Preparing a Phosphate Buffer at Its pKa
A teaching laboratory needs 500.0 mL of 0.100 M sodium phosphate buffer at pH 7.20. The available solids are NaH2PO4·H2O (molar mass 138.0 g/mol) and anhydrous Na2HPO4 (molar mass 142.0 g/mol). For phosphoric acid at 25°C, pKa2 = 7.20. The instructor requires that the total phosphate concentration, counting both forms, be 0.100 M; the buffer is to be made by dissolving both solids in about 400 mL of distilled water and then adjusting the volume to 500.0 mL, with no strong acid or base added. A student argues that because pH equals pKa, only Na2HPO4 is needed and the pH will settle at 7.20. Another student worries that diluting the finished buffer tenfold for an assay will raise the pH because the basic form is diluted. A calibrated pH meter is available. Assume activities may be replaced by concentrations.
1. Calculate the mass of each solid needed and explain why using only Na2HPO4 would not give pH 7.20.
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Henderson–Hasselbalch gives [\text{pH}=\text{p}K_a+\log\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}.] At pH 7.20 = pKa2, the ratio of base to acid must be 1, so the two forms must be equimolar. Total phosphate moles: [0.5000\ \text{L}\times0.100\ \text{mol/L}=0.0500\ \text{mol}.] Each form needs 0.0250 mol.
Mass of NaH2PO4·H2O: [0.0250\times138.0=3.45\ \text{g}.] Mass of Na2HPO4: [0.0250\times142.0=3.55\ \text{g}.] Dissolve both solids and make up to 500.0 mL.
Using only Na2HPO4 would not create an equimolar buffer pair. That salt is amphiprotic and its solution pH is set by the relevant equilibria, roughly near ((\text{p}K_{a2}+\text{p}K_{a3})/2), not simply equal to pKa2. The buffer pair must both be present in comparable amounts.
2. After adding 1.00 mL of 1.00 M HCl to the 500.0 mL buffer, calculate the new pH.
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Moles of HCl added: [0.00100\ \text{L}\times1.00\ \text{mol/L}=0.00100\ \text{mol}.] The strong acid converts 0.00100 mol of (\text{HPO}_4^{2-}) into (\text{H}2\text{PO}4^-). New amounts: [n{\text{acid}}=0.0250+0.0010=0.0260\ \text{mol},] [n{\text{base}}=0.0250-0.0010=0.0240\ \text{mol}.] The volume is common to both, so it cancels in the ratio: [\text{pH}=7.20+\log\frac{0.0240}{0.0260}=7.20-0.035=7.165.] The pH falls by only about 0.035 units, showing buffer action. The added (\text{H}^+) is consumed by conversion of base to acid rather than remaining free.
3. Does tenfold dilution raise the pH? Justify your answer.
Read the worked answer
Dilution reduces both ([\text{HPO}_4^{2-}]) and ([\text{H}_2\text{PO}_4^-]) by the same factor. Their ratio is unchanged, so the Henderson–Hasselbalch pH remains essentially 7.20 when activities are approximated by concentrations. The diluted buffer is 0.0100 M in total phosphate; it still buffers at about the same pH, but its buffer capacity per litre is one-tenth as large.
The student's concern confuses concentration with ratio. A tenfold dilution does not preferentially remove the basic form. In precise work, activity coefficients can cause a small shift, but dilution would not systematically raise the pH for the reason stated.