IChO 2000 · Theory problem 6

Hard Water

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In Denmark the subsoil consists mainly of limestone. In contact with ground water containing carbon dioxide some of the calcium carbonate dissolves as calcium hydrogen carbonate. As a result, such ground water is hard, and when used as tap water the high content of calcium hydrogen carbonate causes problems due to precipitation of calcium carbonate in, for example, kitchen and bathroom environments.

Carbon dioxide, \(\ce{CO2}\), is a diprotic acid in aqueous solution. The \(pK_a\)-values at \(0~^\circ\mathrm{C}\) are:

Figure 1

The liquid volume change associated with dissolution of \(\ce{CO2}\) may be neglected for all of the following problems. The temperature is to be taken as being \(0~^\circ\mathrm{C}\).

6.1 The total concentration of carbon dioxide in water which is saturated with carbon dioxide at a carbon dioxide partial pressure of \(1.00~\mathrm{bar}\) is \(0.0752~\mathrm{mol\,dm^{-3}}\). Calculate the volume of carbon dioxide gas which can be dissolved in one litre of water under these conditions.

The gas constant \[ R=8.314~\mathrm{J\,mol^{-1}\,K^{-1}} =0.08314~\mathrm{L\,bar\,mol^{-1}\,K^{-1}} \]

6.2 Calculate the equilibrium concentration of hydrogen ions and the equilibrium concentration of \(\ce{CO2}\) in water saturated with carbon dioxide at a carbon dioxide partial pressure of \(1.00~\mathrm{bar}\).

6.3 Calculate the equilibrium concentration of hydrogen ions in a \(0.0100~\mathrm{M}\) aqueous solution of sodium hydrogen carbonate saturated with carbon dioxide at a carbon dioxide partial pressure of \(1.00~\mathrm{bar}\).

6.4 Calculate the equilibrium concentration of hydrogen ions in a \(0.0100~\mathrm{M}\) aqueous solution of sodium carbonate saturated with carbon dioxide at a carbon dioxide partial pressure of \(1.00~\mathrm{bar}\). Ignore water dissociation effects.

6.5 The solubility of calcium carbonate in water at \(0~^\circ\mathrm{C}\) is \(0.0012~\mathrm{g}\) per \(100~\mathrm{cm^3}\) of water. Calculate the concentration of calcium ions in a saturated solution of calcium carbonate in water.

The hard groundwater in Denmark is formed via contact of water with limestone in the subsoil which reacts with carbon dioxide dissolved in the groundwater according to the equilibrium equation:

Figure 2

The equilibrium constant, \(K\), for this reaction is \(10^{-4.25}\) at \(0~^\circ\mathrm{C}\).

6.6 Calculate the concentration of calcium ions in water in equilibrium with calcium carbonate in an atmosphere with a partial pressure of carbon dioxide of \(1.00~\mathrm{bar}\).

6.7 A \(0.0150~\mathrm{M}\) solution of calcium hydroxide is saturated with carbon dioxide gas at a partial pressure of \(1.00~\mathrm{bar}\). Calculate the concentration of calcium ions in the solution by considering the equilibrium equation given above in connection with problem 6.6.

6.8 The calcium hydroxide solution referred to in problem 6.7 is diluted to twice the volume with water before saturation with carbon dioxide gas at a partial pressure of \(1.00~\mathrm{bar}\). Calculate the concentration of calcium ions in the resulting solution saturated with \(\ce{CO2}\).

6.9 Calculate the solubility product constant for calcium carbonate from the data given above.

Official English text of the problem.

Answers

Here, \(\mathrm{M}=\mathrm{mol\,dm^{-3}}\).

