/L = 0.075 M.

Since, we have excess NaOH, we find the pOH as pOH = -log [OH^-] = -log (0.075) = -(-1.1249) ≈ 1.125.

Since pH + pOH = 14, we have pH = 14 – pOH = 14 – 1.125 = 12.875 (ans).

Use the Henderson-Hasslebach equation to calculate the ratio of the concentrations of acetic acid/acetate in the buffer.

pH = pK_a + log [acetate]/[acetic acid]

===> 4.8 = 4.8 + log [acetate]/[acetic acid]

===> 0.0 = log [acetate]/[acetic acid]

===> [acetate]/[acetic acid] = 1 [log (1) = 0.0]

===> [acetate] = [acetic acid] ……(1)

Again, [acetate] + [acetic acid] = 0.9 M

===> 2*[acetic acid] = 0.9 M [from (1) above]

===> [acetic acid] = [acetate] = ½*(0.9 M) = 0.45 M.

The volume of acetate buffer added is 350 µL; hence mole acetate = mole acetic acid = (350 µL)*(1 L/10⁶ L)*(0.45 M)*(1 mol/L/1 M) = 1.575*10^-4 mole.

The acetic acid in the buffer will neutralize the NaOH as per the reaction

HAc (aq) + NaOH (aq) --------> NaAc (aq) + H₂O (l) [HAc = acetic acid; NaAc = sodium acetate]

As per the stoichiometry of the reaction,

1 mole HAc = 1 mole NaOH = 1 mole NaAc

Therefore, 3.75*10^-5 mole NaOH = 3.75*10^-5 mole NaAc formed = 3.75*10^-5 mole HAc neutralized.

Therefore, mole of unreacted HAc at equilibrium = (1.575*10^-4 – 3.75*10^-5) mole = 1.2*10^-4 mole; mole of NaAc at equilibrium = (1.575*10^-4 + 3.75*10^-5) mole = 1.95*10^-4 mole.

Since the total volume of the solution is the same (500 + 350) µL = 850 µL for both HAc and NaAc, hence, we can express the molar concentrations of HAc and NaAc at equilibrium as the respective number of moles.

Use the Henderson-Hasslebach equation again.

pH = pK_a + log [acetate]/[acetic acid] = 4.8 + log (1.95*10^-4 mole)/(1.2*10^-4 mole) = 4.8 + log (1.625) = 4.8 + 0.2108 = 5.0108 ≈ 5.011 (ans).