## Modern Experimental Biochemistry

Chemistry

## Quiz 2 :

Structural Analysis of a Dipeptide

_{2}PO 4 ) and dibasic sodium phosphate (Na 2 HPO 4 ) necessary to make 1 liter of the solution.

Substitute the above obtained value in equation (1), So, 0.5 M = Thus, the number of grams of . 0.308 M = 0.308 mol /L, so Thus, the number of grams of .

_{2}PO 4 a solution of NaOH, and a pH meter. You need to prepare a buffer for biochemistry lab. The required solution is 0.5 M sodium phosphate, pH 7.0. Use the Henderson-Hasselbalch equation to calculate the number of moles and grams of monobasic sodium phosphate (NaH

_{2}PO 4 ) and dibasic sodium phosphate (Na 2 HPO 4 ) necessary to make 1 liter of the solution.

In this problem, we are asked to prepare a liter of 0.5 M sodium phosphate buffer, pH 7.0, using only monobasic sodium phosphate (NaH_{2} PO 4 ), a sodium hydroxide (NaOH) solution, and a pH meter.
By using the monobasic phosphate, we can measure out directly 0.5 moles of this material and dissolve in around 900mL of water (we need a liter total, but we want to make sure the pH is correct before we fill the solution to completion). After the phosphate is dissolved in the water, we use our pH meter to determine the pH, and then adjust drop-wise with NaOH until the pH is 7.0; the salt is acidic, so the pH will initially be below 7.0. Once the pH is in range, we fill the vessel to one liter.

In this problem, we are asked to explain the procedures to prepare a 0.1M glycine buffer at pH 10.0. We will use the Henderson-Hasselbalch equation to calculate the moles of isoelectric glycine and sodium glycinate we will need. The Henderson-Hasselbalch equation is as follows: pH = -log (Ka) + log Where: pH the pH of the solution Ka the acid dissociation constant for HA [A-] concentration of the anion of the acid [HA] concentration of the acid The acid in this buffer is the glycine, and the anion of this combination is sodium glycinate. The Ka for isoelectric glycine ion is 2.51x10 -10. We calculate the ratio of our anion to the acid as follows: pH = -log (Ka) + log We also know that the [A-] + [HA] = 0.1M. We can say that [HA] = 0.1 - [A-]. Thus, [A-] equals 0.07 moles, and because the two must add up to 0.1M, the [HA] must be 0.03 moles. This would be added into a 1L volumetric flask and then filled to the mark.

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