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Chemical Equilibria and Sample Preparation (Q1 – Q13)

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    199368
  • Q1. Which of the cationic solutes in Lithia water are main group ions? Which of the cationic solutes are transition metal ions? Evaluate whether main group and transition metal cations have lost all of their valence electrons. Can main group and/or transition metal cations potentially be oxidized by dissolved oxygen?

    The only transition metal that is recognized as an inorganic constituent in Lithia water is iron. All of the other inorganic cations of historical significance determined in this course (K+, Na+, Li+, Ca2+, Mg2+) are main group ions. Main group cations tend to exist in their most oxidized state. Since there are no additional valence electrons, main group cations are unable to function as reductants and cannot reduce elemental oxygen to the oxide. Ferrous ion (Fe2+) can theoretically be oxidized to the ferric ion (Fe3+) and act as a reductant for species such as elemental oxygen.

    Q2. Using the standard reduction tables in Analytical Chemistry 2.0, draw a ladder diagram for the redox couple of dissolved oxygen and Fe3+/2+ under standard conditions. Predict the predominant oxidation state for iron under these conditions.

    For this redox reaction:

    \[\begin{align}
    &\ce{O2 (g) + 4 H+ (aq) + 4 e- \rightleftharpoons 2 H2O (l)} &&\mathrm{E° = 1.23\: V}\nonumber\\
    &\ce{Fe^3+ (aq) + e- \rightleftharpoons Fe^2+ (aq)} &&\mathrm{E° = 0.771\: V}\nonumber
    \end{align}\nonumber\]

    Q2LadderDiagram.png

    Based on this ladder diagram, the predominant species in this solution under standard conditions would be H2O and Fe3+.

    Q3. Using the standard reduction tables in Analytical Chemistry 2.0, draw a ladder diagram for the redox couple of dissolved oxygen and Fe3+/2+ assuming the following non-standard conditions: assume that the dissolved oxygen concentration is 10 mg L-1, the total iron concentration is 10 mg L-1, and that the initial ratio of [Fe2+]/[Fe3+] is 1000. Predict the predominant oxidation state for iron under these conditions and compare the result to your answer in question 2.

    For the redox couple of dissolved oxygen and ferric/ferrous ion, the following calculations need to be performed before constructing the ladder diagram:

    Potential of the dissolved oxygen half-reaction:

    \[\ce{O2 (g) + 4 H+ (aq) + 4 e- \rightleftharpoons 2 H2O (l)}\nonumber\]

    \[[H^+ ]=10^{-pH}=10^{-6.4}=3.98 \times 10^{-7}\: M \nonumber\]

    \[[O_2 ]= \left(\dfrac{10.0\: mg\: O_2}{L}\right)\left(\dfrac{1.00\: mmol\: O_2}{32.0\: mg\: O_2}\right)\left(\dfrac{1\: mol\: O_2}{1000\: mmol\: O_2}\right)=3.1 \times 10^{-4}\: M\nonumber\]

    \[\begin{align}
    E_{O_2} &= E_{O_2}^o-\dfrac{0.05915}{n} \log \dfrac{1}{[O_2 ] [H^+ ]^4}\nonumber\\
    &= 1.23- \dfrac{0.05915}{4} \log \dfrac{1}{(3.1 \times 10^{-4} ) (3.98 \times 10^{-7} )^4}\nonumber\\
    &= 0.800\: V\nonumber
    \end{align}\nonumber\]

    \[\begin{align}
    E_{Fe^{3+}/Fe^{2+}} &= E_{Fe^{3+}/Fe^{2+}}^o-\dfrac{0.05915}{n} \log \dfrac{[Fe^{2+} ]}{[Fe^{3+} ]} \nonumber\\
    &= 0.771- \dfrac{0.05915}{1} \log \dfrac{1000}{1}\nonumber\\
    &=0.594\: V\nonumber
    \end{align}\nonumber\]

    Q3LadderDiagram.png

    The predominant oxidation state for iron under these conditions is Fe3+.

