Question

What masses of monobasic and dibasic sodium phosphate will you use to make 250 mL of 0.1 M sodium phosphate buffer, pH = 7?

Answer 1

• Mass of monobasic sodium phosphate = 1.857 g
• Mass of dibasic sodium phosphate = 1.352 g

Step-by-step explanation:

The equilibrium that takes place is:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺ pka= 7.21 (we know this from literature)

To solve this problem we use the Henderson–Hasselbalch (H-H) equation:

pH = pka +
`log([A^(-) ])/([HA])`

In this case [A⁻] is [HPO₄⁻²], [HA] is [H₂PO₄⁻], pH=7.0, and pka = 7.21

If we use put data in the H-H equation, and solve for [HPO₄⁻²], we're left with:

`7.0=7.21+log([HPO4^(-2) ])/([H2PO4^(-) ])\\ -0.21=log([HPO4^(-2) ])/([H2PO4^(-) ])\\\\10^(-0.21) =([HPO4^(-2) ])/([H2PO4^(-) ])\\0.616 * [H2PO4^(-)] = [HPO4^(-2)]`

From the problem, we know that [HPO₄⁻²] + [H₂PO₄⁻] = 0.1 M

We replace the value of [HPO₄⁻²] in this equation:

0.616 * [H₂PO₄⁻] + [H₂PO₄⁻] = 0.1 M

1.616 * [H₂PO₄⁻] = 0.1 M

[H₂PO₄⁻] = 0.0619 M

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

[HPO₄⁻²] + 0.0619 M = 0.1 M

[HPO₄⁻²] = 0.0381 M

With the concentrations, the volume and the molecular weights, we can calculate the masses:

• Molecular weight of monobasic sodium phosphate (NaH₂PO₄)= 120 g/mol.
• Molecular weight of dibasic sodium phosphate (Na₂HPO₄)= 142 g/mol.

• mass of NaH₂PO₄ = 0.0619 M * 0.250 L * 120 g/mol = 1.857 g

• mass of Na₂HPO₄ = 0.0381 M * 0.250 L * 142 g/mol = 1.352 g