  • 6.1: \(V=1.71~\mathrm{L}\) of carbon dioxide gas at \(0~^\circ\mathrm{C}\) and \(1.00~\mathrm{bar}\).
  • 6.2: \([\ce{H+}]=1.33\times10^{-4}~\mathrm{M}\); \([\ce{CO2(aq)}]=0.0751~\mathrm{M}\).
  • 6.3: \([\ce{H+}]=1.76\times10^{-6}~\mathrm{M}\).
  • 6.4: \([\ce{H+}]=8.80\times10^{-7}~\mathrm{M}\).
  • 6.5: \([\ce{Ca^2+}]=1.2\times10^{-4}~\mathrm{M}\).
  • 6.6: \([\ce{Ca^2+}]=1.02\times10^{-2}~\mathrm{M}\).
  • 6.7: \([\ce{Ca^2+}]=1.02\times10^{-2}~\mathrm{M}\); some calcium carbonate precipitates.
  • 6.8: \([\ce{Ca^2+}]=7.50\times10^{-3}~\mathrm{M}\); all the calcium remains dissolved at final equilibrium.
  • 6.9: \(K_{\mathrm{sp}}=10^{-8.26}\approx5.5\times10^{-9}\), using the \(1~\mathrm{M}\) standard state.

Solution

The species and their SMILES are listed below. Dot-separated salt SMILES describe ionic composition; they do not encode crystal lattices or aqueous solvation.

Species SMILES
Carbon dioxide, \(\ce{CO2}\) O=C=O
Water, \(\ce{H2O}\) O
Hydrogen ion, \(\ce{H+}\) [H+]
Hydrogen carbonate ion, \(\ce{HCO3-}\) O=C([O-])O
Carbonate ion, \(\ce{CO3^2-}\) O=C([O-])[O-]
Hydroxide ion, \(\ce{OH-}\) [OH-]
Sodium ion, \(\ce{Na+}\) [Na+]
Calcium ion, \(\ce{Ca^2+}\) [Ca+2]
Sodium hydrogen carbonate, \(\ce{NaHCO3}\) [Na+].O=C([O-])O
Sodium carbonate, \(\ce{Na2CO3}\) [Na+].[Na+].O=C([O-])[O-]
Calcium carbonate, \(\ce{CaCO3}\) [Ca+2].O=C([O-])[O-]
Calcium hydroxide, \(\ce{Ca(OH)2}\) [Ca+2].[OH-].[OH-]
Calcium hydrogen carbonate, \(\ce{Ca(HCO3)2}\) [Ca+2].O=C([O-])O.O=C([O-])O

We use the concentration approximation for dissolved species and unit activities for pure liquid water and solid calcium carbonate. Standard-state concentration factors are suppressed in the equilibrium expressions.

6.1

The amount of carbon dioxide absorbed by one litre of water is \[ n=(0.0752~\mathrm{mol\,L^{-1}})(1.00~\mathrm{L}) =0.0752~\mathrm{mol}. \]

Using the ideal-gas equation at \(T=273.15~\mathrm{K}\), \[ V=\frac{nRT}{p} =\frac{(0.0752)(0.08314)(273.15)}{1.00}~\mathrm{L} =1.70777~\mathrm{L}. \]

Thus, \[ \boxed{V=1.71~\mathrm{L}}. \]

6.2

The two acid equilibria are \[ \ce{CO2(aq) + H2O(l) <=> HCO3-(aq) + H+(aq)}, \qquad K_{a1}=10^{-6.630}=2.344\times10^{-7}, \] \[ \ce{HCO3-(aq) <=> CO3^2-(aq) + H+(aq)}, \qquad K_{a2}=10^{-10.640}=2.291\times10^{-11}. \]

Define \[ h=[\ce{H+}],\qquad c=[\ce{CO2(aq)}],\qquad b=[\ce{HCO3-}],\qquad d=[\ce{CO3^2-}]. \]

Then \[ K_{a1}=\frac{hb}{c}, \qquad K_{a2}=\frac{hd}{b}. \]

The total dissolved carbon concentration is \[ C_T=c+b+d=0.0752~\mathrm{M}. \]

Water ionization and the second dissociation are negligible in this acidic solution. Consequently, charge balance gives \(b\simeq h\), while carbon balance gives \(c\simeq C_T-h\). Therefore, \[ K_{a1}=\frac{h^2}{C_T-h}, \] and \[ h=\frac{-K_{a1}+\sqrt{K_{a1}^2+4K_{a1}C_T}}{2} =1.32656\times10^{-4}~\mathrm{M}. \]

