IChO 1992 · Theory problem 7

Determination of chloride in Chesapeake Bay water

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When the fresh-water rivers that run into the Chesapeake Bay flood after heavy rains in the spring, the increase in fresh water in the Bay causes a decrease in the salinity in the areas where oysters grow. The minimum concentration of chloride ions needed in oyster beds for normal growth is 8 ppm (\(8\ \mathrm{mg\,dm^{-3}}\)).

After one week of heavy rain, the following analysis is done on water from the bay. To a \(50.00\ \mathrm{cm^3}\) sample of bay water a few drops of a \(\ce{K2CrO4}\) solution are added. The sample is then titrated with \(16.16\ \mathrm{cm^3}\) of a 0.00164 M \(\ce{AgNO3}\) solution. After \(\ce{AgNO3}\) solution has been added to the sample a bright red-orange precipitate forms.

7.1 What is the molar concentration of chloride in the sample?

7.2 Does the water contain sufficient chloride for the normal growth of oysters? Show your calculation.

7.3 Write a balanced equation for the reaction of the analyte with the titrant.

7.4 Write a balanced net-ionic equation that describes the reaction responsible for the colour change at the endpoint of the titration. Which compound produces the brick-red colour?

7.5 The concentration of chromate at the endpoint is 0.020 M. Calculate the concentration of chloride ions in the solution when the red precipitate forms.

7.6 For this titration to work most effectively, the solution being titrated must be neutral or slightly basic. Write a balanced equation for the competing reaction that would occur in acidic medium that would influence the observed endpoint of this titration.

Typically, a buffer is added to the solution being titrated to control the pH if the initial sample is acidic. Suppose the pH of the sample of bay water was 5.10, thus too acidic to perform the analysis accurately.

7.7 Select a buffer from the list that would enable you to establish and maintain a pH of 7.20 in aqueous medium. Show the calculations which lead to your choice.

Buffer systems \(K_\mathrm{a}\) of weak acid
1. 0.10 M lactic acid / 0.10 M sodium lactate \(1.4\times10^{-4}\)
2. 0.10 M acetic acid / 0.10 M sodium acetate \(1.8\times10^{-5}\)
3. 0.10 M sodium dihydrogen phosphate /
/ 0.10 M sodium hydrogen phosphate		 \(6.2\times10^{-8}\)
4. 0.10 M ammonium chloride / 0.10 M ammonia \(5.6\times10^{-10}\)

7.8 Using the selected buffer system, calculate the mass (in g) of weak acid and of conjugated base you would need to dissolve in distilled water to prepare \(500\ \mathrm{cm^3}\) of a stock solution buffered at a pH of 7.2.

7.9 The chloride concentration in another \(50.00\ \mathrm{cm^3}\) sample of bay water was determined by the Volhard method. In this method an excess of \(\ce{AgNO3}\) is added to the sample. The excess \(\ce{Ag+}\) is titrated with standardized \(\ce{KSCN}\), forming a precipitate of \(\ce{AgSCN}\). The endpoint is signalled by the formation of the reddish-brown \(\ce{FeSCN^2+}\) complex that forms when \(\ce{Ag+}\) is depleted. If the excess \(\ce{Ag+}\) from the addition of \(50.00\ \mathrm{cm^3}\) of 0.00129 M \(\ce{AgNO3}\) to the water sample required \(27.46\ \mathrm{cm^3}\) of \(1.41\,10^{-3}\) M \(\ce{KSCN}\) for titration, calculate the concentration of chloride in the bay water sample.

In natural waters with much higher concentration of \(\ce{Cl-}\), the \(\ce{Cl-}\) can be determined gravimetrically by precipitating the \(\ce{Cl-}\) as \(\ce{AgCl}\). A complicating feature of this method is the fact that \(\ce{AgCl}\) is susceptible to photodecomposition as shown by the reaction:

Figure 1

Furthermore, if this photodecomposition occurs in the presence of excess \(\ce{Ag+}\), the following additional reaction occurs:

Figure 2

If 0.010 g of a 3.000 g sample of \(\ce{AgCl}\) contaminated with excess \(\ce{Ag+}\) undergoes photodecomposition by the above equations

7.10 Will the apparent weight of \(\ce{AgCl}\) be too high or too low? Explain your answer showing by how many grams the two values will differ.

Data:

\[ K_{\mathrm{sp}}(\ce{AgCl})=1.78\times10^{-10} \]

\[ K_{\mathrm{sp}}(\ce{Ag2CrO4})=1.00\times10^{-12} \]

Official English text of the problem.