    Q4. Refer to the plaque shown in Fig. 2 of the Major Inorganic Constituents section. What are the two predominant anions in Lithia water? Using the standard reduction tables in Analytical Chemistry 2.0, is there any reason to expect these anions to react with dissolved oxygen at concentrations expected in Lithia water? Provide appropriate support for your answer.

    The two predominant anions in Lithia water are chloride and bicarbonate. Chloride ion may be oxidized to chlorine, and the Nernst equation plus a ladder diagram may be used to determine whether dissolved oxygen has sufficient capability as an oxidant under conditions found in Lithia water. Atmospheric chlorine is extremely reactive and is removed efficiently from the troposphere by chemical reactions. The smell of chlorine can be detected at 3 – 5 ppm. The determination of atmospheric chlorine in Ashland, OR has not been undertaken but a conservative estimate would be 1 pptv (part per trillion by volume) which is 1 pmol Cl2 per mol of air. Assuming air is an ideal gas:

    \[\ce{Cl2 (g) + 2e- \rightleftharpoons 2 Cl- } \hspace{30px} \mathrm{E^o = 1.396\: V}\nonumber\]

    \[[Cl_2 ]= \left(\dfrac{1 \times 10^{-12}\: mol\: Cl_2}{mol\: air}\right)\left(\dfrac{1\: mol\: air}{22.4\: L}\right)=4.5 \times 10^{-14}\: M\: Cl_2\nonumber\]

    \[\begin{align}
    E_{Cl_2/Cl^-} &= E_{Cl_2/Cl^-}^o-\dfrac{0.05915}{n} \log \dfrac{[Cl^- ]^2}{[Cl_2 ]}\nonumber\\
    &= 1.396- \dfrac{0.05915}{2} \log \dfrac{(7.777\times10^{-2} )^2}{4.5 \times 10^{-14}} \nonumber\\
    &=1.067\: V\nonumber
    \end{align}\nonumber\]

    Q4LadderDiagram.png

    According to the ladder diagram, dissolved oxygen and chloride ion do coexist in Lithia water. Under these conditions, dissolved oxygen is not a sufficient oxidant to oxidize chloride ion to chlorine.

    In the bicarbonate ion, the oxidation number for carbon is +4, which is the highest oxidation state for carbon since elemental carbon has only four valence electrons. Therefore, dissolved oxygen cannot oxidize the bicarbonate ion.

    Q5. Using the standard reduction tables in Analytical Chemistry 2.0, is there any reason to expect main group cations to react with transition metal cations present in Lithia water? Provide appropriate support for your answer.

    The predominant main group cations in Lithia water are Li+, Na+, K+, Mg2+, and Ca2+. The predominant transition metal in Lithia water is Fe2+. The standard reduction potentials for the main group cations to their zero valent state and the standard reduction potential for the ferric-ferrous redox couple are listed in the table below:

    Standard Reduction Half-Reaction

    Eo (V)

    \(\ce{Li+ + e- ⇌ Li (s)}\)

    -3.04

    \(\ce{Na+ + e- ⇌ Na (s)}\)

    -2.713

    \(\ce{K+ + e- ⇌ K (s)}\)

    -2.93

    \(\ce{Mg^2+ + 2e- ⇌ Mg (s)}\)

    -2.356

    \(\ce{Ca^2+ + 2e- ⇌ Ca (s)}\)

    -2.84

    \(\ce{Fe^3+ + e- ⇌ Fe^2+ (aq)}\)

    +0.771

    All of the main group cations exist in their fully oxidized state and are extremely weak oxidants as evidenced by the extremely negative (unfavorable) standard reduction potentials. The oxidation of ferrous ion to ferric ion is unfavorable except in the presence of a strong oxidant. Although an instructor may want students to construct a ladder diagram and use the Nernst equation to calculate the electrochemical potential for each half-cell, one can determine by inspection that the reduction of Group IA and IIA cations by the oxidation of ferrous ions to ferric ions is extremely unfavorable. Therefore, no redox reaction should take place among the main group ions and ferrous ions.

