A buffer is prepared by combining 10.0 g of NaH₂PO₄ with 150 mL of 0.20 M NaOH and diluting to 2.0 L.What is the pH of the buffer? (K_a (H₂PO₄−)= 6.2 × 10−⁸)

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To find the pH of the buffer, we use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the base form and [HA] is the concentration of the acid form. First, calculate the moles of NaH2PO4 and NaOH. The molar mass of NaH2PO4 is approximately 120 g/mol, so 10.0 g corresponds to 10.0 g / 120 g/mol = 0.0833 mol. The moles of NaOH are 150 mL * 0.20 mol/L = 0.030 mol. When NaOH is added to NaH2PO4, it converts some of the NaH2PO4 to its conjugate base, Na2HPO4, by the reaction: NaH2PO4 + NaOH -> Na2HPO4 + H2O. This reaction consumes NaOH and NaH2PO4 in a 1:1 ratio, so the final moles of NaH2PO4 are 0.0833 mol - 0.030 mol = 0.0533 mol, and the moles of Na2HPO4 (the base form) are equal to the moles of NaOH added, 0.030 mol. The total volume of the solution is 2.0 L, so the concentrations are [NaH2PO4] = 0.0533 mol / 2.0 L = 0.02665 M and [Na2HPO4] = 0.030 mol / 2.0 L = 0.015 M. The pKa is -log(6.2 × 10^-8) ≈ 7.21. Applying the Henderson-Hasselbalch equation: pH = 7.21 + log(0.015 / 0.02665) ≈ 7.21 - 0.25 ≈ 6.96.

A Buffer Is Prepared by Combining 10