Preparation of a Carbonate Buffer at pH 6.20 Using Carbonate Stock and Hydrochloric Acid
Introduction
Buffer preparation often requires careful consideration of acid–base equilibria, especially for polyprotic systems such as carbonate. In this example, a carbonate buffer is prepared at pH 6.20 using a carbonate stock solution initially at pH 11.30 and concentrated hydrochloric acid. Because carbonate chemistry involves two relevant dissociation equilibria, both acid dissociation constants (\(pK_a\)) must be incorporated into the calculation.
The objective is to prepare:
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500 mL of 0.20 M carbonate buffer
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Target pH: 6.20
Available reagents:
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1.0 M carbonate stock solution, pH 11.30
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2.0 M HCl
Step 1 — Determine Total Carbonate Required
The total number of moles of carbonate species needed is determined from the desired concentration and final volume.
\[n = MV\] \[n = (0.20\ \mathrm{mol,L^{-1}})(0.500\ \mathrm{L})\] \[n = 0.100\ \mathrm{mol}\]Thus, the final buffer must contain:
\[\boxed{0.100\ \mathrm{mol\ total\ carbonate}}\]Step 2 — Calculate Volume of Carbonate Stock Solution
The carbonate stock concentration is 1.0 M, so the required volume is:
\[V = \frac{n}{M}\] \[V = \frac{0.100}{1.0}\] \[V = 0.100\ \mathrm{L}\] \[\boxed{100\ \mathrm{mL\ carbonate\ stock}}\]Step 3 — Determine the Desired Buffer Composition at pH 6.20
At pH 6.20, the relevant carbonate equilibrium is:
\[\mathrm{H_2CO_3 \rightleftharpoons H^+ + HCO_3^-}\]with:
\[pK_{a1} = 6.35\]Using the Henderson–Hasselbalch equation:
\[pH = pK_a + \log\left(\frac{\mathrm{base}}{\mathrm{acid}}\right)\]Substituting values:
\[6.20 = 6.35 + \log\left(\frac{\mathrm{HCO_3^-}}{\mathrm{H_2CO_3}}\right)\]Solving for the ratio:
\[\frac{\mathrm{HCO_3^-}}{\mathrm{H_2CO_3}} = 10^{-0.15} = 0.708\]The total carbonate concentration equals the sum of acid and base forms:
\[\mathrm{acid + base} = 0.100\]Let the amount of carbonic acid be $x$:
\[x + 0.708x = 0.100\] \[1.708x = 0.100\] \[x = 0.0585\ \mathrm{mol}\]Therefore:
\[\boxed{0.0585\ \mathrm{mol\ H_2CO_3}}\]and
\[0.708(0.0585) = 0.0415\ \mathrm{mol}\] \[\boxed{0.0415\ \mathrm{mol\ HCO_3^-}}\]Step 4 — Determine the Initial Carbonate Speciation
The carbonate stock solution has pH 11.30, meaning the second dissociation equilibrium dominates:
\[\mathrm{HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}}\]with:
\[pK_{a2} = 10.33\]Using the Henderson–Hasselbalch equation:
\[6.20 = 6.35 + \log\left(\frac{\mathrm{HCO_3^-}}{\mathrm{H_2CO_3}}\right)\]Solving for the ratio:
\[\frac{\mathrm{CO_3^{2-}}}{\mathrm{HCO_3^-}} = 10^{0.97} = 9.33\]Let:
\[\mathrm{HCO_3^-} = x\]Then:
\[\mathrm{CO_3^{2-}} = 9.33x\]Since total carbonate equals 0.100 mol:
\[x + 9.33x = 0.100\] \[10.33x = 0.100\] \[x = 0.00968\ \mathrm{mol}\]Therefore, the stock solution initially contains:
\[\boxed{0.00968\ \mathrm{mol\ HCO_3^-}}\]and
\[\boxed{0.0903\ \mathrm{mol\ CO_3^{2-}}}\]Step 5 — Calculate the Amount of HCl Required
The acid must convert the initial carbonate species into the desired final mixture of carbonic acid and bicarbonate.
Initial composition:
\[0.0903\ \mathrm{mol\ CO_3^{2-}}\] \[0.00968\ \mathrm{mol\ HCO_3^-}\]Desired final composition:
\[0.0585\ \mathrm{mol\ H_2CO_3}\] \[0.0415\ \mathrm{mol\ HCO_3^-}\]Conversion to Carbonic Acid
Each mole of carbonate requires two moles of hydrogen ion to form carbonic acid:
\[\mathrm{CO_3^{2-} + 2H^+ \rightarrow H_2CO_3}\]To produce 0.0585 mol of carbonic acid:
\[2(0.0585) = 0.1170\ \mathrm{mol\ H^+}\]Conversion to Bicarbonate
Remaining carbonate:
\[0.0903 - 0.0585 = 0.0318\ \mathrm{mol}\]Each mole requires one proton:
\[\mathrm{CO_3^{2-} + H^+ \rightarrow HCO_3^-}\]Thus:
\[0.0318\ \mathrm{mol\ H^+}\]Total Acid Required
\[0.1170 + 0.0318 = 0.1488\ \mathrm{mol\ H^+}\] \[\boxed{0.1488\ \mathrm{mol\ H^+}}\]Step 6 — Calculate Volume of 2.0 M HCl
Using:
\[V = \frac{n}{M}\] \[V = \frac{0.1488}{2.0}\] \[V = 0.0744\ \mathrm{L}\] \[\boxed{74.4\ \mathrm{mL\ of\ 2.0\ M\ HCl}}\]Final Buffer Preparation Procedure
To prepare the buffer:
- Measure:
- Add:
- Dilute the solution with distilled water to a final volume of:
Final Result
The resulting solution is expected to be:
\[\boxed{500\ \mathrm{mL\ of\ 0.20\ M\ carbonate\ buffer\ at\ pH\ 6.20}}\]This calculation demonstrates the importance of accounting for both carbonate equilibria when preparing buffers from strongly basic carbonate stock solutions.
Recap
Protonation sequence
\[\mathrm{CO_3^{2-} \xrightarrow{+H^+} HCO_3^- \xrightarrow{+H^+} H_2CO_3}\]| Species | Initial (mol) | Change from H+ addition | Final (mol) |
|---|---|---|---|
| CO32− | 0.0903 | -0.0903 | 0 |
| HCO3− | 0.00968 | +0.0903 - 0.0585 | 0.0415 |
| H2CO3 | 0 | +0.0585 | 0.0585 |
H+ accounting
| Protonation step | H+ used (mol) |
|---|---|
| CO32− → HCO3− | 0.0903 |
| HCO3− → H2CO3 | 0.0585 |
