Adjusting an Acetate Buffer While Keeping Fe2+ Available
A biochemistry student needs 100.0 mL of 0.100 M acetate buffer at pH 5.00 for an enzyme assay. The stock solutions are 0.100 M acetic acid (Ka = 1.8 × 10−5) and 0.100 M sodium acetate. She mixes 50.0 mL of each and measures pH 4.74. To raise the pH, she plans to add 1.00 M NaOH. The assay also requires 0.010 M FeSO4 as an activator. At the working ionic strength, Fe2+ begins to precipitate as Fe(OH)2 when [OH−] exceeds about 1.0 × 10−6 M. The student wonders whether she should adjust the buffer first and then add FeSO4, or add FeSO4 before pH adjustment. A colleague suggests using a citrate buffer instead because citrate binds Fe2+ strongly and might prevent precipitation. The student has no pH meter after the initial reading and must rely on calculation.
1. Calculate the volume of 1.00 M NaOH needed to bring the 100.0 mL mixture from pH 4.74 to pH 5.00 using the Henderson–Hasselbalch equation. State the assumptions and comment on whether the buffer capacity is adequate.
Read the worked answer
pKa = −log(1.8 × 10−5) = 4.74. Initially the mixture has equal moles of acetic acid and acetate: 0.100 M × 0.0500 L = 5.00 × 10−3 mol each, which is why the measured pH equals pKa.
At pH 5.00: 5.00 = 4.74 + log([A−]/[HA]) log([A−]/[HA]) = 0.26 [A−]/[HA] = 10^0.26 = 1.82
Let x = moles of NaOH added. NaOH converts HA to A−: [A−] = (5.00 × 10−3 + x) / V [HA] = (5.00 × 10−3 − x) / V
The volume V cancels: (5.00 × 10−3 + x) / (5.00 × 10−3 − x) = 1.82 5.00 × 10−3 + x = 9.10 × 10−3 − 1.82x 2.82x = 4.10 × 10−3 x = 1.45 × 10−3 mol
Volume of 1.00 M NaOH = 1.45 × 10−3 L = 1.45 mL.
Assumptions: volumes are additive, activity coefficients are ignored, and no other acid–base equilibria interfere. After addition, HA ≈ 3.55 × 10−3 mol and A− ≈ 6.45 × 10−3 mol, so both buffer components remain present in comparable amounts and the buffer capacity is adequate for the planned assay.
2. Decide whether FeSO4 should be added before or after the pH adjustment. Use the hydroxide precipitation criterion and also evaluate the citrate suggestion.
Read the worked answer
At the final pH 5.00, [H+] = 1.0 × 10−5 M, so [OH−] = Kw/[H+] = 1.0 × 10−14 / 1.0 × 10−5 = 1.0 × 10−9 M. This is far below the stated threshold of 1.0 × 10−6 M, so Fe(OH)2 is not favoured at equilibrium in the final buffer. Adding FeSO4 after the pH adjustment is therefore safer.
If FeSO4 is added before the NaOH adjustment, local droplets of 1.00 M NaOH produce temporarily high pH near the point of addition. There [OH−] can exceed 1.0 × 10−6 M, causing transient Fe(OH)2 precipitation and possible air oxidation of Fe2+ to Fe3+. The precipitate may not redissolve quickly, so the free Fe2+ concentration can be lost even though the final bulk pH is acceptable. Add the FeSO4 after the buffer is adjusted, with gentle stirring and without local excess of base.
The citrate suggestion is chemically reasonable for pH control because citrate has a pKa2 near 4.76, close to the desired pH. However, citrate chelates Fe2+ and especially Fe3+ strongly. If the enzyme requires free Fe2+ as an activator, citrate can reduce the available metal ion and inhibit the assay. Acetate is a weaker complexing agent for Fe2+. Recommendation: keep the acetate buffer, adjust pH first, then add FeSO4; if citrate must be used, increase the Fe2+ concentration or use a non-complexing buffer and verify free Fe2+ independently.
3. The student cannot recheck pH after adding FeSO4. What calculation or observation should be recorded to show that the buffer was still at pH 5.00 within about ±0.1?
Read the worked answer
Record the calculated amounts and the expected ratio after FeSO4 addition. The buffer ratio depends on moles of HA and A−; adding 0.010 M FeSO4 does not consume either directly, except for possible weak complexation and dilution. For a ±0.1 pH tolerance around 5.00, the acceptable ratio is:
pH = 4.74 + log(r) For pH 4.90: log r = 0.16, r = 1.45 For pH 5.10: log r = 0.36, r = 2.29
After NaOH addition the calculated ratio is about 1.82, inside the range 1.45–2.29. The record should state: initial moles HA = A− = 5.00 × 10−3 mol; NaOH added = 1.45 × 10−3 mol; final calculated HA = 3.55 × 10−3 mol and A− = 6.45 × 10−3 mol; ratio = 1.82; expected pH = 4.74 + log(1.82) = 5.00. A visual indicator such as bromocresol green can provide a rough check, but the defensible evidence is the calculated ratio plus the measured initial pH.