Adjusting an Acetate Buffer for an Enzyme Assay
A teaching laboratory is preparing an enzyme assay that works best at pH 4.70. The technician has two stock solutions: 0.100 M acetic acid and 0.100 M sodium acetate. For acetic acid, pKa = 4.74 at the assay temperature. A student first mixes 50.0 mL of the acid with 50.0 mL of the acetate solution, checks the pH meter, and obtains 4.74. The student then takes a 10.0 mL portion of this mixture, dilutes it to 100.0 mL with distilled water, and adds 1.00 mL of 0.100 M hydrochloric acid. The student expects the dilution to have no effect on pH but is unsure about the acid addition. A second student argues that the buffer is now ruined because the pH will fall to about 3. The technician asks for a quantitative check, and then for a procedure to adjust the original 50:50 mixture to exactly pH 4.70 by adding 0.100 M HCl. The lab has a calibrated pH meter, but the technician wants the volumes predicted before touching the stocks.
1. Calculate the pH after mixing 50.0 mL of each stock and explain whether diluting 10.0 mL to 100.0 mL changes it.
Read the worked answer
Moles of acetic acid: [n_{\mathrm{HA}}=0.100\times 0.0500=0.00500\text{ mol}.] Moles of acetate: [n_{\mathrm{A^-}}=0.100\times 0.0500=0.00500\text{ mol}.] The Henderson–Hasselbalch equation gives [\mathrm{pH}=\mathrm{p}K_a+\log\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}} =4.74+\log(1)=4.74.] This agrees with the meter reading.
Diluting 10.0 mL to 100.0 mL multiplies both concentrations by the same factor, so the ratio ([\mathrm{A^-}]/[\mathrm{HA}]) is unchanged. In the ideal buffer model, the pH therefore remains 4.74. A real meter might read slightly differently because of activity coefficients, but the dilution itself does not change the acid-to-base ratio.
2. Calculate the pH after the 10.0 mL portion is diluted to 100.0 mL and 1.00 mL of 0.100 M HCl is added. Compare with the pH if the same HCl were added to 100.0 mL of pure water.
Read the worked answer
The 10.0 mL portion contains [n_{\mathrm{HA}}=0.100\times 0.0100=0.00100\text{ mol},] [n_{\mathrm{A^-}}=0.100\times 0.0100=0.00100\text{ mol}.] The HCl added is [n_{\mathrm{H^+}}=0.100\times 0.00100=0.000100\text{ mol}.] The strong acid converts acetate to acetic acid: [\mathrm{A^-}+\mathrm{H^+}\rightarrow \mathrm{HA}.] After reaction: [n_{\mathrm{HA}}=0.00100+0.000100=0.00110\text{ mol},] [n_{\mathrm{A^-}}=0.00100-0.000100=0.000900\text{ mol}.] Thus [\mathrm{pH}=4.74+\log\frac{0.000900}{0.00110} =4.74+\log(0.818) =4.74-0.087=4.65.] So the pH falls only from 4.74 to about 4.65.
In 100.0 mL of pure water, the same (0.000100) mol of (\mathrm{H^+}) in about 0.101 L gives [[\mathrm{H^+}]=\frac{0.000100}{0.101}=9.90\times10^{-4}\text{ M},] [\mathrm{pH}=-\log(9.90\times10^{-4})\approx 3.00.] The buffer prevents the large pH drop that would occur in pure water.
3. How many mL of 0.100 M HCl must be added to the original 100.0 mL 50:50 mixture to reach pH 4.70? State any assumption.
Read the worked answer
Use the target pH to find the required mole ratio: [4.70=4.74+\log\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}},] [\log\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}=-0.04,] [\frac{n_{\mathrm{A^-}}}{n_{\mathrm{HA}}}=10^{-0.04}=0.912.] The total buffer amount is still (0.0100) mol because adding HCl converts acetate to acetic acid but does not change the sum: [n_{\mathrm{HA}}+n_{\mathrm{A^-}}=0.0100.] Let (n_{\mathrm{A^-}}=0.912n_{\mathrm{HA}}). Then [n_{\mathrm{HA}}(1+0.912)=0.0100,] [n_{\mathrm{HA}}=0.00523\text{ mol},\qquad n_{\mathrm{A^-}}=0.00477\text{ mol}.] Initially there were 0.00500 mol of each, so the HCl must convert [0.00500-0.00477=0.00023\text{ mol}] of acetate into acetic acid. The required volume of 0.100 M HCl is [V=\frac{0.00023}{0.100}=0.0023\text{ L}=2.3\text{ mL}.] Assumptions: HCl reacts completely, the volumes are additive, and activity effects are negligible. The prediction should be checked with the pH meter.