Question

You are working on a DNA extraction, when your assistant accidentally adds 250.0 μL of 0.200 M HCl to your sample. Your sample was originally 2.000 mL of a 0.100 M “tris” buffer at pH 8.072. The DNA extraction will not work if the pH goes below 6.50 or above 9.50. For trisH+, Ka = 8.472274 x 10^–9.

What is the pH of the solution after the HCl was added? Will you be able to continue doing your DNA extraction?

Answer 1

The acid ionization constant is written as

TrisH⁺(aq) <=====> Tris(aq) + H⁺(aq)

K_a = 8.472274*10^-9; therefore,

pK_a = -log K_a

= -log (8.472274*10^-9)

= 8.072.

Determine the ratio of TrisH⁺ and Tris in the buffer by using the Henderson-Hasslebach equation as below.

pH = pK_a + log [Tris]/[TrisH⁺]

====> 8.072 = 8.072 + log [Tris]/[TrisH⁺]

====> log [Tris]/[TrisH⁺] = 0 = log (1)

====> [Tris]/[TrisH⁺] = 1

====> [Tris] = [TrisH⁺] …….(1)

Again, the total Tris concentration is 0.100 M; therefore,

[Tris] + [TrisH⁺] = 0.100 M

====> [Tris] + [Tris] = 0.100 M

====> 2*[Tris] = 0.100 M

====> [Tris] = ½*(0.100 M) = 0.050 M.

Therefore,

[Tris] = [TrisH⁺] = 0.050 M.

The total volume of the buffer is 2.000 mL; therefore,

millimoles of TrisH⁺ in the buffer = (volume of buffer in mL)*(molarity of TrisH⁺)

= (2.000 mL)*(0.050 M)

= 0.100 mmole.

millimoles of Tris in the buffer = (volume of buffer in mL)*(molarity of Tris)

= (2.000 mL)*(0.050 M)

= 0.100 mmole.

HCl reacts with Tris as below.

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

As per the stoichiometric equation,

1 mole Tris = 1 mole HCl = 1 mole TrisH⁺.

Millimoles of HCl added = millimoles of Tris neutralized = millimoles of TrisH⁺ formed = (250.0 μL)*(0.200 M)

= (250.0μL)*(1 mL)/(1000 μL)*(0.200 M)

= 0.050 mmole.

Millimoles of Tris retained = (0.100 – 0.050) mmole = 0.050 mmole.

Millimoles of TrisH⁺ = (0.100 + 0.050) mmole = 0.150 mmole.

Since the volume of the solution stays constant, hence, we can approximate the ratio of the concentrations of Tris and TrisH⁺ to the number of moles of each, i.e,

[Tris]/[TrisH⁺] = (millimoles of Tris)/(millimoles of TrisH⁺) = (0.050 mmole)/(0.150 mmole).

Use the Henderson-Hasslebach equation to find out the new pH as

pH = pK_a + log [Tris]/[TrisH⁺]

= 8.072 + log (0.050 mmole)/(0.150 mmole)

= 8.072 + log (0.333)

= 8.072 + (-0.477)

= 7.595 ≈ 7.60 (ans).

The pH of the buffer solution stays between 6.50 and 9.50 and hence, the extraction of DNA will work at the pH of 7.60 (ans).