Answers

  • 7.1 \(\boxed{[\ce{Cl-}]=5.30\times10^{-4}\ \mathrm{mol\,dm^{-3}}}\), using the usual stoichiometric endpoint approximation.
  • 7.2 Yes. The chloride concentration is \(\boxed{18.8\ \mathrm{mg\,dm^{-3}}}\), exceeding \(8\ \mathrm{mg\,dm^{-3}}\).
  • 7.3 \(\boxed{\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}}\).
  • 7.4 \(\boxed{\ce{2Ag+(aq) + CrO4^2-(aq) -> Ag2CrO4(s)}}\). Silver chromate produces the brick-red colour.
  • 7.5 \(\boxed{[\ce{Cl-}]\approx2.5\times10^{-5}\ \mathrm{mol\,dm^{-3}}}\).
  • 7.6 \(\boxed{\ce{2CrO4^2-(aq) + 2H+(aq) <=> Cr2O7^2-(aq) + H2O(l)}}\).
  • 7.7 Buffer 3, the dihydrogen phosphate/hydrogen phosphate pair: \[ \boxed{pK_\mathrm{a}=7.208,\qquad \frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]}=0.983}. \]
  • 7.8 For anhydrous salts, fixing the acid component at \(0.100\ \mathrm M\), use approximately \(\boxed{6.00\ \mathrm g\ \ce{NaH2PO4}}\) and \(\boxed{6.97\ \mathrm g\ \ce{Na2HPO4}}\), diluted to \(500\ \mathrm{cm^3}\). The nominal equal-\(0.10\ \mathrm M\) recipe uses \(6.00\ \mathrm g\) and \(7.10\ \mathrm g\), respectively, and gives pH \(7.208\approx7.2\).
  • 7.9 \(\boxed{[\ce{Cl-}]=5.16\times10^{-4}\ \mathrm{mol\,dm^{-3}}}\), equivalent to \(18.3\ \mathrm{mg\,dm^{-3}}\).
  • 7.10 The apparent mass is too high, by \(\boxed{0.0059\ \mathrm g}\); the measured solid has mass approximately \(\boxed{3.006\ \mathrm g}\).

Solution

Dot-disconnected SMILES below specify the ionic composition of salts. Their crystal lattices are not represented by these strings. Oxyanion SMILES give one formal bond and charge representation; the actual bonding includes resonance delocalization.

7.1

The chloride ion, [Cl-], reacts with the silver(I) ion, [Ag+], in a \(1:1\) molar ratio. Using the usual approximation that the observed endpoint represents stoichiometric equivalence,

\[ \begin{aligned} n(\ce{Cl-}) &=n(\ce{Ag+})\\ &=(0.00164\ \mathrm{mol\,dm^{-3}}) (16.16\times10^{-3}\ \mathrm{dm^3})\\ &=2.65024\times10^{-5}\ \mathrm{mol}. \end{aligned} \]

The original sample volume is \(50.00\times10^{-3}\ \mathrm{dm^3}\), so

\[ [\ce{Cl-}] =\frac{2.65024\times10^{-5}} {50.00\times10^{-3}} =\boxed{5.30\times10^{-4}\ \mathrm{mol\,dm^{-3}}}. \]

7.2

Using \(M(\ce{Cl})=35.45\ \mathrm{g\,mol^{-1}}\),

\[ \begin{aligned} \rho(\ce{Cl-}) &=(5.30048\times10^{-4}\ \mathrm{mol\,dm^{-3}}) (35.45\ \mathrm{g\,mol^{-1}})\\ &=0.0187902\ \mathrm{g\,dm^{-3}}\\ &=\boxed{18.8\ \mathrm{mg\,dm^{-3}}}. \end{aligned} \]

Since \(18.8>8\), the sample contains sufficient chloride for normal oyster growth under the criterion stated in the problem.

7.3

The analyte is chloride, and the reacting titrant species is silver(I). They form insoluble silver chloride, [Ag+].[Cl-]:

\[ \boxed{\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}}. \]

This net-ionic equation applies irrespective of the counterions originally accompanying chloride in the sample.

7.4

The indicator supplies chromate ions, [O-][Cr](=O)(=O)[O-]. Chromate has an approximately tetrahedral arrangement of oxygen atoms around chromium.

As the free silver(I) concentration rises, silver chromate precipitates:

\[ \boxed{\ce{2Ag+(aq) + CrO4^2-(aq) -> Ag2CrO4(s)}}. \]

The brick-red solid is silver chromate, [Ag+].[Ag+].[O-][Cr](=O)(=O)[O-].

