Question

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).

Answer 1

Answer:

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 $\frac{A^{-} }{HA}$

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 $\frac{HPO4^{2-} }{H2PO4^{-} }$

If desire pH is 7,0 you will obtain:

0,617 =  $\frac{HPO4^{2-} }{H2PO4^{-} }$ (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× $\frac{0,0382mol}{1L}$× $\frac{141,96g}{1mol}$ = 0,542 g of Na₂HPO₄

NaH₂PO₄ = 0,1 L× $\frac{0,0618mol}{1L}$× $\frac{119,96g}{1mol}$ = 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 $\frac{Tris-base }{Tris-H^{+} }$

If desire pH is 8,0 you will obtain:

0,841 =  $\frac{Tris-base }{TrisH^{+} }$ (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× $\frac{0,0457mol}{1L}$× $\frac{121,1g}{1mol}$ = 0,553 g of Tris-base

Tris-HCl = 0,1 L× $\frac{0,0543mol}{1L}$× $\frac{157,6g}{1mol}$ = 0,856 g of Tris-HCl

I hope it helps!