Phosphate Buffer Design

Introduction

Phosphate is the classic buffer for work near neutral pH. Its second dissociation, H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻, has a thermodynamic pKa₂ of 7.20 at 25 °C, placing it right in the range needed by most biological and environmental systems. Designing a phosphate buffer looks like a simple Henderson–Hasselbalch exercise, but the charges on the ions make activity effects significant. A good design combines the ideal calculation with an honest account of these realistic corrections.

Core explanation

The ideal calculation. For a target pH, the required ratio is:

[HPO₄²⁻]/[H₂PO₄⁻] = 10^(pH − 7.20)

If the total phosphate concentration is C, then [H₂PO₄⁻] = C/(1 + r) and [HPO₄²⁻] = C × r/(1 + r), where r is the ratio. The salts NaH₂PO₄ (M = 119.98 g mol⁻¹) and Na₂HPO₄ (M = 141.96 g mol⁻¹) are the usual sources, or potassium salts where sodium must be avoided.

Ionic strength matters. The pair involves a singly charged and a doubly charged ion. At any real concentration their activity coefficients differ, and the Henderson–Hasselbalch equation written in concentrations becomes:

pH = pKa₂ + log(γ(HPO₄²⁻)/γ(H₂PO₄⁻)) + log([HPO₄²⁻]/[H₂PO₄⁻])

Because the doubly charged ion has the smaller activity coefficient, the correction term is negative. The combination pKa₂ + log(γ₂/γ₁) is called the apparent pKa, pKa′. For a 0.1 mol dm⁻³ phosphate buffer the ionic strength is roughly 0.2 mol dm⁻³, and pKa′ falls to about 6.8. A design based on 7.20 therefore gives a pH about 0.3–0.4 units lower than intended.

Other realistic factors.

- Temperature: pKa₂ decreases slightly as temperature rises, by roughly 0.003 units per kelvin. - Dilution: diluting a phosphate stock raises its pH, because ionic strength falls and pKa′ rises back towards 7.20. - Precipitation: phosphate forms sparingly soluble salts with Ca²⁺ and Mg²⁺, so it is unsuitable where these ions must stay in solution. - Hydrates: commercial salts are often hydrates, such as Na₂HPO₄·7H₂O, so the correct molar mass must be used.

The practical conclusion. Calculate the ratio as a starting point, prepare the solution, then check it with a calibrated pH meter at the working temperature and adjust with small additions of acid or base. The calculation tells you where to start; the meter tells you where you are.

Formulae

r = 10^(pH − pKa′); n(H₂PO₄⁻) = C V/(1 + r); n(HPO₄²⁻) = C V r/(1 + r). Ionic strength: I = ½ Σ cᵢ zᵢ².

Step-by-step reasoning

1. Choose the H₂PO₄⁻/HPO₄²⁻ pair because pKa₂ is closest to neutral. 2. Decide whether to use pKa₂ = 7.20 or an apparent pKa′ for the expected ionic strength. 3. Calculate the ratio r from the target pH. 4. Split the total phosphate between the two forms. 5. Convert moles to masses using the correct molar masses. 6. Prepare, measure and fine-tune the pH.

Visual explanation

Plot the fraction of HPO₄²⁻ against pH: an S-shaped curve centred on 7.20 for infinite dilution. Draw a second, identical curve shifted about 0.4 units to the left for 0.1 mol dm⁻³ buffer; the gap between the curves is the activity correction.

Real-world analogy

Designing with the ideal pKa is like planning a road trip using a map without traffic. The route is right, but real conditions add delays, so you still check progress on the way and adjust.

Real-world example

Phosphate-buffered saline, widely used to wash and dilute cells, contains sodium and potassium phosphates alongside sodium chloride. Its high ionic strength is one reason its measured pH is set by adjustment rather than by calculation alone.

Why?

Why does ionic strength affect HPO₄²⁻ more than H₂PO₄⁻? Activity coefficients depend on the square of the ionic charge, so the doubly charged ion is shielded by its ionic atmosphere about four times as strongly, lowering its effective concentration more.

Common misconception

"A phosphate buffer made with a 1:1 ratio always has pH 7.20." At realistic concentrations it is nearer 6.8–6.9, because activity effects lower the apparent pKa. The value 7.20 applies only at infinite dilution.

Worked example

Question: Design 1.00 L of 0.100 mol dm⁻³ phosphate buffer at pH 7.40 using pKa₂ = 7.20, then comment on the result.

Reasoning: r = 10^(7.40 − 7.20) = 10^0.20 = 1.58. n(H₂PO₄⁻) = 0.100/2.58 = 0.0388 mol; n(HPO₄²⁻) = 0.0612 mol. Masses: NaH₂PO₄ = 0.0388 × 119.98 = 4.65 g; Na₂HPO₄ = 0.0612 × 141.96 = 8.69 g. The ionic strength is I = ½(0.1612 × 1 + 0.0388 × 1 + 0.0612 × 4) ≈ 0.22 mol dm⁻³, so pKa′ ≈ 6.8.

Answer: 4.65 g NaH₂PO₄ and 8.69 g Na₂HPO₄ per litre; the measured pH will be near 7.0, so extra base (or a ratio nearer 4 : 1) is needed to reach 7.40.

Quick check

1. Why does a phosphate buffer designed with pKa₂ = 7.20 usually read lower than the target pH on a meter? Answer: At real ionic strengths the apparent pKa is about 6.8, so the actual pH is lower than calculated.

Exam focus

Show the ratio, the split of total concentration and the conversion to masses clearly. Higher-level questions may ask for the ionic-strength calculation or for a qualitative explanation of why the apparent pKa is lower; mention the charge of HPO₄²⁻ explicitly.

Advanced insight

Using the Davies equation at I ≈ 0.22 mol dm⁻³ gives log γ ≈ −0.13 for singly charged ions and ≈ −0.52 for doubly charged ions. The correction log(γ₂/γ₁) ≈ −0.39, taking pKa₂ from 7.20 to about 6.81, in good agreement with laboratory experience.

Summary

Phosphate buffers use the H₂PO₄⁻/HPO₄²⁻ pair near pKa₂ = 7.20. The ratio follows from 10^(pH − pKa), and the total concentration is split accordingly. Because HPO₄²⁻ is doubly charged, ionic strength lowers the apparent pKa to about 6.8 at 0.1 mol dm⁻³. Temperature, dilution, hydrates and precipitation with calcium also matter, so a final pH adjustment is standard practice.

Practice questions

1. Calculate the ideal [HPO₄²⁻]/[H₂PO₄⁻] ratio for pH 6.80 using pKa₂ = 7.20. Answer: 10^(6.80 − 7.20) = 10^−0.40 = 0.40. 2. For 500 cm³ of 0.0500 mol dm⁻³ phosphate at the ratio in question 1, find the moles of each form. Answer: Total 0.0250 mol; H₂PO₄⁻ = 0.0250/1.40 = 0.0179 mol; HPO₄²⁻ = 0.0071 mol. 3. Explain why diluting a concentrated phosphate stock tenfold raises its pH slightly. Answer: Ionic strength falls, activity coefficients rise towards 1 (especially for HPO₄²⁻), and the apparent pKa rises back towards 7.20. 4. Give one situation in which phosphate is a poor choice of buffer. Answer: In solutions containing Ca²⁺ or Mg²⁺, because insoluble calcium or magnesium phosphates can precipitate.