IChO 2004 · Theory problem 8

Colloids

Download PDFIChO 2004

The combination of an inorganic and an organic component on a nanometer scale yields materials with excellent properties. Thus the synthesis of hybrid nanoparticles is of interest.

(\(T = 298.15\ \mathrm{K}\) throughout whole problem)

Solution A is an aqueous solution of \(\ce{CaCl2}\) with a concentration of \(1.780\ \mathrm{g\,dm^{-3}}\).

Solution B is an aqueous solution of \(\ce{Na2CO3}\) with a concentration of \(1.700\ \mathrm{g\,dm^{-3}}\).

\[ pK_{a1}(\ce{H2CO3}) = 6.37 \qquad pK_{a2}(\ce{HCO3-}) = 10.33 \]

8.1 Calculate the \(pH\) of solution B using reasonable assumptions.

\(100\ \mathrm{cm^3}\) of solution A and \(100\ \mathrm{cm^3}\) of solution B are mixed to form solution C. Solution C is adjusted to \(pH\) 10. A precipitate forms.

\[ K_{sp}(\ce{Ca(OH)2}) = 6.46\times10^{-6} \qquad K_{sp}(\ce{CaCO3}) = 3.31\times10^{-9} \]

8.2 Show by calculation for each of the compounds \(\ce{Ca(OH)2}\) and \(\ce{CaCO3}\) whether it can be found in the precipitate or not.

In a similar experiment \(100\ \mathrm{cm^3}\) of solution A additionally contain \(2\ \mathrm{g}\) of a copolymer consisting of two water soluble blocks: a poly(ethylene oxide) block and a poly(acrylic acid) block:

Figure 1

The polymer does not undergo any chemical reaction (except protolysis of the acid) and yet has a strong effect: after mixing of the two solutions (A+B) no precipitate can be observed. Small calcium carbonate particles with the polymer chains attached to their surface form. The attached polymers prevent further crystal growth and the hybrid particles remain in solution.

8.3 Circle the block of the polymer (on the answer sheet) that attaches to the surface of the growing calcium carbonate crystal.

To characterize the hybrid particles they are separated from the preparation solution and transferred into \(50\ \mathrm{cm^3}\) of an aqueous \(\ce{NaOH}\) solution (\(c(\ce{NaOH}) = 0.19\ \mathrm{mol\,dm^{-3}}\)). The solution is diluted by the addition of \(200\ \mathrm{cm^3}\) of water. Assume that the new solution contains only the hybrid particles and no additional calcium or carbonate ions. All acidic groups participate in the acid-base equilibrium.

  • For the new solution, a \(pH\) of 12.30 is measured.
  • In electron microscopy you only can see the inorganic particles (not the polymer): Spherical particles of \(100\ \mathrm{nm}\) diameter are observed.
  • The molar mass of the hybrid particles (inorganic and organic part together) is determined to be \(M = 8.01\cdot10^8\ \mathrm{g\,mol^{-1}}\).
  • The charge of the particles is found to be \(Z = -800\) (number of unit charges).

(\(pK_a(\ce{COOH},\text{ copolymer}) = 4.88\))

8.4 How much of the initial amount of polymer (\(2\ \mathrm{g}\)) can still be found in the hybrid particles?

8.5 Calculate which modification of calcium carbonate has been formed.

Modification density
Calcite \(2.71\ \mathrm{g\,cm^{-3}}\)
Vaterite \(2.54\ \mathrm{g\,cm^{-3}}\)
Aragonite \(2.95\ \mathrm{g\,cm^{-3}}\)

Official English text of the problem.

Answers

  • 8.1: \(\boxed{pH \approx 11.24}\). The usual small-hydrolysis approximation gives \(11.27\).
  • 8.2: Calcium carbonate, \(\ce{CaCO3}\), [Ca+2].[O-]C(=O)[O-], precipitates. Calcium hydroxide, \(\ce{Ca(OH)2}\), [Ca+2].[OH-].[OH-], does not precipitate.
  • 8.3: Circle the right-hand poly(acrylic acid) block, containing eight repeat units. Its acid-form repeat-unit SMILES is [*]CC(C(=O)O)[*]; its deprotonated form is [*]CC(C(=O)[O-])[*].
  • 8.4: The formal acid–base calculation gives \(\boxed{m_{\mathrm{polymer}}\approx2.0\ \mathrm{g}}\), essentially 100% of the initial polymer.
  • 8.5: The individual-particle data give \(\boxed{\rho\approx2.54\ \mathrm{g\,cm^{-3}}}\), identifying vaterite, a calcium carbonate polymorph.

The results for 8.4 and 8.5 cannot both satisfy the overall material balance with all the supplied data. The titration in 8.4 gives \(4.512\times10^{-3}\ \mathrm{mol}\) of carboxylate charges, that is \(5.640\times10^{-6}\ \mathrm{mol}\) of particles with 800 charges each; at the stated molar mass this would be \(4.52\times10^3\ \mathrm{g}\) of hybrid, whereas the initial calcium and polymer allow only about \(2.16\ \mathrm{g}\).