    Q6. Briefly summarize the redox chemistry that can take place among the major inorganic cations in Lithia water.

    Dissolved oxygen acts as an oxidant to oxidize ferrous ions to ferric ions. The oxidation of ferrous ions to ferric ions leads to the precipitation of ferric hydroxide, which appears as a fine orange precipitate at the bottom of the Lithia water sample. The main group cations are stable in Lithia water as they have large negative standard reduction potentials, which make them extremely weak oxidants.

    Q7. Using the solubility rules provided above and the primary inorganic constituents in Lithia water, which inorganic salts would be most likely to form a precipitate in Lithia water?

    Carbonates and hydroxides would be the most likely inorganic salts to form a precipitate in Lithia water.

    Q8. Why is there no term for the solid calcium carbonate in the Ksp expression?

    Unlike the concentration of dissolved ions in aqueous solution, which is an extensive property and varies based on the number of ions in a given volume of solvent, the concentration of a pure solid (or a pure liquid) is an intensive property and is constant. Equilibrium constant expressions do not contain concentration terms for pure solids or liquids, and they are given a value of 1 in the equilibrium constant expression.

    Q9. Write expressions that relate the concentration of calcium ion and the concentration of phosphate ion to the solubility of calcium phosphate in water.

    \[\begin{array}{llclcl}\ce{
    &Ca3(PO4)2 (s) &⇌ &3 Ca^2+ (aq) &+ &2 PO4^3- (aq)\\
    Initial\: (I) &------ & &0 & &0\\
    Change\: (C) &-X & &+3X & &+2X\\
    Equilibrium\: (E)&------ & &3X & &2X}
    \end{array}\nonumber\]

    If the molar solubility of calcium phosphate is X, then

    \[\mathrm{3X = [Ca^{2+}] \hspace{20px} or \hspace{20px} X = \dfrac{[Ca^{2+}]}{3}}\nonumber\]

    \[\mathrm{2X = [PO_4^{3-}] \hspace{20px} or \hspace{20px} X = \dfrac{[PO_4^{3-}]}{2}}\nonumber\]

    So the molar solubility of Ca3(PO4)2 can be expressed as either ([Ca2+]/3) or ([PO43-]/2).

    Q10. Calculate the solubility of calcium hydroxide in a solution that has a pH of 12. Note: The pH has been adjusted to 12 by some other means than the dissolution of calcium hydroxide. If calcium hydroxide is added to the water with a pH of 12, some of the calcium hydroxide will dissolve to satisfy the Ksp equilibrium expression for calcium hydroxide.

    The solubility equilibrium established in a saturated solution of Ca(OH)2 is:

    \[\ce{Ca(OH)2 (s) ⇌ Ca^2+ (aq) + 2 OH- (aq)}\nonumber\]

    The equilibrium constant expression for Ca(OH)2 is:

    \[K_{sp} = [Ca^{2+}][OH^-]^2\nonumber\]

    The molar solubility of Ca(OH)2 can be defined as [Ca2+] because dissolving one mole of Ca(OH)2 provides one mole of calcium ions. Solving for the molar solubility of Ca(OH)2 may be accomplished using the following steps:

    \[[H^+ ]= 10^{-pH}= 1.0 \times 10^{-12}\: M \nonumber\]

    \[[OH^-] = \dfrac{K_w}{[H^+]}=\dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-12} } = 1.0 \times 10^{-2}\: M \nonumber\]

    \[[Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(0.010)^2} =0.065\: M \nonumber\]

    Q11. What is the molar solubility and mass solubility (i.e. solubility in grams of solid per liter of solution) of each of the metal hydroxides listed in Table 1 at pH values of 10.0, 6.0, and 2.0? How is the solubility of a metal hydroxide affected by pH? Offer an explanation for this behavior using LeChatelier’s principle.

