You wish to prepare 250. mL of a 0.300 M Tris buffer at room temperature (25 C)

Question

You wish to prepare 250. mL of a 0.300 M Tris buffer at room temperature (25 C) such that it has a pH of 7.80 at 10.0 C.
Starting with Tris base (F.W. 121.2, 99.9 % purity), answer the following questions regarding the preparation of this buffer. Refer to Table 2-1 in your coursepack for information regarding the temperature coefficient of Tris (pH/T).

-What mass of Tris is needed?

-To what pH (at room temperature, 25 C) would you adjust the buffer so that it had a pH of 7.80 at 10.0 C?

-Which would be the correct solution to use to adjust the pH?

-What volume of the 2 M solution identified above will be needed to reach the correct pH?

Explanation / Answer

(i) According to Henderson-Hasselbulch equation: pH = pKa + Log(nTris/nTrisH+)

i.e. 7.8 = 8.1 + Log(nTris/nTrisH+)

i.e. Log(nTrisH+/nTris) = 8.1 - 7.8 = 0.3

i.e. nTrisH+/nTris = 100.3 = 2 ....... Equaiton 1

nTrisH+ + nTris = 0.3 mmol/mL * 250 mL = 75 mmol

i.e. nTrisH+ = 75 - nTris .......... Equation 2

From equations 1 and 2, you can write as follows.

(75 - nTris)/nTris = 2

i.e. nTris = 75/3 = 25 mmol

i.e. nTrisH+ = 75-25 = 50 mmol

The mass of Tris needed = 25*10-3 mol * 121.2 g mol-1 = 3.03 g

(ii) If the temperature is increased from 10 oC to 25 oC, [H+] increases, as a result, pH decreases since pH = -Log[H+]

i.e. pH < 7.8 at 25 oC

(iii) If you add some strong acid such as HCl to the Tris buffer, the pH can be adjusted to less than 7.8.

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