Question

1. In lab, we have been using PBS. This is phosphate buffered saline. Phosphate Buffered Saline is usually 25mM to 100mM Phosphate Buffer with 137mM NaCl added and often at pH 7.4. You need to make 1L of 50mM PBS. The buffer is 50mM and the salt remains at 137mM.

You have the following reagents on the shelf:                                                            

- NaCl (MW = 58.44 g/mole);   

- Sodium phosphate dibasic heptahydrate (Na₂HPO₄ · 7H₂O; MW = 268 g/mole)

- Sodium phosphate monobasic (NaH₂PO₄; MW = 120g/mole)

How many grams for each of those three reagents is needed to make 1L of a 50mM PBS at pH 7.4? (use the pKa value from part #2 above)?

The pKa value in part #2 was 7.09. Here is what I have done so far:

known ea: [ acid] + [Conj base 1 = 50mm H-4 equation: 7.4 pH = pka + log[con; base]/I weak acid] 7.4 = 7.09 + 109 [26]/CAJ 0.31 = 109 [66] 1 CAJ 2.0417=CC6J1CAJ [con; base 2.0417 [weak acid] plug into other (conj.base=50mm-c wear acid] 2.0417 (weak asid ] = 50 mM-[weak acid] [wear acid] =16.438 mm (con; base] = 33.566 mm ak

I don't know what to do next. Please help :(

Answer 1

The pH of a ph of a buffer from the solution is determined Henderson - Hasselback equation: - pH = pka & log [conjugate base] 0 [weak acid ] Now, the desired pH = 764 & pka = 7.09 (given) [con jugate base] Y: 7.4-7.09 - 0.31 7.2 . [weak acid] 0.31 conjugate base] [weak a cid No Now, molarity of buffer= [conjugate base ] . - + [weak acid ] ☆ 50 mm = [Conjugate base 7+ [weak acid] → [Conjugate base) = 50 mM – (weak acid] Now, playing this in ega 0 50 mM - [weak acid ] - = 2.042 1 [weak a cid ] D 50m M = 2.042 - Tweak a cid ] - - 50mm = 16.436 mm » [weak a cid] = 2,042+1 = 16,436 mmol/l 4 [Conjugate base 7 = (50-16.436) = 33.564 mmol/l

Now, to prepare 12 of solution containing 16.436 mm Na Ho POL (monobasic) & 33.564 mM Naz HPO (dibasic) we have to take. 76.436 mmol of Na H2 PO= 16.636 *10^3 mol - Na H₂ PO = 16.436x10-3x molar mass of Na H₂PO4 = 16.436 x 10%x 120 g 1 of Narz POL = 1,972 g of Na H2 POL 33.564 mmol of Nag HPO2 = 33.564X 10 -3 268 g of Naz HPO4.7H2O = 8.995 g of Naz HPO4. TH2O We also have to take Na Cl with 137 mM concentration. Se 137 mmol of NaCl = 1378103 x 58.44 g of NaCl = 8.006 g of NaCl - - Now, in a water to the dieagents add these quantities of the reagents little less than 1L distilled and mix well. Then make it up mark, ie 1L volume IL mix well.

Note: It is always better to check the pH while preparing a buffer and if needed adjust the pH accordingly by adding some strong acid or base (e.g. HCl or NaOH).