IChO 2005 · Theory problem 6

Alkalinity of Water and Solubility of CO2

Alkalinity of Water and Solubility of \(\ce{CO2}\)

The capacity of water to accept \(\ce{H+}\) ions is called alkalinity. Alkalinity is important in water treatment and in the chemistry and biology of natural waters. Generally, the basic species responsible for alkalinity in water are \(\ce{HCO3^-}\), \(\ce{CO3^2-}\), and \(\ce{OH^-}\). At pH values below 7, \(\ce{H+}\) in water detracts significantly from alkalinity. Therefore, the complete equation for alkalinity in a medium where \(\ce{HCO3^-}\), \(\ce{CO3^2-}\), and \(\ce{OH^-}\) are the only contributors to alkalinity can be expressed as

\[ \text{alkalinity} =[\ce{HCO3^-}]+2[\ce{CO3^2-}]+[\ce{OH^-}]-[\ce{H+}] \]

The contributions made by different species to alkalinity depend upon pH. Relevant chemical equations and equilibrium constants (at 298 K) are shown below:

Figure 1

Note: Calculations must be shown.

6.1 Natural waters (river or lake water) generally contain dissolved \(\ce{CO2}\). The ratio of \([\ce{H2CO3}]:[\ce{HCO3^-}]:[\ce{CO3^2-}]\) in a water at \([\ce{H+}]=1.00\times10^{-7}\ \mathrm{M}\) will be:

\[ \underline{(a)}:1.00:\underline{(b)}. \]

Calculate (a) and (b).

6.2 Gaseous \(\ce{CO2}\) in the atmosphere can be regarded as a contributor to the alkalinity of water in equilibrium with air. Calculate the concentration of \(\ce{CO2(aq)}\) \((\mathrm{mol\,dm^{-3}})\) in pure water that is in equilibrium with the unpolluted air at \(1.01\times10^5\ \mathrm{Pa}\) and 298 K containing 0.0360 % (molar ratio) \(\ce{CO2}\). (assuming standard pressure \(=1.01\times10^5\ \mathrm{Pa}\)).

If you are unable to solve this problem, assume that concentration of \(\ce{CO2(aq)}\) is equal to \(1.11\times10^{-5}\ \mathrm{M}\) for further calculations.

The solubility (\(S\)) of \(\ce{CO2}\) in water can be defined as

\[ S=[\ce{CO2(aq)}]+[\ce{H2CO3}]+[\ce{HCO3^-}]+[\ce{CO3^2-}]. \]

The solubility of atmospheric \(\ce{CO2}\) in water that is in equilibrium with the unpolluted air at 298 K and \(1.01\times10^5\ \mathrm{Pa}\) will vary with alkalinity.

6-3 Find the solubility of atmospheric \(\ce{CO2}\) in pure water \((\mathrm{mol\,dm^{-3}})\). Neglect dissociation of water.

6.4 Find the solubility of atmospheric \(\ce{CO2}\) \((\mathrm{mol\,dm^{-3}})\) in a \(1.00\times10^{-3}\ \mathrm{mol\,dm^{-3}}\) \(\ce{NaOH}\) solution.

At 298 K, \(1.01\times10^5\ \mathrm{Pa}\) unpolluted air is in equilibrium with natural water saturated with \(\ce{CaCO3}\). The following main equilibrium may exist:

Figure 2

6.5 Calculate the equilibrium constant for the above equation.

If you are unable to solve this problem, assume that equilibrium constant \(K_{\mathrm{eq}}=5.00\times10^{-5}\) for further calculations.

6.6 Calculate the concentration of \(\ce{Ca^2+}\) \((\mathrm{mg\,dm^{-3}})\) in \(\ce{CaCO3}\)-saturated natural water that is in equilibrium with atmospheric \(\ce{CO2}\).

If you are unable to solve this problem, assume that concentration of \(\ce{Ca^2+(aq)}=40.1\ \mathrm{mg\,dm^{-3}}\) for further calculations.

