Q.42. The second pKa of phosphoric acid is 6.8. The ratio of 𝐍𝐚𝟐𝐇𝐏𝐎𝟒 to 𝐍𝐚𝐇𝟐𝐏𝐎𝟒 required to obtain
a buffer of pH 7.0 is ____ (rounded off to two decimal places).

Phosphate buffers are essential in biochemistry and exams due to phosphoric acid’s multiple pKa values. The second pKa of 6.8 corresponds to the H₂PO₄^– / HPO₄^2- equilibrium, ideal for pH near 7. For a pH 7.0 buffer, the ratio of Na₂HPO₄ (base, [A^–]) to NaH₂PO₄ (acid, [HA]) is 1.58.

Henderson-Hasselbalch Calculation

The Henderson-Hasselbalch equation is pH = pK_a + log₁₀ ( [A^–] / [HA] ). Rearranging gives [Na₂HPO₄] / [NaH₂PO₄] = 10^{(pH – pK_a)}.

Substitute pH 7.0 and pKa 6.8: 10^{7.0 – 6.8} = 10^0.2 ≈ 1.5849, rounded to 1.58.This means 1.58 moles of Na₂HPO₄ per mole of NaH₂PO₄ yields pH 7.0.

Why Second pKa for This Buffer

Phosphoric acid (H₃PO₄) has pKa1 ≈ 2.1, pKa2 = 6.8, pKa3 ≈ 12.4; the second controls buffers around pH 6-8. NaH₂PO₄ provides H₂PO₄^– (acid), Na₂HPO₄ provides HPO₄^2- (conjugate base).Effective buffering occurs within ±1 of pKa, so pH 7.0 fits perfectly.

Common Exam Mistakes Explained

• Wrong ratio direction: Question asks Na₂HPO₄ / NaH₂PO₄ ([A^–]/[HA]), not inverse (0.63); pH > pKa means base > acid.
• pKa confusion: Using pKa1 (2.1) gives absurd ratio; always match ionization step.
• Rounding error: 10^0.2 = 1.58489319246, rounds to 1.58 (two decimals).
• No options listed: Fill-in assumes direct calc; verifies via log check: log(1.58) ≈ 0.20, pH = 6.8 + 0.2 = 7.0.

Practical Applications

Phosphate buffers mimic physiological pH 7.4 in labs and cells. For 1 L 0.1 M buffer, mix ~38.7 g NaH₂PO₄ and 61.3 g Na₂HPO₄ (MW 120/142 g/mol).Adjust volumes for equal concentration stocks using same ratio.

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