You need 1.00 L of 0.05 M ammonium bicarbonate buffer pH 6.8. Describe the steps to prepare this buffer including the amount of solid ammonium bicarbonate you need. The pKa values for ammonium bicarbonate are 6.35 and 9.25.
The Correct Answer and Explanation is:
To prepare 1.00 L of 0.05 M ammonium bicarbonate buffer at pH 6.80, follow these steps:
🧪 Step‑by‑Step Preparation
- Calculate total moles of ammonium bicarbonate needed
- Desired concentration: 0.05 mol/L → need 0.05 mol in 1 L.
- Molar mass of NH₄HCO₃ ≈ 79.06 g/mol → 0.05×79.06=3.953 g 0.05 \times 79.06 = 3.953\;g0.05×79.06=3.953g
- Weigh out ≈ 3.95 g ammonium bicarbonate.
- Dissolve and adjust volume
- Dissolve the 3.95 g NH₄HCO₃ in ~800 mL deionized water.
- Once fully dissolved, transfer to a 1 L volumetric flask and make up to the mark with water.
- Adjust pH to 6.80
- The effective buffer system is bicarbonate (HCO3−\mathrm{HCO_3^-}HCO3−) / carbonic acid (H2CO3\mathrm{H_2CO_3}H2CO3), with pKa ≈ 6.35.
- Using the Henderson–Hasselbalch equation: pH=pKa+log[HCO3−][H2CO3] \text{pH} = \text{p}K_a + \log\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]}pH=pKa+log[H2CO3][HCO3−] 6.80=6.35+log[HCO3−][H2CO3]⇒[HCO3−][H2CO3]=100.45≈2.82 6.80 = 6.35 + \log\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]} \quad\Rightarrow\quad \frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]} = 10^{0.45} \approx 2.826.80=6.35+log[H2CO3][HCO3−]⇒[H2CO3][HCO3−]=100.45≈2.82
- In your solution, most bicarbonate comes from the solid; to reach this ratio, slowly add dilute HCl (e.g., 0.1 M) until the pH meter reads 6.80.
- Stir thoroughly and let equilibrate.
- Final volume check
- After pH adjustment, recheck the volume and top up to 1 L if necessary.
- Re‑check the pH and fine‑tune with tiny increments of HCl or NaOH if needed.
- Label and store
- Label as “0.05 M NH₄HCO₃ buffer, pH 6.80.”
- Store at 4 °C if needed, though bicarbonate buffers are best used fresh due to CO₂ loss.
🧬 Explanation
Ammonium bicarbonate (NH₄HCO₃) dissociates into ammonium (NH4+\mathrm{NH_4^+}NH4+) and bicarbonate (HCO3−\mathrm{HCO_3^-}HCO3−). Its two relevant pKa values are 6.35 (carbonic acid ⇌ bicarbonate) and 9.25 (ammonium ⇌ ammonia). At pH 6.8—which is near 6.35—the primary buffering action comes from the bicarbonate/carbonic acid pair, not ammonium/ammonia. So we focus on the first pKa.
The buffer’s total molarity (0.05 M) sets the sum [HCO3−]+[H2CO3]=0.05 M[\mathrm{HCO_3^-}] + [\mathrm{H_2CO_3}] = 0.05\;M[HCO3−]+[H2CO3]=0.05M. The Henderson–Hasselbalch equation relates pH to the acid/base ratio:6.80=6.35+log([HCO3−][H2CO3])6.80 = 6.35 + \log\left(\frac{[\mathrm{HCO_3^-}]}{[\mathrm{H_2CO_3}]}\right)6.80=6.35+log([H2CO3][HCO3−])
Solving gives a ratio [HCO3−]:[H2CO3]≈2.82[\mathrm{HCO_3^-}]:[\mathrm{H_2CO_3}] \approx 2.82[HCO3−]:[H2CO3]≈2.82. Let x = [H2CO3][\mathrm{H_2CO_3}][H2CO3], then [HCO3−]=2.82x[\mathrm{HCO_3^-}] = 2.82x[HCO3−]=2.82x, and x+2.82x=0.05x + 2.82x = 0.05x+2.82x=0.05. This yields x≈0.0131 Mx ≈ 0.0131\;Mx≈0.0131M and bicarbonate ≈ 0.0369 M.
By dissolving 0.05 mol NH₄HCO₃ (3.95 g) in 1 L, you introduce 0.05 M bicarbonate, which is higher than the target 0.0369 M. To adjust, you must shift some bicarbonate into the acid form (H2CO3\mathrm{H_2CO_3}H2CO3) by adding a small amount of strong acid (HCl), thus lowering [HCO3−][\mathrm{HCO_3^-}][HCO3−] and raising [H2CO3][\mathrm{H_2CO_3}][H2CO3] until the correct ratio is reached.
Using a calibrated pH meter, you titrate until the pH stabilizes at 6.80. Because carbonic species can lose CO₂ over time, it’s best to seal the buffer or prepare fresh each time. This approach ensures an accurate 0.05 M buffer at pH 6.8 with minimal error.
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