6.7 Find the alkalinity (mol/L) of the above solution.

6.8 In an underground lake saturated with \(\ce{CaCO3}\), the water has a high content of \(\ce{CO2}\). The concentration of \(\ce{Ca^2+}\) in this lake was found to be as high as \(100\ \mathrm{mg\,dm^{-3}}\). Assume the lake and the air above is a closed system, calculate the effective pressure of \(\ce{CO2}\) (Pa) in air which is in equilibrium with this \(\ce{Ca^2+}\) content.

Official English text of the problem.

Answers

The results retain all given equilibria, except water dissociation in 6-3, as instructed. Parenthetical alternatives use the approximation \([\ce{HCO3^-}]\approx2[\ce{Ca^2+}]\).

  • 6.1: \(a=4.48\times10^{-4}\), \(b=4.69\times10^{-4}\); hence the ratio is \(4.48\times10^{-4}:1.00:4.69\times10^{-4}\).
  • 6.2: \([\ce{CO2(aq)}]=1.24\times10^{-5}\ \mathrm{mol\,dm^{-3}}\).
  • 6-3: \(S=1.48\times10^{-5}\ \mathrm{mol\,dm^{-3}}\).
  • 6.4: \(S=1.00\times10^{-3}\ \mathrm{mol\,dm^{-3}}\).
  • 6.5: \(K_{\mathrm{eq}}=4.28\times10^{-5}\).
  • 6.6: \([\ce{Ca^2+}]=20.7\ \mathrm{mg\,dm^{-3}}\) (\(20.4\ \mathrm{mg\,dm^{-3}}\) with the approximation).
  • 6.7: Alkalinity \(=1.03\times10^{-3}\ \mathrm{mol\,L^{-1}}\) (\(1.02\times10^{-3}\ \mathrm{mol\,L^{-1}}\) with the approximation).
  • 6.8: \(p_{\ce{CO2}}=4.25\times10^3\ \mathrm{Pa}\) (\(4.26\times10^3\ \mathrm{Pa}\) with the approximation).

Solution

The species used below have the following names and SMILES:

Species and name SMILES
\(\ce{CO2}\), carbon dioxide O=C=O
\(\ce{H2O}\), water O
\(\ce{H2CO3}\), carbonic acid O=C(O)O
\(\ce{HCO3^-}\), hydrogen carbonate (bicarbonate) ion O=C(O)[O-]
\(\ce{CO3^2-}\), carbonate ion O=C([O-])[O-]
\(\ce{H+}\), hydrogen ion (proton) [H+]
\(\ce{OH^-}\), hydroxide ion [OH-]
\(\ce{Ca^2+}\), calcium ion [Ca+2]
\(\ce{Na+}\), sodium ion [Na+]
\(\ce{CaCO3}\), calcium carbonate [Ca+2].[O-]C(=O)[O-]
\(\ce{NaOH}\), sodium hydroxide [Na+].[OH-]

The disconnected-ion SMILES for the salts do not describe their crystal lattices. SMILES also do not specify gas versus aqueous state or aqueous solvation; \(\ce{H+}\) is the conventional shorthand for the solvated proton.

Use ideal-solution equilibrium expressions, with concentrations in \(\mathrm{mol\,dm^{-3}}\) and unit standard concentration understood. Pure water and solid calcium carbonate have unit activity. For mass conversions, take \(M(\ce{Ca})=40.1\ \mathrm{g\,mol^{-1}}\).

6.1

The two acid dissociations give

\[ K_{a1}=\frac{[\ce{HCO3^-}][\ce{H+}]}{[\ce{H2CO3}]}, \qquad K_{a2}=\frac{[\ce{CO3^2-}][\ce{H+}]}{[\ce{HCO3^-}]}. \]

Normalizing the bicarbonate concentration to 1.00,

\[ a=\frac{[\ce{H2CO3}]}{[\ce{HCO3^-}]} =\frac{1.00\times10^{-7}}{2.23\times10^{-4}} =\boxed{4.48\times10^{-4}}, \]

\[ b=\frac{[\ce{CO3^2-}]}{[\ce{HCO3^-}]} =\frac{4.69\times10^{-11}}{1.00\times10^{-7}} =\boxed{4.69\times10^{-4}}. \]

Thus bicarbonate overwhelmingly predominates among these three species at the specified hydrogen-ion concentration.