Solution

Use \(K_w=1.00\times10^{-14}\) at \(298.15\ \mathrm{K}\), \(N_A=6.022\times10^{23}\ \mathrm{mol^{-1}}\), and standard atomic molar masses. Activities are approximated by concentrations, and volume changes other than the stated mixing and dilution are neglected.

8.1 — pH of solution B

The analytical concentration of sodium carbonate, [Na+].[Na+].[O-]C(=O)[O-], is

\[ C_B=\frac{1.700\ \mathrm{g\,dm^{-3}}} {105.99\ \mathrm{g\,mol^{-1}}} =0.01604\ \mathrm{mol\,dm^{-3}}. \]

Carbonate, [O-]C(=O)[O-], reacts with water, O, to give hydrogen carbonate, OC(=O)[O-], and hydroxide, [OH-]:

\[ \ce{CO3^2- + H2O <=> HCO3- + OH-}. \]

The corresponding base constant is

\[ K_b=\frac{K_w}{K_{a2}} =\frac{10^{-14}}{10^{-10.33}} =2.138\times10^{-4}. \]

Further hydrolysis to carbonic acid, OC(=O)O,

\[ \ce{HCO3- + H2O <=> H2CO3 + OH-}, \]

has \(K_b'=K_w/K_{a1}=2.34\times10^{-8}\) and is negligible here. Water autoionization also contributes negligibly to the hydroxide concentration.

Let \(x=[\ce{OH-}]\approx[\ce{HCO3-}]\). Then

\[ K_b=\frac{x^2}{C_B-x}, \]

so

\[ x=\frac{-K_b+\sqrt{K_b^2+4K_bC_B}}{2} =1.748\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

Therefore,

\[ pOH=2.7575, \qquad \boxed{pH=11.2425\approx11.24}. \]

If carbonate depletion is also neglected,

\[ x\approx\sqrt{K_bC_B}, \qquad pH\approx \frac{pK_w+pK_{a2}+\log C_B}{2} =11.27. \]

The quadratic calculation avoids this last approximation: about \(10.9\%\) of the carbonate undergoes hydrolysis.

8.2 — Composition of the precipitate

For calcium chloride, [Ca+2].[Cl-].[Cl-],

\[ c_A=\frac{1.780}{110.98} =0.01604\ \mathrm{mol\,dm^{-3}}. \]

Mixing equal volumes halves both analytical concentrations:

\[ C_{\mathrm{Ca}}=8.019\times10^{-3}\ \mathrm{mol\,dm^{-3}}, \qquad C_{\mathrm{carbonate}}=8.020\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

At \(pH=10\),

\[ [\ce{OH-}]=10^{-4}\ \mathrm{mol\,dm^{-3}}. \]

For calcium hydroxide, the dissolution equilibrium is

\[ \ce{Ca(OH)2(s) <=> Ca^2+ + 2OH-}. \]

The concentration of calcium(II) ions, [Ca+2], cannot exceed \(C_{\mathrm{Ca}}\). Thus the largest possible ion product is

\[ Q_{\ce{Ca(OH)2}} \leq (8.019\times10^{-3})(10^{-4})^2 =8.019\times10^{-11}. \]

Since

\[ 8.019\times10^{-11}\ll6.46\times10^{-6}, \]

\[ \boxed{\ce{Ca(OH)2}\text{ cannot precipitate}.} \]

For calcium carbonate, carbonate protonation must be considered. At \(pH=10\),

\[ \frac{[\ce{HCO3-}]}{[\ce{CO3^2-}]} =10^{pK_{a2}-pH} =10^{0.33}=2.138. \]

Also,

\[ \frac{[\ce{H2CO3}]}{[\ce{HCO3-}]} =10^{6.37-10} =2.34\times10^{-4}, \]

so carbonic acid makes a negligible contribution. Before precipitation, therefore,

\[ [\ce{CO3^2-}] \approx\frac{8.020\times10^{-3}}{1+2.138} =2.56\times10^{-3}\ \mathrm{mol\,dm^{-3}}. \]

The initial ion product is

\[ Q_{\ce{CaCO3}} =(8.019\times10^{-3})(2.56\times10^{-3}) =2.05\times10^{-5}. \]

Because

\[ 2.05\times10^{-5}\gg3.31\times10^{-9}, \]

calcium carbonate precipitates:

\[ \boxed{\ce{Ca^2+ + CO3^2- -> CaCO3(s)}}. \]

Its precipitation further lowers the calcium concentration, reinforcing the exclusion of calcium hydroxide.

8.3 — Polymer block attached to the crystal

Circle the entire right-hand bracketed block with subscript 8, containing the pendant \(\ce{COOH}\) groups.

This is the poly(acrylic acid) block:

  • Acid-form repeat unit: [*]CC(C(=O)O)[*].
  • Deprotonated polyacrylate repeat unit: [*]CC(C(=O)[O-])[*].

