A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of 1.0 M NaH₂PO₄. How many moles of HCl must be added to this buffer solution to change the pH by 0.18 units? A) 0.025 mol HCl B) 0.063 mol HCl C) 0.082 mol HCl D) 0.50 mol HCl E) 1.0 mol HCl

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The correct answer is A.The Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution:pH = pKa + log([A-]/[HA])where pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.In this case, the weak acid is NaH2PO4 and the conjugate base is Na2HPO4. The pKa of NaH2PO4 is 7.21.The initial concentration of NaH2PO4 is 1.0 M and the initial concentration of Na2HPO4 is 0.0 M. Therefore, the initial pH of the buffer solution is:pH = 7.21 + log(0.0/1.0) = 7.21To change the pH by 0.18 units, the new pH must be 7.21 + 0.18 = 7.39.Using the Henderson-Hasselbalch equation, we can solve for the new concentration of Na2HPO4:7.39 = 7.21 + log([A-]/[HA])0.18 = log([A-]/[HA])[A-]/[HA] = 10^0.18 = 1.51Therefore, the new concentration of Na2HPO4 is 1.51 times the new concentration of NaH2PO4.The total volume of the buffer solution is 150 mL + 250 mL = 400 mL. Therefore, the total number of moles of NaH2PO4 and Na2HPO4 in the buffer solution is:(1.0 M)(0.250 L) + (1.51)(1.0 M)(0.250 L) = 0.378 molTo change the pH by 0.18 units, we need to add enough HCl to convert 0.378 mol of Na2HPO4 to NaH2PO4. The balanced chemical equation for this reaction is:Na2HPO4 + HCl → NaH2PO4 + NaClTherefore, we need to add 0.378 mol of HCl to the buffer solution to change the pH by 0.18 units.

A Buffer Is Prepared by Adding 150 ML of 1