Methods: part a: preparation of buffers make two buffers starting with solid material, which is the most common way to make buffers. you will be given a desired ph, and your task is to prepare 100 ml of two appropriate buffers at a concentration of 0.10 m. one of the buffers will be a phosphate buffer (ph 7.0) and the other will be a tris buffer (ph 8.0). 1. calculate the weight of the buffer you would need to make 100 ml of a 0.10 m solution. weigh out the correct amount and dissolve in 50 ml water. you will need the following compound molecular weights: na2hpo4 (141.96 g/mol), nah2po4 (119.96 g/mol), and tris base (121.1 g/mol).

0,542 g of Na₂HPO₄ and 0,741 g of NaH₂PO₄.

0,856 g of Tris-HCl and 0,553 g of Tris-base

Explanation:

It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:

pH = pka + log₁₀

Where A⁻ is conjugate base and HA is conjugate acid

The equilibrium of phosphate buffer is:

H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺    Kₐ₂ = 6,20x10⁻⁸; pka=7,21

Thus, Henderson–Hasselbalch equation for phosphate buffer is:

pH = 7,21 + log₁₀

If desire pH is 7,0 you will obtain:

0,617 =  (1)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [HPO₄²⁻] + [H₂PO₄⁻] (2)

Replacing (1) in (2) you will obtain:

[H₂PO₄⁻] = 0,0618 M

And with this value:

[HPO₄²⁻] = 0,0382 M

As desire volume is 100mL -0,1L- the weight of both Na₂HPO₄ and NaH₂PO₄ is:

Na₂HPO₄ = 0,1 L× × = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× × = 0,741 g of NaH₂PO₄

For tris buffer the equilibrium is:

Tris-base + H⁺ ⇄ Tris-H⁺ pka = 8,075

Henderson–Hasselbalch equation for tris buffer is:

pH = 8,075 + log₁₀

If desire pH is 8,0 you will obtain:

0,841 =  (3)

Then, if desire concentration of buffers is 0,10 M:

0,10 M = [Tris-base] + [Tris-H⁺] (4)

Replacing (3) in (4) you will obtain:

[Tris-HCl] = 0,0543 M

[Tris-base] = 0,0457 M

As desire volume is 100mL -0,1L- the weight of both Tris-base and Tris-HCl is:

Tris-base = 0,1 L× × = 0,553 g of Tris-base

Tris-HCl = 0,1 L× × = 0,856 g of Tris-HCl

I hope it helps!

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