6.2

The carbon dioxide mole fraction and partial pressure are

\[ y_{\ce{CO2}}=\frac{0.0360}{100}=3.60\times10^{-4}, \]

\[ p_{\ce{CO2}} =y_{\ce{CO2}}P =(3.60\times10^{-4})(1.01\times10^5) =36.36\ \mathrm{Pa}. \]

For

\[ \ce{CO2(g) <=> CO2(aq)}, \]

the given equilibrium constant uses pressure relative to the standard pressure:

\[ K_{\ce{CO2}} =\frac{[\ce{CO2(aq)}]/c^\circ}{p_{\ce{CO2}}/p^\circ}, \qquad c^\circ=1\ \mathrm{mol\,dm^{-3}}. \]

Therefore,

\[ [\ce{CO2(aq)}] =(3.44\times10^{-2}) \frac{36.36}{1.01\times10^5} (1\ \mathrm{mol\,dm^{-3}}) =1.2384\times10^{-5}\ \mathrm{mol\,dm^{-3}}. \]

\[ \boxed{[\ce{CO2(aq)}]=1.24\times10^{-5}\ \mathrm{mol\,dm^{-3}}} \]

This is the concentration of molecular dissolved carbon dioxide, rather than the sum of all dissolved carbon species.

6-3

Hydration gives

\[ \ce{CO2(aq) + H2O(l) <=> H2CO3(aq)}, \]

so

\[ [\ce{H2CO3}] =K_{\ce{H2CO3}}[\ce{CO2(aq)}] =(2.00\times10^{-3})(1.2384\times10^{-5}) =2.4768\times10^{-8}\ \mathrm{M}. \]

Define

\[ h=[\ce{H+}], \qquad A=K_{a1}[\ce{H2CO3}] =5.523264\times10^{-12}\ \mathrm{M^2}. \]

Then

\[ [\ce{HCO3^-}]=\frac{A}{h}, \qquad [\ce{CO3^2-}]=\frac{K_{a2}A}{h^2}. \]

Neglecting water dissociation, electroneutrality requires

\[ h=[\ce{HCO3^-}]+2[\ce{CO3^2-}] =\frac{A}{h}+\frac{2K_{a2}A}{h^2}. \]

The second dissociation is very small, so initially

\[ h\simeq\sqrt{A}=2.3502\times10^{-6}\ \mathrm{M}. \]

Indeed,

\[ \frac{[\ce{CO3^2-}]}{[\ce{HCO3^-}]} =\frac{K_{a2}}{h}\simeq2.00\times10^{-5}, \]

which verifies that approximation. Retaining the small carbonate term gives

\[ [\ce{HCO3^-}]=2.35012\times10^{-6}\ \mathrm{M}, \qquad [\ce{CO3^2-}]=4.68981\times10^{-11}\ \mathrm{M}. \]

Hence

\[ \begin{aligned} S &=1.2384\times10^{-5} +2.4768\times10^{-8} +2.35012\times10^{-6} +4.68981\times10^{-11}\\ &=1.47589\times10^{-5}\ \mathrm{mol\,dm^{-3}}. \end{aligned} \]

\[ \boxed{S=1.48\times10^{-5}\ \mathrm{mol\,dm^{-3}}} \]

6.4

Sodium hydroxide dissociates completely:

\[ \ce{NaOH(aq) -> Na+(aq) + OH^-(aq)}. \]

Atmospheric carbon dioxide consumes hydroxide, principally through

\[ \ce{CO2(aq) + OH^-(aq) <=> HCO3^-(aq)}. \]

Consequently, the final hydroxide concentration is not the initial analytical sodium hydroxide concentration. The conserved sodium concentration is

\[ [\ce{Na+}]=1.00\times10^{-3}\ \mathrm{M}. \]

Equilibrium with the same atmosphere fixes \([\ce{CO2(aq)}]\), \([\ce{H2CO3}]\), and \(A\) at the values found above. Electroneutrality gives

\[ [\ce{Na+}]+h =\frac{A}{h}+\frac{2K_{a2}A}{h^2}+\frac{K_w}{h}. \]

Thus

\[ 1.00\times10^{-3} =\frac{A+K_w}{h} +\frac{2K_{a2}A}{h^2}-h. \]

