Question

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.

Answer 1

• Mass of NaH₂PO₄ = 4.707 g

• Mass of Na₂HPO₄ = 8.627 g

Step-by-step 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 +
`log([A^(-) ])/([HA])`

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⁻]:

`7.4=7.21+log([HPO4^(-2) ])/([H2PO4^(-) ]) \\0.19=log([HPO4^(-2) ])/([H2PO4^(-) ]) \\10^(0.19)= ([HPO4^(-2) ])/([H2PO4^(-) ]) \\1.549*[H2PO4^(-) ]=[HPO4^(-2) ]`

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:

• Mass of NaH₂PO₄ = 3.923 * 10⁻⁴ M * 100 L * 119.98 g/mol = 4.707 g

• Mass of Na₂HPO₄ = 6.077 * 10⁻⁴ M * 100 L * 141.96 g/mol = 8.627 g