IChO 2006 · Theory problem 5

Acid-Base Chemistry

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5.1 Calculate \([\ce{H+}]\), \([\ce{OH-}]\), \([\ce{HSO4-}]\), and \([\ce{SO4^2-}]\) in a \(1.0\times10^{-7}\ \mathrm{M}\) solution of sulfuric acid (\(K_w=1.0\times10^{-14}\), \(K_2=1.2\times10^{-2}\) at \(25\ ^\circ\mathrm{C}\)). In your work you may use mass- and charge-balance equations. Answer with two significant figures.

5.2 Calculate the volume of \(0.80\ \mathrm{M}\) \(\ce{NaOH}\) solution that should be added to a \(250\ \mathrm{cm^3}\) aqueous solution containing \(3.48\ \mathrm{cm^3}\) of concentrated phosphoric acid in order to prepare a pH 7.4 buffer. Answer with three significant figures. (\(\ce{H3PO4 (aq)}\), purity = 85 mass %, density = \(1.69\ \mathrm{g/cm^3}\), \(M_r=98.00\)) (\(pK_1=2.15\), \(pK_2=7.20\), \(pK_3=12.44\)).

5.3 The efficacy of a drug is greatly dependent on its ability to be absorbed into the blood stream. Acid-base chemistry plays an important role in drug absorption.

Figure 1

Assume that the ionic form (\(\ce{A-}\)) of a weakly acidic drug does not penetrate the membrane, whereas the neutral form (\(\ce{HA}\)) freely crosses the membrane. Also assume that equilibrium is established so that the concentration of \(\ce{HA}\) is the same on both sides. Calculate the ratio of the total concentration (\([\ce{HA}]+[\ce{A-}]\)) of aspirin (acetylsalicylic acid, \(pK=3.52\)) in the blood to that in the stomach.

Official English text of the problem.

Answers

  • 5.1: \([\ce{H+}]=2.4\times10^{-7}\ \mathrm{M}\); \([\ce{OH-}]=4.1\times10^{-8}\ \mathrm{M}\); \([\ce{HSO4-}]=2.0\times10^{-12}\ \mathrm{M}\); \([\ce{SO4^2-}]=1.0\times10^{-7}\ \mathrm{M}\).
  • 5.2: \(1.03\times10^2\ \mathrm{cm^3}\) of \(0.80\ \mathrm{M}\) sodium hydroxide solution.
  • 5.3: \(\displaystyle \frac{C_{\mathrm{blood}}}{C_{\mathrm{stomach}}}\approx7.4\times10^3\).

Solution

5.1

The chemical species and their SMILES are:

Species SMILES
Sulfuric acid, \(\ce{H2SO4}\) O=S(=O)(O)O
Hydrogen ion (proton), \(\ce{H+}\) [H+]
Hydroxide ion, \(\ce{OH-}\) [OH-]
Hydrogen sulfate ion, \(\ce{HSO4-}\) O=S(=O)(O)[O-]
Sulfate ion, \(\ce{SO4^2-}\) O=S(=O)([O-])[O-]
Water, \(\ce{H2O}\) O

The oxyanion SMILES show particular Lewis representations; resonance is not fully conveyed by the localized bond orders and charges.

The first dissociation of sulfuric acid is treated as complete:

\[ \ce{H2SO4 -> H+ + HSO4-}. \]

The remaining equilibria are

\[ \ce{HSO4- <=> H+ + SO4^2-}, \qquad K_2=\frac{[\ce{H+}][\ce{SO4^2-}]}{[\ce{HSO4-}]}, \]

\[ \ce{H2O <=> H+ + OH-}, \qquad K_w=[\ce{H+}][\ce{OH-}]. \]

Because the analytical acid concentration is comparable to \(10^{-7}\ \mathrm{M}\), water’s contribution must be included.

Let \(C=1.0\times10^{-7}\ \mathrm{M}\) and \(h=[\ce{H+}]\). Sulfate mass balance gives

\[ [\ce{HSO4-}]+[\ce{SO4^2-}]=C, \]

and charge balance gives

\[ h=[\ce{OH-}]+[\ce{HSO4-}]+2[\ce{SO4^2-}]. \]

Combining mass balance with the equilibrium expressions,

\[ [\ce{HSO4-}]=\frac{Ch}{K_2+h}, \qquad [\ce{SO4^2-}]=\frac{CK_2}{K_2+h}, \qquad [\ce{OH-}]=\frac{K_w}{h}. \]

Thus,

\[ h=\frac{K_w}{h}+C\frac{h+2K_2}{h+K_2}. \]

Since \(h\ll K_2\), an excellent approximation is

\[ h=2C+\frac{K_w}{h}, \]

whose positive solution is

\[ h=C+\sqrt{C^2+K_w} \approx2.4142\times10^{-7}\ \mathrm{M}. \]

Solving the full balance equation gives \(h=2.4141964\times10^{-7}\ \mathrm{M}\), confirming the approximation. Substitution gives

\[ \begin{aligned} [\ce{OH-}]&=4.14217\times10^{-8}\ \mathrm{M},\\ [\ce{HSO4-}]&=2.01179\times10^{-12}\ \mathrm{M},\\ [\ce{SO4^2-}]&=9.99980\times10^{-8}\ \mathrm{M}. \end{aligned} \]

