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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

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(NH4)2SO4=4.63 g(NH_4)_2SO_4 = \text{4.63 g}

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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 (NH3NH_3) as the weak base and ammonium ion (NH4+NH_4^+) as its conjugate acid, which comes from ammonium sulfate ((NH4)2SO4(NH_4)_2SO_4).

Given:

  • Target pH = 9.45
  • Volume of NH3NH_3 solution = 425 mL = 0.425 L
  • Concentration of NH3NH_3 = 0.258 M
  • The base dissociation constant (KbK_b) for NH3NH_3 is 1.8×10−51.8 \times 10^{-5}.
  • Molar mass of (NH4)2SO4(NH_4)_2SO_4:
    • N: 2×14.01=28.022 \times 14.01 = 28.02 g/mol
    • H: 8×1.008=8.0648 \times 1.008 = 8.064 g/mol
    • S: 1×32.07=32.071 \times 32.07 = 32.07 g/mol
    • O: 4×16.00=64.004 \times 16.00 = 64.00 g/mol
    • Total molar mass = 28.02+8.064+32.07+64.00=132.15428.02 + 8.064 + 32.07 + 64.00 = 132.154 g/mol
  1. Calculate the pOHpOH from the given pHpH.

    pOH=14−pHpOH = 14 - pH pOH=14−9.45pOH = 14 - 9.45 pOH=4.55pOH = 4.55
  2. Calculate the pKbpK_b for ammonia (NH3NH_3).

    pKb=−log⁡(Kb)pK_b = -\log(K_b) pKb=−log⁡(1.8×10−5)pK_b = -\log(1.8 \times 10^{-5}) pKb≈4.7447pK_b \approx 4.7447
  3. Use the Henderson-Hasselbalch equation to find the required concentration of the conjugate acid ([NH4+][NH_4^+]).

    The Henderson-Hasselbalch equation for a basic buffer is:

    pOH=pKb+log⁡([conjugate acid][weak base])pOH = pK_b + \log\left(\frac{[\text{conjugate acid}]}{[\text{weak base}]}\right) pOH=pKb+log⁡([NH4+][NH3])pOH = pK_b + \log\left(\frac{[NH_4^+]}{[NH_3]}\right)

    Substitute the known values:

    4.55=4.7447+log⁡([NH4+]0.258 M)4.55 = 4.7447 + \log\left(\frac{[NH_4^+]}{0.258 \text{ M}}\right)

    Rearrange to solve for the logarithm term:

    log⁡([NH4+]0.258 M)=4.55−4.7447\log\left(\frac{[NH_4^+]}{0.258 \text{ M}}\right) = 4.55 - 4.7447 log⁡([NH4+]0.258 M)=−0.1947\log\left(\frac{[NH_4^+]}{0.258 \text{ M}}\right) = -0.1947

    Take the antilog (10 to the power of both sides):

    [NH4+]0.258 M=10−0.1947\frac{[NH_4^+]}{0.258 \text{ M}} = 10^{-0.1947} [NH4+]0.258 M≈0.6387\frac{[NH_4^+]}{0.258 \text{ M}} \approx 0.6387

    Solve for [NH4+][NH_4^+]:

    [NH4+]=0.6387×0.258 M[NH_4^+] = 0.6387 \times 0.258 \text{ M} [NH4+]≈0.1648 M[NH_4^+] \approx 0.1648 \text{ M}
  4. Calculate the total moles of NH4+NH_4^+ needed.

    Moles of NH4+=Concentration×Volume\text{Moles of } NH_4^+ = \text{Concentration} \times \text{Volume} Moles of NH4+=0.1648 mol/L×0.425 L\text{Moles of } NH_4^+ = 0.1648 \text{ mol/L} \times 0.425 \text{ L} Moles of NH4+≈0.07004 mol\text{Moles of } NH_4^+ \approx 0.07004 \text{ mol}
  5. Determine the moles of (NH4)2SO4(NH_4)_2SO_4 required.

    Ammonium sulfate, (NH4)2SO4(NH_4)_2SO_4, dissociates to produce two NH4+NH_4^+ ions for every one formula unit:

    (NH4)2SO4(s)→2NH4+(aq)+SO42−(aq)(NH_4)_2SO_4(s) \rightarrow 2NH_4^+(aq) + SO_4^{2-}(aq)

    Therefore, the moles of (NH4)2SO4(NH_4)_2SO_4 needed are half the moles of NH4+NH_4^+:

    Moles of (NH4)2SO4=Moles of NH4+2\text{Moles of } (NH_4)_2SO_4 = \frac{\text{Moles of } NH_4^+}{2} Moles of (NH4)2SO4=0.07004 mol2\text{Moles of } (NH_4)_2SO_4 = \frac{0.07004 \text{ mol}}{2} Moles of (NH4)2SO4≈0.03502 mol\text{Moles of } (NH_4)_2SO_4 \approx 0.03502 \text{ mol}
  6. Calculate the mass of (NH4)2SO4(NH_4)_2SO_4 needed.

    Mass of (NH4)2SO4=Moles×Molar Mass\text{Mass of } (NH_4)_2SO_4 = \text{Moles} \times \text{Molar Mass} Mass of (NH4)2SO4=0.03502 mol×132.154 g/mol\text{Mass of } (NH_4)_2SO_4 = 0.03502 \text{ mol} \times 132.154 \text{ g/mol} Mass of (NH4)2SO4≈4.628 g\text{Mass of } (NH_4)_2SO_4 \approx 4.628 \text{ g}

    Rounding to three significant figures (consistent with 0.258 M and 425 mL):

    Mass of (NH4)2SO4=4.63 g\text{Mass of } (NH_4)_2SO_4 = \boxed{\text{4.63 g}}
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