Difference in calculated pH and the real pH of a phosphate buffer

Question

I want to prepare a buffer system with $\mathrm{pH} = 7.00$ and a total concentration $c_\mathrm{tot} = \pu{0.10 mol l^-1}$ using sodium hydrogen phosphate $(\ce{Na2HPO4})$ and sodium dihydrogen phosphate $(\ce{NaH2PO4})$ salts. I am calculating the amounts to prepare a phosphate buffer as follows:

$$\ce{H2PO4- <=> HPO4^2- + H+} \qquad \mathrm{p}K_\mathrm{a} = 7.20 \tag{R1}$$

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]} \tag{1}$$

$$\frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}} = 10^{7.00-7.20} \approx 0.632 \tag{2}$$

$$ \begin{align} c_\mathrm{tot} &= [\ce{H2PO4-}] + [\ce{HPO4^2-}] \\ &= [\ce{H2PO4-}] + [\ce{H2PO4-}]\times 0.632 \\ &= [\ce{H2PO4-}]\times 1.632 \tag{3} \end{align} $$

$$[\ce{H2PO4-}] = \frac{c_\mathrm{tot}}{1.632} = \frac{\pu{0.10 mol l^-1}}{1.632} \approx \pu{0.061 mol l^-1}\tag{4}$$

$$[\ce{HPO4^2-}] = c_\mathrm{tot} - [\ce{H2PO4-}] = \pu{0.10 mol l^-1} - \pu{0.061 mol l^-1} = \pu{0.039 mol l^-1} \tag{5}$$

For the preparation of $V = \pu{500 ml}$ buffer solution I used the following masses of corresponding salts:

$$ \begin{align} m(\ce{NaH2PO4}) &= [\ce{H2PO4-}]\cdot V\cdot M(\ce{NaH2PO4}) \\ &= (\pu{0.061 mol l^-1})(\pu{0.500 l})(\pu{119.97 g mol^-1}) \\ &\approx\pu{3.66 g} \tag{6} \end{align} $$

$$ \begin{align} m(\ce{Na2HPO4}) &= [\ce{HPO4^2-}]\cdot V\cdot M(\ce{Na2HPO4}) \\ &= (\pu{0.039 mol l^-1})(\pu{0.500 l})(\pu{141.97 g mol^-1}) \\ &\approx\pu{2.77 g} \tag{7} \end{align} $$

When I mixed the calculated amounts, the obtained $\mathrm{pH} = 6.5$ deviated quite a lot from the desired value. I thought it was my mistake in calculations or an error when I weighed the amounts, but I repeated everything several times and still not get the $\mathrm{pH} = 7.00,$ and the value always was $6.4$ or $6.5.$

I would like to know the reason. I think if the concentration of the monobasic salt is greater than the one of the dibasic salt, the $\mathrm{pH}$ tends to decrease, but in the calculations the relation [dibasic]/[monobasic] is always less than 1 and this again causes [monobasic] > [dibasic].

Answer 1

The flaw in your calculations is the assumption that the $\mathrm{p}K_\mathrm{a}$ is a constant independent of concentration. In reality, $\mathrm{p}K_\mathrm{a}$ is a function of the ionic strength $I$.

At infinite dilution $I = 0$ and $\mathrm{p}K_\mathrm{a} = 7.20$ as you assumed. However at the concentration $\pu{0.10 mol l^-1}$ of phosphate buffer $\mathrm{p}K_\mathrm{a} = 6.81.$ See Florida State University — Phosphate Buffers and the paper by Green [1].

Reference

1. Green, A. A. The Preparation of Acetate and Phosphate Buffer Solutions of Known PH and Ionic Strength. J. Am. Chem. Soc. 1933, 55 (6), 2331–2336. DOI: 10.1021/ja01333a018.