The negatively charged carboxylate groups adsorb at calcium sites on the growing crystal surface. The hydrated poly(ethylene oxide) block, repeat-unit SMILES [*]CCO[*], extends into the surrounding water and provides steric stabilization.

Here [*] marks continuation of the polymer chain; the repeat counts are eight for the acrylic acid block and 68 for the ethylene oxide block. The drawing specifies no tacticity. These fragment SMILES do not describe the spatial arrangement of the adsorbed chains.

8.4 — Amount of retained polymer

The intended titration balance attributes the consumption of added hydroxide to the polymer’s acidic groups:

\[ \ce{RCOOH + OH- -> RCOO- + H2O}, \]

where \(R\) denotes the polymer backbone.

At the measured pH,

\[ \frac{[\ce{RCOO-}]}{[\ce{RCOOH}]} =10^{12.30-4.88} =2.63\times10^7. \]

Consequently, the fraction of deprotonated acid groups is

\[ \alpha= \frac{1}{1+10^{4.88-12.30}} =0.999999962\approx1. \]

The added sodium hydroxide, [Na+].[OH-], supplies

\[ n_{\ce{OH-},0} =(0.050\ \mathrm{dm^3}) (0.19\ \mathrm{mol\,dm^{-3}}) =9.50\times10^{-3}\ \mathrm{mol}. \]

The final volume is

\[ V=0.050+0.200=0.250\ \mathrm{dm^3}. \]

Since \(pOH=14.00-12.30=1.70\),

\[ [\ce{OH-}]_{\mathrm{final}} =10^{-1.70} =1.995\times10^{-2}\ \mathrm{mol\,dm^{-3}}, \]

and hence

\[ n_{\ce{OH-},\mathrm{final}} =(0.250)(1.995\times10^{-2}) =4.988\times10^{-3}\ \mathrm{mol}. \]

Thus the amount of deprotonated carboxyl groups inferred from this balance is

\[ n_{\mathrm{carboxylate}} =9.50\times10^{-3}-4.988\times10^{-3} =4.512\times10^{-3}\ \mathrm{mol}. \]

Each polymer chain has eight acidic groups, so

\[ n_P=\frac{n_{\mathrm{carboxylate}}}{8\alpha} =5.640\times10^{-4}\ \mathrm{mol}. \]

The depicted chain contains 68 ethylene oxide repeat units, eight acrylic acid repeat units, and two terminal hydrogen atoms. Its acid-form molar mass is approximately

\[ M_P =68(44.052)+8(72.062)+2(1.008) =3574\ \mathrm{g\,mol^{-1}}. \]

Therefore,

\[ m_P=n_PM_P =(5.640\times10^{-4})(3574) =2.016\ \mathrm{g}. \]

With the precision of the supplied measurements,

\[ \boxed{m_P\approx2.0\ \mathrm{g},\qquad \text{retained fraction}\approx100\%.} \]

The small calculated excess over the initial \(2\ \mathrm{g}\) is consistent with rounding: complete neutralization of exactly \(2.000\ \mathrm{g}\) would give \(pH\approx12.303\).

8.5 — Calcium carbonate modification

For a neutral, stoichiometric calcium carbonate core, the negative particle charge is attributed to the polymer’s deprotonated carboxyl groups. Each chain contributes approximately eight negative unit charges. Thus the number of attached chains per particle is

\[ N_P=\frac{|Z|}{8\alpha} \approx\frac{800}{8} =100. \]

Allowing for loss of the acidic hydrogen atoms, the organic contribution per mole of hybrid particles is

\[ M_{\mathrm{organic}} \approx100\bigl(3574-8(1.008)\bigr) =3.57\times10^5\ \mathrm{g\,mol^{-1}}. \]

The inorganic contribution is therefore

\[ M_{\mathrm{inorganic}} =8.01\times10^8-3.57\times10^5 =8.0064\times10^8\ \mathrm{g\,mol^{-1}}. \]

Electron microscopy gives the inorganic diameter. Hence

\[ r=50\ \mathrm{nm}=5.00\times10^{-6}\ \mathrm{cm}, \]

and the inorganic volume per particle is

\[ V_{\mathrm{inorganic}} =\frac{4\pi r^3}{3} =5.236\times10^{-16}\ \mathrm{cm^3}. \]

The inorganic mass per particle is

\[ m_{\mathrm{inorganic}} =\frac{M_{\mathrm{inorganic}}}{N_A} =1.330\times10^{-15}\ \mathrm{g}. \]

Therefore,

\[ \rho =\frac{m_{\mathrm{inorganic}}}{V_{\mathrm{inorganic}}} =\frac{1.330\times10^{-15}}{5.236\times10^{-16}} \approx2.54\ \mathrm{g\,cm^{-3}}. \]

This matches the tabulated density of vaterite:

\[ \boxed{\text{The individual-particle data identify vaterite.}} \]

Vaterite, calcite, and aragonite all have the ionic composition represented by [Ca+2].[O-]C(=O)[O-]. This SMILES does not distinguish their crystal lattices.

The solution was checked against the official solution.