Question

Phosphoric acid is a triprotic acid (Ka₁=6.9×10⁻³ M, Ka₂=6.2×10⁻⁸ M, and Ka₃=4.8×10⁻¹³ M)

To find the pH of a buffer composed of H₂PO⁻⁴(aq) and HPO₂⁻⁴(aq), which pKa value would you use in the Henderson–Hasselbalch equation?
pKa₁=2.16
pKa₂=7.21
pKa₃=12.32
Calculate the pH of a buffer solution obtained by dissolving 29.029.0 g KH₂PO₄(s and 46.0 g Na₂HPO₄(s) in water and then diluting to 1.00 L.

Answer 1

The pH of the buffer solution composed of H2PO4- and HPO42- is calculated using the pKa2 value of 7.21 and the Henderson-Hasselbalch equation, resulting in a pH of 7.392.

Step-by-step explanation:

To find the pH of a buffer composed of H2PO4-(aq) and HPO42-(aq), you would use the pKa value that corresponds to the deprotonation step between these two species. This is pKa2 which is 7.21. The buffer solution has been prepared by mixing KH2PO4 and Na2HPO4, which correspond to the acid and its conjugate base, respectively. Therefore, the Henderson-Hasselbalch equation is applicable:

pH = pKa + log([A-]/[HA])

To calculate the pH of the buffer solution, first determine the number of moles of each compound using their molar masses (KH2PO4: 136.09 g/mol, Na2HPO4: 141.96 g/mol). For 29.0 g of KH2PO4, we have 29.0 g / 136.09 g/mol = 0.213 mol of KH2PO4 (acid, HA). For 46.0 g of Na2HPO4, we have 46.0 g / 141.96 g/mol = 0.324 mol of Na2HPO4 (base, A-). The total volume of the solution is 1.00 L. Plugging the numbers into the Henderson-Hasselbalch equation gives:

pH = 7.21 + log(0.324/0.213) = 7.21 + log(1.521) = 7.21 + 0.182 = 7.392

Therefore, the calculated pH of the buffer solution is 7.392.