    To solve for the molar solubility of each metal hydroxide at each pH, the hydroxide ion concentration must be calculated at each pH and substituted into the equilibrium constant expression to solve for the molar solubility of the salt. The molar solubility can be converted to mass solubility by multiplying the molar solubility by the molar mass of the salt.

    Step 1: Calculate the hydroxide ion concentration at pH 10.0, 6.0, and 2.0

    \[[H^+ ]= 10^{-pH}= 10^{-10.0}= 1.0 \times 10^{-10}\: M \nonumber\]

    \[[OH^-] = \dfrac{K_w}{[H^+]}=\dfrac{1.0 \times 10^{-14}}{1.0 \times 10^{-10}} = 1.0 \times 10^{-4}\: M \nonumber\]

    \[At\: pH\: 6.0,\: [OH^- ]=1.0 \times 10^{-8}\: M \nonumber\]

    \[At\: pH\: 2.0,\: [OH^- ]=1.0 \times 10^{-12}\: M \nonumber\]

    Step 2: Calculate the molar solubility of calcium hydroxide at each pH level

    \[At\: pH\: 10.0,\:[Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-4}\: M)^2} =6.5 \times 10^2\: M \nonumber\]

    \[At\: pH\: 6.0,\:[Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-8}\: M)^2} =6.5 \times 10^{10}\: M \nonumber\]

    \[At\: pH\: 2.0,\: [Ca^{2+} ]= \dfrac{K_{sp}}{[OH^- ]^2} =\dfrac{6.5 \times 10^{-6}}{(1.0 \times 10^{-12}\: M)^2} =6.5 \times 10^{18}\: M \nonumber\]

    Step 3: Calculate the mass solubility (S) of calcium hydroxide at each pH level

    \[\begin{align}
    At\: pH\: 10.0,\: S_{Ca(OH)_2} &= [Ca^{2+} ] MW_{Ca(OH)_2 }\nonumber\\
    &= (6.5 \times 10^2\: M\: Ca^{2+} )\left(\dfrac{1\: mol\: Ca(OH)_2}{1\: mol\: Ca^{2+}}\right)\left(\dfrac{74.093\: g\: Ca(OH)_2}{1\: mol\: Ca(OH)_2}\right)\nonumber\\
    &=\dfrac{4.8 \times 10^4\: g\: Ca(OH)_2}{L\: solution}\nonumber
    \end{align}\nonumber\]

    \[At\: pH\: 6.0,\: S_{Ca(OH)_2} =\dfrac{4.8 \times 10^{12}\: g\: Ca(OH)_2}{L\: solution}\nonumber\]

    \[At\: pH\: 2.0,\: S_{Ca(OH)_2} =\dfrac{4.8 \times 10^{19}\: g\: Ca(OH)_2}{L\: solution}\nonumber\]

    The table below contains the molar solubility and mass solubility for all the hydroxide salts from Table 1 in the Chemical Equilibria document.

    Molar Solubility (M)

    Mass Solubility (g/L)

    Salt

    Ksp

    pH 10

    pH 6

    pH 2

    pH 10

    pH 6

    pH 2

    Ca(OH)2

    6.5 x 10-6

    6.5 x 102

    6.5 x 1010

    6.5 x 1018

    4.8 x 104

    4.8 x 1012

    4.8 x 1018

    Mg(OH)2

    7.1 x 10-12

    7.1 x 10-4

    7.1 x 104

    7.1 x 1012

    4.1 x 10-2

    4.1 x 106

    4.1 x 1014

    Al(OH)3

    4.6 x 10-33

    4.6 x 10-21

    4.6 x 10-9

    4.6 x 103

    3.6 x 10-19

    3.6 x 10-7

    3.6 x 105

    Fe(OH)2

    8.0 x 10-16

    8.0 x 10-8

    8.0 x 100

    8.0 x 108

    7.2 x 10-6

    7.2 x 102

    7.2 x 1010

    Fe(OH)3

    1.6 x 10-39

    1.6 x 10-27

    1.6 x 10-15

    1.6 x 10-3

    1.7 x 10-25

    1.7 x 10-13

    1.7 x 10-1

    Note that for salts having a relatively large Ksp or under acidic conditions, the predicted molar solubility and mass solubility are unrealistically large. In these cases, the solubility is no longer dependent on the solubility equilibrium but on the number of moles of solvent required to interact with each mole of solute. The molarity of water is 55.6 M, and each ion would be surrounded by several molecules of water. It is, therefore, unreasonable to expect that the solute concentration under low pH conditions would exceed several moles per liter.

