Biol 3360 Spring 2008 - Kinetics of Papain

Tue, 06/04/2013 - 13:57 -- sewm02

Biol 3360 Spring 2008 - Kinetics of Papain

This Wiki will be used by the participants of the Biochemistry Laboratory Biol 3360 during Spring 2008 to collaborate in the pre-laboratory work of the "wet" portion of the course.

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Making a buffer with a pH of 7 - Jordan and Greg

Buffer with pH of 5 - Aaron and Ian

Benzoate Buffer C6H5CO2H (122.12g/mol) has a pKa of 4.19

We want to make a buffer solution with a pH of 5.0

pH = pKa + log ([base]/[acid]

5 = 4.19 + log ([base]/[acid])

Log ([base]/[acid]) = 0.81

Therefore, adding 1mole of benzoic acid per liter for every 6.3 moles of sodium benzoate (its conjugate base) will yield a buffer solution with an approximate pH of 5.0.

Making a citrate buffer with a pH of 6.0 - Leslie & Tiffany

This is the pH that I want:

pH = 6.0

Citric acid is polyprotic, it has 3 pKa values:

pKa1 = 3.10

pKa2 = 4.75

pKa3 = 6.40

This is the Henderson-Hasselbalch equation, where [D] is the concentration of the acid (proton donor) and [A] is the concentration of the base

(proton acceptor) :

pH = pKa + log ([D]/[A])

This is the Henderson-Hasselbalch equation with the pH that I want (6.0) and pKa3 (6.40) substituted into it.

6.0 = 6.40 + log ([D]/[A])

This is the above equation simplified.

[D]/[A] = 10-0.40

[D]/[A] = 0.398

This equation is just saying that the total concentration of acid and base is 0.01 M.

[A] + [D] = 0.01

This is the above equation rearranged to solve for [A].

[A] = 0.01 - [D]

This equation was formed by substituting the previous equation ([A] = 0.01 - [D]) into the equation [D]/[A] = 3.98.

[D]/(0.01-[D]) = 0.398

This is the previous equation with both sides multiplied by (0.01 -[D]).

[D] = 0.398(0.01-[D])

This is the previous equation with the 0.398 multiplied through.

[D] = 0.00398 - 0.398[D]

This is the previous equation with 0.398[D] added to both sides.

[D] + 0.398[D] = 0.00398

This is the previous equation with the two [D] terms combined

1.398[D] = 0.00398

This is the above equation solved for [D], which is the amount of citric acid needed (in moles).

[D] = 0.0028 moles = moles of citric acid

This equation is used to find [A] needed for the buffer.

[A] + [D] = 0.01

This is the previous equation with the calculated [D] substituted into it.

[A] + 0.0028 = 0.01

This is the above equation rearranged to solve for [A].

[A] = 0.01 - 0.0028

This is the above equation solved for [A], which is the amount of sodium citrate needed (in moles).

[A] = 0.0072 moles = moles of sodium citrate

This is the amount of citric acid needed (in grams) for each liter of buffer.

(0.0028 mol)(192.13g/mol) = 0.538 g citric acid

This is the amount of sodium citrate needed (in grams) for each liter of buffer.

(0.0072 mol)(258.07g/mol) = 1.858 g sodium citrate

Weigh out 0.530 g of citric acid and 1.858 g of sodium citrate. Place into a 1.0 L volumetric flask. Dilute to the mark with

distilled water. Adjust the pH to 6.0 using concentrated NaOH or HCl.

Making a acetic acid buffer of a pH of 4.0 - Megan Smith

Making a Carbonate buffer of 9 - Alyssa

(1 mol Na2HCO3/ 1L) (105.989 g Na2HCO3/ mol) = 105.989 g Na2HCO3 in 1 L dH2O

Adjust the pH with a pH probe and NaOH

Making a Carbonate buffer of 10 - J.P. Johnson

 

 

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