Therefore, to two significant figures,

\[ \boxed{ \begin{aligned} [\ce{H+}]&=2.4\times10^{-7}\ \mathrm{M},\\ [\ce{OH-}]&=4.1\times10^{-8}\ \mathrm{M},\\ [\ce{HSO4-}]&=2.0\times10^{-12}\ \mathrm{M},\\ [\ce{SO4^2-}]&=1.0\times10^{-7}\ \mathrm{M}. \end{aligned}} \]

5.2

The additional chemical species are:

Species SMILES
Sodium hydroxide, \(\ce{NaOH}\) [Na+].[OH-]
Phosphoric acid, \(\ce{H3PO4}\) O=P(O)(O)O
Dihydrogen phosphate ion, \(\ce{H2PO4-}\) O=P(O)(O)[O-]
Hydrogen phosphate ion, \(\ce{HPO4^2-}\) O=P(O)([O-])[O-]
Phosphate ion, \(\ce{PO4^3-}\) O=P([O-])([O-])[O-]

The amount of phosphoric acid is

\[ n= \frac{ (3.48\ \mathrm{cm^3}) (1.69\ \mathrm{g\,cm^{-3}}) (0.85) }{ 98.00\ \mathrm{g\,mol^{-1}} } =0.0510104\ \mathrm{mol}. \]

At pH 7.4, the principal phosphate species are \(\ce{H2PO4-}\) and \(\ce{HPO4^2-}\), because this pH is close to \(pK_2=7.20\). The other forms are negligible:

\[ \frac{[\ce{H3PO4}]}{[\ce{H2PO4-}]} =10^{pK_1-\mathrm{pH}} =10^{-5.25}, \]

\[ \frac{[\ce{PO4^3-}]}{[\ce{HPO4^2-}]} =10^{\mathrm{pH}-pK_3} =10^{-5.04}. \]

The relevant buffer equilibrium is

\[ \ce{H2PO4- <=> H+ + HPO4^2-}. \]

Consequently,

\[ \frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]} =10^{\mathrm{pH}-pK_2} =10^{7.40-7.20} =1.584893. \]

Adding hydroxide first neutralizes one proton from each phosphoric acid molecule:

\[ \ce{H3PO4 + OH- -> H2PO4- + H2O}. \]

Further hydroxide converts part of the dihydrogen phosphate into hydrogen phosphate:

\[ \ce{H2PO4- + OH- -> HPO4^2- + H2O}. \]

Let \(x\) mol undergo this second neutralization. The final amounts are approximately \(x\) mol of \(\ce{HPO4^2-}\) and \(n-x\) mol of \(\ce{H2PO4-}\). Their common final volume cancels:

\[ \frac{x}{n-x}=1.584893. \]

Therefore,

\[ x=\frac{1.584893}{1+1.584893}\,n =0.0312764\ \mathrm{mol}. \]

The total amount of hydroxide required is

\[ n_{\ce{OH-}}=n+x =0.0822868\ \mathrm{mol}, \]

so

\[ V_{\ce{NaOH}} =\frac{0.0822868\ \mathrm{mol}} {0.80\ \mathrm{mol\,L^{-1}}} =0.102858\ \mathrm{L} =102.858\ \mathrm{cm^3}. \]

Thus, to the requested three significant figures,

\[ \boxed{V_{\ce{NaOH}}=1.03\times10^2\ \mathrm{cm^3}}. \]

The initial \(250\ \mathrm{cm^3}\) affects the final concentrations but cancels from the buffer ratio. Including all phosphate species and the residual free hydrogen and hydroxide ions does not change the rounded answer.

5.3

Here \(\ce{HA}\) is aspirin (acetylsalicylic acid), CC(=O)Oc1ccccc1C(=O)O, and \(\ce{A-}\) is the acetylsalicylate ion, CC(=O)Oc1ccccc1C(=O)[O-]. Ionization removes the carboxylic acid proton; the ester group remains intact.

For the equilibrium

\[ \ce{HA <=> H+ + A-}, \]

\[ K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}, \]

so

\[ \frac{[\ce{A-}]}{[\ce{HA}]} =10^{\mathrm{pH}-pK_a}. \]

The total drug concentration in either compartment is therefore

\[ C=[\ce{HA}]+[\ce{A-}] =[\ce{HA}]\left(1+10^{\mathrm{pH}-pK_a}\right). \]

At membrane equilibrium,

\[ [\ce{HA}]_{\mathrm{blood}} =[\ce{HA}]_{\mathrm{stomach}}. \]

Using the blood pH of 7.4 and stomach pH of 2.0 shown in the figure,

\[ \begin{aligned} \frac{C_{\mathrm{blood}}}{C_{\mathrm{stomach}}} &=\frac{1+10^{7.4-3.52}} {1+10^{2.0-3.52}}\\ &=\frac{1+7585.7758}{1+0.0301995}\\ &=7364.375\ldots. \end{aligned} \]

Hence,

\[ \boxed{\frac{C_{\mathrm{blood}}}{C_{\mathrm{stomach}}} \approx7.4\times10^3}. \]

The higher blood pH favors the ionized form, which cannot cross the membrane. This produces a much larger total drug concentration in blood even though the neutral-form concentrations are equal.

The solution was checked against the official solution.