# 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.

It is recommended that the students add this page to their watchlist to make sure they are informed when information is added to the page. The instructor will post comments in the corresponding discussion tabs.

**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