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

Answer:

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

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 + $log\frac{[A^{-} ]}{[HA]}$

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:

$7.0=7.21+log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\ -0.21=log\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\\\10^{-0.21} =\frac{[HPO4^{-2} ]}{[H2PO4^{-} ]}\\0.616 * [H2PO4^{-}] = [HPO4^{-2}]$

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