Hence \[ c=0.0752-0.000132656 =0.0750673~\mathrm{M}. \]

The neglected carbonate fraction is indeed tiny: \[ \frac{d}{b}=\frac{K_{a2}}{h}=1.73\times10^{-7}. \]

Thus, \[ \boxed{[\ce{H+}]=1.33\times10^{-4}~\mathrm{M}}, \qquad \boxed{[\ce{CO2(aq)}]=0.0751~\mathrm{M}}. \]

At fixed temperature and carbon dioxide partial pressure, Henry’s law fixes the concentration of molecular \(\ce{CO2(aq)}\). We therefore use \(c=0.0750673~\mathrm{M}\) in the subsequent solutions saturated with carbon dioxide at \(1.00~\mathrm{bar}\). Their total dissolved carbon concentrations can differ.

6.3

Sodium hydrogen carbonate dissociates: \[ \ce{NaHCO3(aq) -> Na+(aq) + HCO3-(aq)}. \]

Thus \([\ce{Na+}]=0.0100~\mathrm{M}\). Neglecting water ionization, charge balance is \[ 0.0100+h=b+2d. \]

Since \(h\) and \(d\) are small compared with \(0.0100~\mathrm{M}\), \[ b\simeq0.0100~\mathrm{M}. \]

The first acid equilibrium gives \[ h=\frac{K_{a1}c}{b} \simeq\frac{(2.34423\times10^{-7})(0.0750673)}{0.0100}. \]

Therefore, \[ \boxed{[\ce{H+}]=1.76\times10^{-6}~\mathrm{M}}. \]

This gives \(d=K_{a2}b/h\simeq1.30\times10^{-7}~\mathrm{M}\), confirming that both correction terms in the charge balance are small.

6.4

Sodium carbonate supplies two sodium ions per formula unit: \[ \ce{Na2CO3(aq) -> 2Na+(aq) + CO3^2-(aq)}. \]

Carbonate reacts with dissolved carbon dioxide: \[ \ce{CO3^2-(aq) + CO2(aq) + H2O(l) <=> 2HCO3-(aq)}. \]

The equilibrium constant for this conversion is \[ \frac{K_{a1}}{K_{a2}}=10^{4.010}, \] so hydrogen carbonate predominates under the specified conditions.

Here \([\ce{Na+}]=0.0200~\mathrm{M}\). Ignoring water dissociation as instructed, \[ 0.0200+h=b+2d, \] giving \(b\simeq0.0200~\mathrm{M}\).

Thus, \[ h\simeq\frac{K_{a1}c}{0.0200} =\frac{(2.34423\times10^{-7})(0.0750673)}{0.0200}, \] and \[ \boxed{[\ce{H+}]=8.80\times10^{-7}~\mathrm{M}}. \]

The resulting carbonate concentration is approximately \(5.21\times10^{-7}~\mathrm{M}\), sufficiently small to validate the approximation.

6.5

Calcium carbonate dissolves according to \[ \ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-(aq)}. \]

Each mole dissolved supplies one mole of calcium ions. Using \(M(\ce{CaCO3})\simeq100.1~\mathrm{g\,mol^{-1}}\) and \(100~\mathrm{cm^3}=0.100~\mathrm{dm^3}\), \[ [\ce{Ca^2+}] =\frac{0.0012~\mathrm{g}} {(100.1~\mathrm{g\,mol^{-1}})(0.100~\mathrm{dm^3})}. \]

Therefore, \[ \boxed{[\ce{Ca^2+}]=1.2\times10^{-4}~\mathrm{mol\,dm^{-3}}}. \]

Acid–base redistribution of dissolved carbonate does not change this calcium balance.

