To prepare a buffer solution with a specific pH using a weak base and its conjugate acid, we use the Henderson-Hasselbalch equation for bases. The buffer system consists of ammonia (NH3) as the wea

To prepare a buffer solution with a specific pH using a weak base and its conjugate acid, we use the Henderson-Hasselbalch equation for bases. The buffer system consists of ammonia (NH_3) as the weak base and ammonium ion (NH_4^+) as its conjugate acid, which comes from ammonium sulfate ((NH_4)_2SO_4). Given: Target pH = 9.45 Volume of NH_3 solution = 425 mL = 0.425 L Concentration of NH_3 = 0.258 M The base dissociation constant (K_b) for NH_3 is 1.8 × 10^-5. Molar mass of (NH_4)_2SO_4: N: 2 × 14.01 = 28.02 g/mol H: 8 × 1.008 = 8.064 g/mol S: 1 × 32.07 = 32.07 g/mol O: 4 × 16.00 = 64.00 g/mol Total molar mass = 28.02 + 8.064 + 32.07 + 64.00 = 132.154 g/mol Step 1: Calculate the pOH from the given pH. pOH = 14 - pH pOH = 14 - 9.45 pOH = 4.55 Step 2: Calculate the pK_b for ammonia (NH_3). pK_b = -(K_b) pK_b = -(1.8 × 10^-5) pK_b ≈ 4.7447 Step 3: Use the Henderson-Hasselbalch equation to find the required concentration of the conjugate acid ([NH_4^+]). The Henderson-Hasselbalch equation for a basic buffer is: pOH = pK_b + ([conjugate acid][weak base]) pOH = pK_b + (([NH_4^+])/([NH_3])) Substitute the known values: 4.55 = 4.7447 + (([NH_4^+])/(0.258 M)) Rearrange to solve for the logarithm term: (([NH_4^+])/(0.258 M)) = 4.55 - 4.7447 (([NH_4^+])/(0.258 M)) = -0.1947 Take the antilog (10 to the power of both sides): ([NH_4^+])/(0.258 M) = 10^-0.1947 ([NH_4^+])/(0.258 M) ≈ 0.6387 Solve for [NH_4^+]: [NH_4^+] = 0.6387 × 0.258 M [NH_4^+] ≈ 0.1648 M Step 4: Calculate the total moles of NH_4^+ needed. Moles of NH_4^+ = Concentration × Volume Moles of NH_4^+ = 0.1648 mol/L × 0.425 L Moles of NH_4^+ ≈ 0.07004 mol Step 5: Determine the moles of (NH_4)_2SO_4 required. Ammonium sulfate, (NH_4)_2SO_4, dissociates to produce two NH_4^+ ions for every one formula unit: (NH_4)_2SO_4(s) → 2NH_4^+(aq) + SO_4^2-(aq) Therefore, the moles of (NH_4)_2SO_4 needed are half the moles of NH_4^+: Moles of (NH_4)_2SO_4 = Moles of NH_4^+2 Moles of (NH_4)_2SO_4 = 0.07004 mol2 Moles of (NH_4)_2SO_4 ≈ 0.03502 mol Step 6: Calculate the mass of (NH_4)_2SO_4 needed. Mass of (NH_4)_2SO_4 = Moles × Molar Mass Mass of (NH_4)_2SO_4 = 0.03502 mol × 132.154 g/mol Mass of (NH_4)_2SO_4 ≈ 4.628 g Rounding to three significant figures (consistent with 0.258 M and 425 mL): Mass of (NH_4)_2SO_4 = 4.63 g Drop the next question.