I created a 0.1M TRIS buffer in the lab, with a pH of 7.28 and a volume of 50.6m

Question

I created a 0.1M TRIS buffer in the lab, with a pH of 7.28 and a volume of 50.6mL. If I add 3mL of 1.0M NaOH to my buffer, will I still have a valid buffer?

In other words, what would be the resulting pH if I added the TRIS buffer and NaOH? I know that in order to still have a valid buffer, the resulting pH needs to be +/-1 from the original pH (which in this case would be 7.28).

I would really appreciate a throughly explained answer, since I would like to understand the calculations so that I can know what to on my own next time! Thank you.

Explanation / Answer

The pH of Tris buffer is given by the Henderson-Hasslebach equation as

pH = pKa + log [Tris]/[Tris-H+]

Plug in values and write (pKa of Tris is 8.07 at 25C) and obtain

7.28 = 8.07 + log [Tris]/[Tris-H+]

===> -0.79 = log [Tris]/[Tris-H+]

===> [Trsi]/[Tris-H+] = antilog(-0.79) = 0.162

===> [Tris] = 0.162*[Tris-H+] …..(1)

Also, [Tris-H+] + [Tris] = 0.1 M

===> [Tris-H+] + 0.162*[Tris-H+] = 0.1 M (use equation 1 from above)

===> 1.162*[Tris-H+] = 0.1 M

===> [Tris-H+] = (0.1/1.162) M = 0.086 M

Therefore, [Tris] = 0.162*[Tris-H+] = 0.162*(0.086 M) = 0.013932 M 0.014 M

Next, write down the neutralization reaction between Tris-H+ and NaOH as below:

Tris-H+ + OH- ------> Tris + H2O

As per the stoichiometric reaction,

1 mole NaOH = 1 mole Tris = 1 mole Tris-H+

Moles of Tris-H+ in the original buffer = (50.6 mL)*(0.086 M) = 4.3516 mmole.

Moles of Tris in the original buffer = (50.6 mL)*(0.014 M) = 0.7084 mmole.

Moles of NaOH added = (3 mL)*(1.0 M) = 3.0 mmole = millimoles of Tris formed.

Therefore, millimoles of Tris formed after addition of NaOH = (0.7084 + 3.0) mmole = 3.7084 mmole.

Millimoleos of Tris-H+ neutralized = 3.0 mmole; therefore, millimoles of Tris-H+ retained after the addition = (4.3516 – 3.0) mmole = 1.3516 mmole.

Since the volume of the solution stays the same for both Tris-H+ and Tris, we can express the equilibrium concentrations in terms of the number of millimoles.

Therefore, we obtain

pH = pKa + log (millimoles of Tris)/(millimoles of Tris-H+) = 8.07 + log (3.7084/1.3516) = 8.07 + 8.508 8.51

The pH of the solution changes by (8.51 – 7.28) = 1.23; the change in pH is more than +1 unit and hence, the buffering action doesn’t occur. Therefore, the prepared buffer is no longer valid (ans).

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