6.6

The given equilibrium is \[ \ce{CaCO3(s) + CO2(aq) + H2O(l) <=> Ca^2+(aq) + 2HCO3-(aq)}, \] with \[ K=\frac{[\ce{Ca^2+}]b^2}{c}=10^{-4.25}. \]

Let \(s=[\ce{Ca^2+}]\). Charge balance gives \[ 2s+h=b+2d. \]

Hydrogen carbonate is the predominant anion, so \(b\simeq2s\). Consequently, \[ K\simeq\frac{s(2s)^2}{c}=\frac{4s^3}{c}, \] and \[ s=\left(\frac{Kc}{4}\right)^{1/3} =\left(\frac{10^{-4.25}(0.0750673)}{4}\right)^{1/3} =0.0101812~\mathrm{M}. \]

Thus, \[ \boxed{[\ce{Ca^2+}]=1.02\times10^{-2}~\mathrm{M}}. \]

The corresponding hydrogen-ion concentration is about \(8.64\times10^{-7}~\mathrm{M}\), and carbonate is also negligible compared with \(b\simeq0.0204~\mathrm{M}\).

6.7

Absorbed carbon dioxide neutralizes hydroxide, giving the overall conversion \[ \ce{Ca(OH)2(aq) + 2CO2(aq) -> Ca^2+(aq) + 2HCO3-(aq)}. \]

If all the initial calcium remained dissolved, then approximately \[ [\ce{Ca^2+}]=0.0150~\mathrm{M}, \qquad b=0.0300~\mathrm{M}. \]

The reaction quotient for the equilibrium in 6.6 would be \[ Q=\frac{(0.0150)(0.0300)^2}{0.0750673} =1.80\times10^{-4}. \]

Since \[ Q>K=5.62\times10^{-5}, \] the equilibrium shifts toward solid calcium carbonate. Some calcium precipitates, and the remaining solution reaches the saturation concentration calculated in 6.6: \[ \boxed{[\ce{Ca^2+}]=1.02\times10^{-2}~\mathrm{M}}. \]

Approximately \(0.0150-0.01018=0.0048~\mathrm{mol}\) of calcium carbonate precipitates per litre, confirming that a solid phase remains.

6.8

Dilution to twice the volume reduces the total calcium concentration to \[ C_{\mathrm{Ca,tot}}=\frac{0.0150}{2} =0.00750~\mathrm{M}. \]

If all this calcium is dissolved, \(b\simeq0.0150~\mathrm{M}\), and \[ Q=\frac{(0.00750)(0.0150)^2}{0.0750673} =2.25\times10^{-5}<K. \]

The final solution is therefore undersaturated with respect to calcium carbonate. Any calcium carbonate formed during addition of the gas can dissolve completely before the final equilibrium is reached.

All available calcium remains in solution: \[ \boxed{[\ce{Ca^2+}]=7.50\times10^{-3}~\mathrm{M}}. \]

6.9

Combine the following equilibria: \[ \begin{aligned} \ce{CaCO3(s) &<=> Ca^2+(aq) + CO3^2-(aq)} && K_{\mathrm{sp}},\\ \ce{CO2(aq) + H2O(l) &<=> HCO3-(aq) + H+(aq)} && K_{a1},\\ \ce{CO3^2-(aq) + H+(aq) &<=> HCO3-(aq)} && \frac{1}{K_{a2}}. \end{aligned} \]

Their sum is the reaction given before 6.6. Therefore, \[ K=K_{\mathrm{sp}}\frac{K_{a1}}{K_{a2}}, \] so \[ K_{\mathrm{sp}} =K\frac{K_{a2}}{K_{a1}} =10^{-4.25}\frac{10^{-10.640}}{10^{-6.630}} =10^{-8.260}. \]

Thus, \[ \boxed{K_{\mathrm{sp}}\approx5.5\times10^{-9}}. \]

Under the concentration approximation, the corresponding saturation product is \[ [\ce{Ca^2+}][\ce{CO3^2-}] =5.5\times10^{-9}~\mathrm{mol^2\,dm^{-6}}. \]

Carbonate hydrolysis, \[ \ce{CO3^2-(aq) + H2O(l) <=> HCO3-(aq) + OH-(aq)}, \] makes the free carbonate concentration smaller than the total molar solubility in 6.5. Consequently, that solubility cannot simply be squared to obtain \(K_{\mathrm{sp}}\).

The solution was checked against the official solution.