Question

Given 0.1 m solutions of na3po4 and h3po4, describe the preparation of 1 l of a phosphate buffer at a ph of 7.5. what are the molar concentrations of the ions in the final buffer solution, including na+ and h+?

Answer 1

Answer:

The buffer will be prepared with  0.174 L of Na₃PO₄ and 0.286 of H₃PO₄;

[Na⁺] = 0.0522 M

[PO₄³⁻] = 0.0174 M

[H⁺] = 0.0572

[HPO₄⁻] = 0.0286 M

Explanation:

First, notice that H₃PO₄ is a polyprotic acid, and its pKa are:

pK1 = 2.15

pK2 = 7.1

pK3 = 12.4

So, for a pH = 7.5, it's better to use a solution that reaches the pK2. The reaction is:

H₂PO₄⁻ ⇄ H⁺ + HPO₄²⁻

Using the Henderson-Hasselbalch equation we can compute the ratio of the base form (ion) and the acid form:

pH = pKa + log ([HPO₄²⁻]/[H₂PO₄⁻])

7.5 = 7.1 + log ([HPO₄²⁻]/[H₂PO₄⁻])

log ([HPO₄²⁻]/[H₂PO₄⁻]) = 0.4

[HPO₄²⁻]/[H₂PO₄⁻] = $10^{0.4}$

[HPO₄²⁻]/[H₂PO₄⁻] = 2.5

HPO₄²⁻ comes from the salt and H₂PO₄⁻. The final volume must be 1 L, so the concentrations in the ratio can be substituted for the number of moles. Let's call x the number of moles of H₂PO₄⁻, so:

nHPO₄²⁻ = 2.5x

The number of moles is the molar concentration multiplied by the volume. Calling V1 the salt volume, and V2 the acid volume:

V1 + V2 = 1 L --> V2 = 1 - V1

0.1*V1/0.1*V2 = 2.5

0.1*V1 = 0.25*V2

0.1V1 = 0.25*(1 - V1)

0.1V1 = 0.25 - 0.25V1

0.35V1 = 0.25

V1 = 0.714 L

V2 = 0.286 L

So, the buffer will be prepared with 0.174 L of Na₃PO₄ and 0.286 of H₃PO₄.

The number of moles of Na₃PO₄ is 0.1*0.174 = 0.0174 mol, and the concentration in 1 L will be 0.0174 M

Na₃PO₄ → 3Na⁺ + PO₄³⁻

[Na⁺] = 3x0.0174 = 0.0522 M

[PO₄³⁻] = 0.0174 M

The number of moles of H₃PO₄ is: 0.1*0.286 = 0.0286 mol, and the concentration in 1 L: 0.0286 M

H₃PO₄ → 2H⁺ + HPO₄⁻

[H⁺] = 2x0.0286 = 0.0572

[HPO₄⁻] = 0.0286 M