Final answer:
To find the pH of the buffer using the Henderson-Hasselbalch equation, the ratio [K2HPO4]/[KH2PO4] is calculated for a given pH of 7.45 and the pKa of phosphoric acid's second dissociation (approximately 7.2). The calculated ratio comes out to approximately 1.78, although this is not one of the provided answers, suggesting a discrepancy in the question.
Step-by-step explanation:
To determine the pH of the buffer, we use the Henderson-Hasselbalch equation:
pH = pKa + log([A¯]/[HA])
Phosphoric acid (H₃PO₄) is a tri-protic acid, meaning it can donate three protons or hydrogen ions. The second dissociation constant (K₂) for phosphoric acid is applicable here since we're dealing with the conjugate base pairs HPO₄2- and H₂PO₄¯. The pKa value associated with this second dissociation is about 7.2.
Given this, the Henderson-Hasselbalch equation for the desired pH of 7.45 is:
7.45 = 7.2 + log([K₂HPO₄]/[KH₂PO₄])
Solving for the ratio [K₂HPO₄]/[KH₂PO₄]:
log([K₂HPO₄]/[KH₂PO₄]) = 7.45 - 7.2
log([K₂HPO₄]/[KH₂PO₄]) = 0.25
[K₂HPO₄]/[KH₂PO₄] = 10^0.25
[K₂HPO₄]/[KH₂PO₄] ≈ 1.78
This ratio is not an option given in the original set of answers, suggesting a mistake in the question or a possible misunderstanding. However, the logic applied to solve the problem is sound using the Henderson-Hasselbalch equation.