Blood substitute. as noted in this chapter, blood contains a total concentration of phosphate of approximately 1 mm and typically has a ph of 7.4. you wish to make 100 liters of phosphate buffer with a ph of 7.4 from nah 2 po 4 (molecular weight, 119. 98 g mol 1) and na 2 hpo 4 (molecular weight, 141. 96 g mol 1). how much of each (in grams) do you need? berg, jeremy m.. biochemistry (p. 25). w. h. freeman. kindle edition.

Explanation:

The equilibrium relevant for this problem is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

The Henderson–Hasselbalch (H-H) equation is needed to solve this problem:

pH= pka +

In this case, [A⁻] = [HPO₄⁻²], [HA] = [H₂PO₄⁻], pH = 7.4; from literature we know that pka=7.21.

We use the H-H equation to describe [HPO₄⁻²] in terms of  [H₂PO₄⁻]:

The problem tells us that the concentration of phosphate is 1 mM, which means:

[HPO₄⁻²] + [H₂PO₄⁻] = 1 mM = 0.001 M

In this equation we can replace [HPO₄⁻²] with the term expressed in the H-H eq:

1.549 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.001 M

2.549 * [H₂PO₄⁻] = 0.001 M

[H₂PO₄⁻] = 3.923 * 10⁻⁴ M

With the value of [H₂PO₄⁻] we can calculate [HPO₄⁻²]:

[HPO₄⁻²] + 3.923 * 10⁻⁴ M = 0.001 M

[HPO₄⁻²] = 6.077 * 10⁻⁴ M

With the concentrations, the molecular weight, and the volume, we calculate the mass of each reagent:

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.