What Would Be The Expected PH Shift In A Buffer Solution Containing 0.5 M Acetic Acid And 0.25 M Sodium Acetate When 0.05 Moles Of Hydrochloric Acid Are Added To 1 Liter Of The Solution, And How Would This Shift Be Affected If The Buffer Solution Were Diluted By A Factor Of 5 Prior To The Acid Addition?

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To determine the expected pH shift in a buffer solution containing 0.5 M acetic acid and 0.25 M sodium acetate when 0.05 moles of hydrochloric acid are added, we use the Henderson-Hasselbalch equation:

pH=pKa+log⁡([A−][HA]){ \text{pH} = \text{pKa} + \log\left(\frac{[A^-]}{[HA]}\right) }

  1. Initial pH Calculation:

    • Concentrations: 0.5 M acetic acid (HA) and 0.25 M sodium acetate (A-)
    • pKa of acetic acid = 4.76
    • Initial pH: pH=4.76+log⁡(0.250.5)=4.76+log⁡(0.5)≈4.76−0.301=4.459{ \text{pH} = 4.76 + \log\left(\frac{0.25}{0.5}\right) = 4.76 + \log(0.5) \approx 4.76 - 0.301 = 4.459 }

  2. After Adding 0.05 Moles of HCl:

    • Moles of HA: 0.5 moles
    • Moles of A-: 0.25 moles
    • Reaction: H+ + A- → HA
    • New moles:
      • A-: 0.25 - 0.05 = 0.20 moles
      • HA: 0.5 + 0.05 = 0.55 moles
    • Concentrations in 1 liter:
      • [A-] = 0.20 M
      • [HA] = 0.55 M
    • New pH: pH=4.76+log⁡(0.200.55)≈4.76+log⁡(0.3636)≈4.76−0.438=4.322{ \text{pH} = 4.76 + \log\left(\frac{0.20}{0.55}\right) \approx 4.76 + \log(0.3636) \approx 4.76 - 0.438 = 4.322 }
    • pH shift: 4.459−4.322≈0.14{ 4.459 - 4.322 \approx 0.14 }

  3. Effect of Dilution:

    • Diluting the buffer by a factor of 5:
      • Concentrations: 0.1 M HA and 0.05 M A- in 5 liters
      • Initial pH remains the same: 4.46
    • Adding 0.05 moles of HCl:
      • Moles of HA: 0.5 moles
      • Moles of A-: 0.25 moles
      • New moles after reaction:
        • A-: 0.25 - 0.05 = 0.20 moles
        • HA: 0.5 + 0.05 = 0.55 moles
      • Concentrations in 5 liters:
        • [A-] = 0.04 M
        • [HA] = 0.11 M
      • New pH: pH=4.76+log⁡(0.040.11)≈4.76+log⁡(0.3636)≈4.76−0.438=4.322{ \text{pH} = 4.76 + \log\left(\frac{0.04}{0.11}\right) \approx 4.76 + \log(0.3636) \approx 4.76 - 0.438 = 4.322 }
      • pH shift: 4.459−4.322≈0.14{ 4.459 - 4.322 \approx 0.14 }

Thus, the expected pH shift is approximately \boxed{0.14}, and diluting the buffer by a factor of 5 does not change the magnitude of this shift.