    The solubility of metal hydroxides increases with decreasing pH. Metal hydroxides that have a cation with an oxidation state of +2 exhibit a one hundred-fold increase in their solubility for every unit decrease in pH. Metal hydroxides that have a cation with an oxidation state of +3 exhibit a one thousand-fold increase in their solubility for every unit decrease in pH.

    Q12. Integrating the ideas of redox chemistry and solubility expressed in this module, summarize the chemical and physical changes you would expect over time in a sample of Lithia water? What sample preparation step could be used to prevent such changes from occurring?

    Previous questions on redox chemistry (see Q6 in this guide) indicate that dissolved oxygen does not affect the concentration or oxidation state of main group metals, but that dissolved oxygen is predicted to oxidize ferrous ion to ferric ion. Previous questions on analyte solubility as a function of pH indicate that metal hydroxides of interest in the analysis of Lithia water tend to be insoluble under basic to slightly acidic conditions, but that most of these hydroxides are moderately to freely soluble at low pH. The Ksp of ferric hydroxide is orders of magnitude lower than ferrous hydroxide, which results in the precipitation of ferric hydroxide in Lithia water after the oxidation of ferrous ion by dissolved oxygen occurs. Observations in the author’s laboratory indicated that most of the ferrous ion in Lithia water is oxidized and precipitates as ferric hydroxide within the first 24 hours after sample collection. However, at low pH (< 2.0), even the least soluble of the metal hydroxides, ferric hydroxide, is moderately soluble at pH 2.0, with a mass solubility of 170 mg L-1. Therefore, it is suggested that if the pH of the Lithia water is adjusted to a pH of 2.0 or less, analytes that form insoluble hydroxides can be prevented from precipitating by acidification with trace-metal grade or reagent grade acid. Instructors are encouraged to address this question as students are observing changes in their mineral water sample.

    Q13. Collection of samples for trace metal analysis in drinking water is described in the Sample Collection, Preservation, and Storage section of EPA method 200.5, Determination of Trace Elements in Drinking Water by Axially Viewed Inductively Coupled Plasma - Atomic Emission Spectrometry (http://www.epa.gov/nerlcwww/ordmeth.htm). After reading this document, suggest a general approach to preserving Lithia water prior to analysis. Would this work for all analytes of interest in Lithia water? Explain.

    EPA method 200.5 is a protocol for the determination of a large set of cations in drinking water. In the Collection, Preservation, and Storage section of EPA method 200.5, to properly keep all inorganic cations in solution, the analyst is instructed to acidify a one liter water sample with 3.0 mL of (1+1) nitric acid, and after 16 hours of cold storage, verify that the pH of the acidified sample is below 2.0.

    The effect of sample acidification on the solubility of sparingly soluble salts is introduced in the Lithia water module entitled Chemical Equilibria. In the case of Lithia water, students are expected to discover that dissolved oxygen oxidizes the ferrous ion to ferric ion. Because iron (III) hydroxide is significantly less soluble than iron (II) hydroxide, a layer of iron (III) hydroxide settles out of solution as an orange precipitate in 1-2 days unless the sample is acidified. However, while sample acidification keeps metal ions in solution, sample acidification also protonates the bicarbonate ion to form carbonic acid, which subsequently dissociates into carbon dioxide and water. Therefore, students analyzing Lithia water should suggest the collection of two samples; one acidified sample for the determination of iron and another unacidified sample for the remaining analytes. Lithia water students can also suggest, based on EPA method 200.5, that the acidified sample be used for the determination of metal ions, and the unacidified sample be used for the determination of anions.