I. Pre-Lab Determine the quantity of each chemical that you are going to use for

Question

I. Pre-Lab Determine the quantity of each chemical that you are going to use for preparing the buffer Calculate the amount of each chemical you will need to weigh out to make solutions (A) (F) as listed in the "Materials" section below. Also perform the following calculations i) How much of solution (A) and solution (B) would you need to mix to obtain 100 ml of 0.2 M phosphate buffer at pH 6.8 (the relevant pK,is 7.2)? Will you need to add any extra water or just mix solutions (A) and (B)? How much acid (5 and 1 N HCl solutions will be provided) will you need to add to a 1 M Tris (pK, 8.3) base solution to adjust the pH to the indicated pH of 8? ii) iii) What hazards does imidazole present? iv) Why is the pKa of a buffer potentially dependent on temperature? III. Materials Distilled water, monobasic sodium phosphate, dibasic sodium phosphate, sodium chloride, imidazole, magnesium chloride, Tris base and (1 M and 5 M) HCI to make the following "stock" solutions (A) (B) 0.25 L of 0.2 M monobasic sodium phosphate 0.25 L of 0.2 M dibasic sodium phosphate (sometimes precipitation is observed over time so this should be made freshly) (C) 20 mL of 2 M NaCl (D) 20 mL of 2 M imidazole (E) 20 mL of 0.5 M MgCl (F) 40 mL of 1 M Tris base

Explanation / Answer

i) From the Henderson-Hasselbalch equation, we can determine the relationship of the concentration of both substances in solution to form said buffer:

pH = pKa + Log ([Salt] / [Acid])

Clearing the relationship we have:

[Salt] / [Acid] = 10 ^ (pH-pKa) = 10 ^ (6.8-7.2) = 0.3981 = n Salt / n Acid

The requested buffer is 0.2 M concentration in both species, so:

0.2 mol / L * 0.1 L = 0.02 moles Salt + Acid

n Salt + n Acid = 0.02 moles

n Salt = 0.02 - n Acid

We substitute in the previous relation:

0.02 - n Acid / n Acid = 0.3981

and clearing:

n Acid = 0.0143 moles

n Salt = 0.02 - 0.0143 = 0.0057 moles

According to the initial solutions:

Vol Acid = 0.0143 moles * (1000 mL / 0.2 moles) = 71.5 mL

Vol Sal = 0.0057 moles * (1000 mL / 0.2 moles) = 28.5 mL

no need to add more water to the solution.

ii) Assuming 1 Liter of Base solution and using same previous clearance of Henderson-Hasselbalch equation:

n Salt / n Acid = 10 ^ (pKa-pH) = 10 ^ (8-8.3) = 2

n Necessary acid = n Salt / 2 = 1 mol / 2 = 0.5 mol Acid.

Vol Acid (5M) = 0.5 mol * (1000 mL / 5 mol) = 100 mL Acid

Vol Acid (1M) = 0.5 mol (1000 mL / 1 mol) = 500 mL Acid

iii) Acute toxicity (oral), skin corrosion or irritation, serious eye damage or eye irritation, toxicity to reproduction.

iv) The Van't Hoff equation allows us to study qualitatively how the chemical equilibrium shifts with temperature depending on whether we have an endothermic or exothermic reaction:

If a reaction is endothermic, H> 0, increasing the temperature increases the equilibrium constant and the equilibrium moves to the right (towards the formation of products, increasing the numerator). On the other hand, if the temperature drops, the equilibrium constant also decreases and the reaction moves to the left.

If a reaction is exothermic, H <0, increasing the temperature decreases the equilibrium constant and the reaction moves to the left, while if the temperature decreases, the equilibrium constant increases and the equilibrium moves to the right.

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