7.5

At the onset of silver chromate precipitation,

\[ K_{\mathrm{sp}}(\ce{Ag2CrO4}) =[\ce{Ag+}]^2[\ce{CrO4^2-}]. \]

Consequently,

\[ [\ce{Ag+}] =\sqrt{\frac{1.00\times10^{-12}}{0.020}} =7.071\times10^{-6}\ \mathrm{mol\,dm^{-3}}. \]

Solid silver chloride is also present, so its solubility equilibrium requires

\[ K_{\mathrm{sp}}(\ce{AgCl}) =[\ce{Ag+}][\ce{Cl-}]. \]

Thus,

\[ \begin{aligned} [\ce{Cl-}] &=\frac{1.78\times10^{-10}}{7.071\times10^{-6}}\\ &=2.517\times10^{-5}\ \mathrm{mol\,dm^{-3}}\\ &\approx\boxed{2.5\times10^{-5}\ \mathrm{mol\,dm^{-3}}}. \end{aligned} \]

This is the dissolved chloride concentration in the titration mixture when the red precipitate first forms.

7.6

Acid converts chromate into dichromate:

\[ \boxed{\ce{2CrO4^2-(aq) + 2H+(aq) <=> Cr2O7^2-(aq) + H2O(l)}}. \]

The species introduced here are the hydrogen ion (proton), [H+]; water, O; and the dichromate ion, [O-][Cr](=O)(=O)O[Cr](=O)(=O)[O-]. Dichromate consists of two chromium–oxygen tetrahedra sharing one oxygen atom.

Increasing acidity shifts this equilibrium to the right, lowering the chromate concentration. Since precipitation requires

\[ [\ce{Ag+}] =\sqrt{\frac{K_{\mathrm{sp}}(\ce{Ag2CrO4})} {[\ce{CrO4^2-}]}}, \]

less chromate means that a higher silver(I) concentration is needed to produce the indicator precipitate. The endpoint is therefore delayed.

7.7

A buffer works most effectively when its acid and conjugate-base concentrations are comparable, so its \(pK_\mathrm a\) should be close to the desired pH.

From \(pK_\mathrm a=-\log_{10}K_\mathrm a\):

Buffer number \(pK_\mathrm a\) Required base/acid ratio at pH 7.20
1 3.854 \(2.22\times10^3\)
2 4.745 \(2.85\times10^2\)
3 7.208 \(0.983\)
4 9.252 \(8.88\times10^{-3}\)

The ratios follow from the Henderson–Hasselbalch equation,

\[ \mathrm{pH}=pK_\mathrm a+ \log_{10}\frac{[\text{conjugate base}]}{[\text{weak acid}]}. \]

Therefore, select buffer 3. Its weak acid is the dihydrogen phosphate ion, O=P(O)(O)[O-], and its conjugate base is the hydrogen phosphate ion, O=P(O)([O-])[O-]. Both have approximately tetrahedral coordination around phosphorus.

Their equilibrium is

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

At pH 7.20,

\[ \begin{aligned} \frac{[\ce{HPO4^2-}]}{[\ce{H2PO4-}]} &=\frac{K_\mathrm a}{[\ce{H+}]}\\ &=\frac{6.2\times10^{-8}}{10^{-7.20}}\\ &=\boxed{0.9826\approx0.983}. \end{aligned} \]

The nearly equal amounts provide effective buffering. The base component neutralizes added acid through

\[ \ce{HPO4^2-(aq) + H+(aq) -> H2PO4-(aq)}. \]

7.8

Assume anhydrous salts:

  • Sodium dihydrogen phosphate, \(\ce{NaH2PO4}\), [Na+].O=P(O)(O)[O-], with \(M=119.98\ \mathrm{g\,mol^{-1}}\).
  • Disodium hydrogen phosphate, \(\ce{Na2HPO4}\), [Na+].[Na+].O=P(O)([O-])[O-], with \(M=141.96\ \mathrm{g\,mol^{-1}}\).

Take the acid concentration as \(0.100\ \mathrm{mol\,dm^{-3}}\), following the nominal concentration in the table. For \(500\ \mathrm{cm^3}=0.500\ \mathrm{dm^3}\),

\[ n(\ce{NaH2PO4}) =(0.100)(0.500)=0.0500\ \mathrm{mol}. \]

The required base amount is

\[ n(\ce{Na2HPO4}) =0.9826(0.0500) =0.04913\ \mathrm{mol}. \]

Therefore,

\[ m(\ce{NaH2PO4}) =(0.0500)(119.98) \approx\boxed{6.00\ \mathrm g}, \]

\[ m(\ce{Na2HPO4}) =(0.04913)(141.96) \approx\boxed{6.97\ \mathrm g}. \]

Dissolve both salts in distilled water and dilute to a final solution volume of \(500\ \mathrm{cm^3}\).