Equivalently,

\[ h^3+(1.00\times10^{-3})h^2 -(A+K_w)h-2K_{a2}A=0. \]

The positive root is

\[ h=5.62533\times10^{-9}\ \mathrm{M}, \]

giving

\[ \begin{aligned} [\ce{HCO3^-}]&=9.81856\times10^{-4}\ \mathrm{M},\\ [\ce{CO3^2-}]&=8.18602\times10^{-6}\ \mathrm{M},\\ [\ce{OH^-}]&=1.77767\times10^{-6}\ \mathrm{M}. \end{aligned} \]

Therefore,

\[ \begin{aligned} S &=1.2384\times10^{-5} +2.4768\times10^{-8} +9.81856\times10^{-4} +8.18602\times10^{-6}\\ &=1.00245\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \end{aligned} \]

\[ \boxed{S=1.00\times10^{-3}\ \mathrm{mol\,dm^{-3}}} \]

6.5

The required balanced reaction is

\[ \ce{CaCO3(s) + CO2(aq) + H2O(l) <=> Ca^2+(aq) + 2HCO3^-(aq)}. \]

It is obtained by adding the following equilibria:

\[ \begin{array}{ll} \ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-(aq)} &K_{sp},\\[2pt] \ce{CO2(aq) + H2O(l) <=> H2CO3(aq)} &K_{\ce{H2CO3}},\\[2pt] \ce{H2CO3(aq) <=> HCO3^-(aq) + H+(aq)} &K_{a1},\\[2pt] \ce{CO3^2-(aq) + H+(aq) <=> HCO3^-(aq)} &1/K_{a2}. \end{array} \]

Adding reactions multiplies their equilibrium constants, so

\[ K_{\mathrm{eq}} =\frac{K_{sp}K_{\ce{H2CO3}}K_{a1}}{K_{a2}} =\frac{(4.50\times10^{-9})(2.00\times10^{-3})(2.23\times10^{-4})} {4.69\times10^{-11}}. \]

\[ \boxed{K_{\mathrm{eq}}=4.28\times10^{-5}} \]

Its activity expression is

\[ K_{\mathrm{eq}} =\frac{a_{\ce{Ca^2+}}a_{\ce{HCO3^-}}^2}{a_{\ce{CO2(aq)}}}. \]

With the concentration convention used here, its working expression is

\[ K_{\mathrm{eq}} =\frac{[\ce{Ca^2+}][\ce{HCO3^-}]^2}{[\ce{CO2(aq)}]}. \]

6.6

Write \(x=[\ce{Ca^2+}]\). If bicarbonate is the dominant anion, electroneutrality gives

\[ [\ce{HCO3^-}]\simeq2x. \]

The main equilibrium then yields

\[ K_{\mathrm{eq}}\simeq\frac{x(2x)^2}{[\ce{CO2(aq)}]}, \]

so

\[ x\simeq \left(\frac{K_{\mathrm{eq}}[\ce{CO2(aq)}]}{4}\right)^{1/3} = \left(\frac{(4.27932\times10^{-5})(1.2384\times10^{-5})}{4}\right)^{1/3} =5.09791\times10^{-4}\ \mathrm{M}. \]

This approximation gives

\[ [\ce{Ca^2+}] \simeq(5.09791\times10^{-4})(40.1)(10^3) =\boxed{20.4\ \mathrm{mg\,dm^{-3}}}. \]

To retain all the supplied equilibria, use the complete charge balance

\[ 2x+h=[\ce{HCO3^-}]+2[\ce{CO3^2-}]+[\ce{OH^-}], \]

together with

\[ [\ce{HCO3^-}]=\frac{A}{h}, \qquad [\ce{CO3^2-}]=\frac{K_{a2}A}{h^2}, \qquad x=\frac{K_{sp}h^2}{K_{a2}A}. \]

Substitution gives a single equation:

\[ \frac{2K_{sp}h^2}{K_{a2}A}+h =\frac{A+K_w}{h}+\frac{2K_{a2}A}{h^2}. \]

Its positive solution is

\[ h=5.45130\times10^{-9}\ \mathrm{M}, \qquad x=5.16232\times10^{-4}\ \mathrm{M}. \]

