Buffer Preparation in a Crowded Lab
A student needs to prepare 500.0 mL of a buffer at pH 7.40 using 0.100 M KH2PO4 and 0.100 M K2HPO4 solutions. The lab has only 250.0 mL of 0.100 M KH2PO4 and 150.0 mL of 0.100 M K2HPO4 available. The pKa2 of phosphoric acid is 7.20 at 25 °C. The student also has solid K2HPO4 (molar mass 174.18 g/mol) and solid KH2PO4 (molar mass 136.09 g/mol). The student must decide whether to use the solutions as is, or to adjust the amounts by adding solid. The student also needs to consider that the buffer capacity should be at least 0.020 mol/L per pH unit. Assume volumes are additive and the temperature is 25 °C.
1. Calculate the volumes of the two solutions needed to make 500.0 mL of pH 7.40 buffer if only the solutions are used, and determine whether the available volumes are sufficient.
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Using the Henderson–Hasselbalch equation: pH = pKa + log([HPO4^2−]/[H2PO4^−]). 7.40 = 7.20 + log([HPO4^2−]/[H2PO4^−]), so log(ratio) = 0.20, ratio = 10^0.20 = 1.585. Let V1 be the volume of KH2PO4 solution and V2 be the volume of K2HPO4 solution. Since both concentrations are 0.100 M, the ratio of moles equals the ratio of volumes: V2/V1 = 1.585. Also V1 + V2 = 500.0 mL. Solving: V1 + 1.585 V1 = 500.0 → 2.585 V1 = 500.0 → V1 = 193.4 mL, V2 = 306.6 mL. The available volumes are 250.0 mL of KH2PO4 (enough) and 150.0 mL of K2HPO4 (not enough; need 306.6 mL). Therefore, the solutions alone are insufficient.
2. Determine how much solid K2HPO4 must be added to the available 150.0 mL of K2HPO4 solution (and using the maximum available KH2PO4 solution) to achieve the desired buffer, while keeping the total volume at 500.0 mL. Specify the volumes of solutions and mass of solid.
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Use all 250.0 mL of 0.100 M KH2PO4 (0.02500 mol H2PO4^−). To achieve pH 7.40, the required moles of HPO4^2− = 1.585 × 0.02500 = 0.03963 mol. The available K2HPO4 solution provides 0.1500 L × 0.100 M = 0.01500 mol. Additional moles needed = 0.03963 − 0.01500 = 0.02463 mol. Mass of solid K2HPO4 = 0.02463 mol × 174.18 g/mol = 4.290 g. The total volume from solutions is 250.0 + 150.0 = 400.0 mL. To make 500.0 mL, add water to the mark. The solid dissolves in the buffer. The final concentrations: [H2PO4^−] = 0.02500 mol / 0.5000 L = 0.0500 M; [HPO4^2−] = 0.03963 mol / 0.5000 L = 0.07926 M. Check ratio: 0.07926/0.0500 = 1.585, pH = 7.20 + log(1.585) = 7.40. Buffer capacity: total phosphate concentration = 0.0500 + 0.07926 = 0.1293 M. Buffer capacity β ≈ 2.303 × (Ka × [H+])/(Ka + [H+])^2 × C_total? More simply, β ≈ 2.303 × C_total × (Ka[H+]/(Ka+[H+])^2). At pH = pKa, β = 2.303 × C_total / 4 = 0.5766 × C_total. Here pH is 0.20 above pKa, so [H+]/Ka = 10^-0.20 = 0.631. Then Ka[H+]/(Ka+[H+])^2 = (0.631)/(1.631)^2 = 0.237. β = 2.303 × 0.1293 × 0.237 = 0.0706 mol/L per pH unit, which exceeds 0.020. So the buffer capacity requirement is met.