If the tabulated \(0.10\ \mathrm M/0.10\ \mathrm M\) composition is used directly, each component requires \(0.0500\ \mathrm{mol}\). The corresponding masses are \(6.00\ \mathrm g\) and \(7.10\ \mathrm g\), and

\[ \mathrm{pH}=pK_\mathrm a+\log_{10}1=7.208\approx7.2. \]

Thus this nominal recipe also meets the pH stated to one decimal place in 7.8.

7.9

Thiocyanate ion, [S-]C#N, precipitates excess silver(I) as silver thiocyanate, [Ag+].[S-]C#N:

\[ \ce{Ag+(aq) + SCN-(aq) -> AgSCN(s)}. \]

Both the chloride precipitation and this back-titration have \(1:1\) stoichiometry. Hence,

\[ n(\ce{Cl-}) =n(\text{silver added})-n(\text{silver remaining}). \]

The amount of silver added is

\[ n_\mathrm{added} =(50.00\times10^{-3})(0.00129) =6.450\times10^{-5}\ \mathrm{mol}. \]

The thiocyanate titration measures

\[ n_\mathrm{remaining} =(27.46\times10^{-3})(1.41\times10^{-3}) =3.87186\times10^{-5}\ \mathrm{mol}. \]

Therefore,

\[ n(\ce{Cl-}) =6.450\times10^{-5}-3.87186\times10^{-5} =2.57814\times10^{-5}\ \mathrm{mol}. \]

Using the original water-sample volume,

\[ [\ce{Cl-}] =\frac{2.57814\times10^{-5}}{50.00\times10^{-3}} =\boxed{5.16\times10^{-4}\ \mathrm{mol\,dm^{-3}}}. \]

The corresponding mass concentration is

\[ (5.15628\times10^{-4})(35.45)(10^3) =\boxed{18.3\ \mathrm{mg\,dm^{-3}}}. \]

7.10

Photodecomposition produces metallic silver, represented atomically by [Ag], and chlorine, ClCl:

\[ \ce{2AgCl(s) -> 2Ag(s) + Cl2(g)}. \]

The chlorine then reacts with excess silver(I):

\[ \ce{3Cl2(g) + 3H2O(l) + 5Ag+(aq) -> 5AgCl(s) + ClO3-(aq) + 6H+(aq)}. \]

The soluble chlorine-containing product is chlorate ion, [O-][Cl](=O)=O.

For every six moles of silver chloride photodecomposed, three moles of chlorine form. These generate five moles of new silver chloride. The combined bookkeeping equation is

\[ \ce{6AgCl(s) + 3H2O(l) + 5Ag+(aq) -> 6Ag(s) + 5AgCl(s) + ClO3-(aq) + 6H+(aq)}. \]

Both metallic silver and newly precipitated silver chloride remain in the weighed solid.

Using

\[ M(\ce{AgCl})=107.87+35.45=143.32\ \mathrm{g\,mol^{-1}}, \]

the amount initially photodecomposed is

\[ n_\mathrm{decomp} =\frac{0.010}{143.32} =6.98\times10^{-5}\ \mathrm{mol}. \]

The retained metallic silver has mass

\[ m(\ce{Ag}) =\frac{0.010}{143.32}(107.87) =0.00753\ \mathrm g. \]

The newly formed silver chloride has mass

\[ m(\ce{AgCl})_\mathrm{new} =\frac56\left(\frac{0.010}{143.32}\right)(143.32) =0.00833\ \mathrm g. \]

Thus the mass change is

\[ \begin{aligned} \Delta m &=m(\ce{Ag})+m(\ce{AgCl})_\mathrm{new} -m(\ce{AgCl})_\mathrm{decomposed}\\ &=0.007526+0.008333-0.010\\ &=+0.00586\ \mathrm g. \end{aligned} \]

The apparent mass is therefore too high by approximately \(0.0059\ \mathrm g\):

\[ m_\mathrm{apparent} =3.000+0.00586 =3.00586\ \mathrm g \approx\boxed{3.006\ \mathrm g}. \]

The gain occurs because additional silver from solution becomes incorporated into the collected solid.

The solution was checked against the official solution.