The corresponding anion concentrations are

\[ \begin{aligned} [\ce{HCO3^-}]&=1.01320\times10^{-3}\ \mathrm{M},\\ [\ce{CO3^2-}]&=8.71702\times10^{-6}\ \mathrm{M},\\ [\ce{OH^-}]&=1.83442\times10^{-6}\ \mathrm{M}. \end{aligned} \]

Thus the complete equilibrium calculation gives

\[ \boxed{ [\ce{Ca^2+}] =(5.16232\times10^{-4})(40.1)(10^3) =20.7\ \mathrm{mg\,dm^{-3}} }. \]

The dominant-bicarbonate approximation underestimates this result by about \(1.25\%\).

6.7

Rearranging the charge balance from 6.6,

\[ [\ce{HCO3^-}]+2[\ce{CO3^2-}]+[\ce{OH^-}]-[\ce{H+}] =2[\ce{Ca^2+}]. \]

Therefore, for the specified calcium carbonate–carbon dioxide–water system,

\[ \text{alkalinity}=2[\ce{Ca^2+}]. \]

Using the complete equilibrium result,

\[ \text{alkalinity} =2(5.16232\times10^{-4}) =1.03246\times10^{-3}\ \mathrm{mol\,L^{-1}}, \]

\[ \boxed{\text{alkalinity}=1.03\times10^{-3}\ \mathrm{mol\,L^{-1}}}. \]

Using the approximate calcium concentration instead gives

\[ \boxed{\text{alkalinity}\simeq1.02\times10^{-3}\ \mathrm{mol\,L^{-1}}}. \]

6.8

Convert the measured calcium concentration:

\[ x=[\ce{Ca^2+}] =\frac{100\times10^{-3}\ \mathrm{g\,dm^{-3}}} {40.1\ \mathrm{g\,mol^{-1}}} =2.49377\times10^{-3}\ \mathrm{M}. \]

Calcium carbonate saturation fixes the carbonate concentration:

\[ D=[\ce{CO3^2-}] =\frac{K_{sp}}{x} =\frac{4.50\times10^{-9}}{2.49377\times10^{-3}} =1.80450\times10^{-6}\ \mathrm{M}. \]

From the second acid dissociation,

\[ [\ce{HCO3^-}]=\frac{Dh}{K_{a2}}. \]

The complete charge balance is consequently

\[ 2x+h=\frac{Dh}{K_{a2}}+2D+\frac{K_w}{h}, \]

or

\[ \left(\frac{D}{K_{a2}}-1\right)h^2 -(2x-2D)h+K_w=0. \]

The root corresponding to the high-\(\ce{CO2}\) water specified in the question gives

\[ h=1.29536\times10^{-7}\ \mathrm{M}, \qquad [\ce{HCO3^-}]=4.98397\times10^{-3}\ \mathrm{M}. \]

The equilibrium from 6.5 now determines molecular dissolved carbon dioxide:

\[ [\ce{CO2(aq)}] =\frac{x[\ce{HCO3^-}]^2}{K_{\mathrm{eq}}} =\frac{(2.49377\times10^{-3})(4.98397\times10^{-3})^2} {4.27932\times10^{-5}} =1.44755\times10^{-3}\ \mathrm{M}. \]

Applying Henry’s law,

\[ p_{\ce{CO2}} =p^\circ\frac{[\ce{CO2(aq)}]/c^\circ}{K_{\ce{CO2}}} =(1.01\times10^5) \frac{1.44755\times10^{-3}}{3.44\times10^{-2}} =4.25007\times10^3\ \mathrm{Pa}. \]

\[ \boxed{p_{\ce{CO2}}=4.25\times10^3\ \mathrm{Pa}} \]

With the simpler approximation \([\ce{HCO3^-}]\simeq2x\),

\[ p_{\ce{CO2}} \simeq p^\circ\frac{4x^3}{K_{\mathrm{eq}}K_{\ce{CO2}}} =\boxed{4.26\times10^3\ \mathrm{Pa}}. \]

The headspace volume is unnecessary: the aqueous equilibrium composition determines the required equilibrium partial pressure.

The solution was checked against the official solution.