# Major Inorganic Constituents

Q1. Which cations and anions are represented by a single salt and which are represented by multiple salts?

From the summary of the 1915 Lithia water analysis on the engraved stone:

Cations: Na+ is represented by multiple salts, and K+, Li+, Ca2+, and Mg2+ are represented by single salts.

Anions: HCO3- is represented by multiple salts, and Cl-, SO42-, and BO2- are represented by single salts.

Q2. Of those analytes listed on the plaque, are there any cation or anion concentrations that are unknowns based on the information provided? How can you tell?

Iron and aluminum oxides are represented by a single concentration, and they would both be unknowns because it is impossible to calculate an analyte concentration when there are two unknowns and one concentration. Although it is not indicated on the plaque, based on recent analyses for the City of Ashland, instructors can assume that aluminum does not contribute significantly to the ionic content of Lithia water, and iron represents the vast majority of this Lithia water constituent. Instructors may also assume that all measured silica exists in Lithia water as SiO32-, the silicate ion.

Q3. Based on the analysis engraved on the plaque above, how would you calculate the concentration of each ion represented by a single salt? Complete these calculations, and express the concentration in milligrams per liter (mg L-1) and in moles per liter (M).

Listed below are the conversions of single salt concentrations to concentrations of individual cations and anions in Lithia water. Instructors may use these questions to provide students with examples of stoichiometric calculations.

1. Potassium
1. $$\left(\dfrac{279.5 \: mg \: KHCO_{3}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: {KHCO}_{3}}{100.12\: {g}}\right)\left(\dfrac{1\: {mol}\: K^{+}}{1\: {mol}\: {KHCO}_{3}}\right)=2.792 \times 10^{-3}\: {M}\: {K}^{+}$$
2. $$\left(2.792 \times 10^{-3}\: M\: K^{+}\right)\left(\dfrac{39.098\: g\: K^{+}}{mol\: K^{+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{109.1\: mg\: K^{+}}{L}$$
2. Lithium
1. $$\left(\dfrac{153.82\: {mg}\: {LiHCO}_{3}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {LiHCO}_{3}}{67.96\: {g}}\right)\left(\dfrac{1\: {mol}\: {Li}^{+}}{1\: {mol}\: {LiHCO}_{3}}\right)=2.263 \times 10^{-3}\: {M}\: Li^{+}$$

2. $$\left(2.263 \times 10^{-3}\: M\: L i^{+}\right)\left(\dfrac{6.941\: g\: L i^{+}}{mol\: Li^{+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{15.7\: mg\: Li^{+}}{L}$$

3. Calcium
1. $$\left(\dfrac{1404\: mg\: Ca\left(H C O_{3}\right)_{2}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: C a\left(HCO_{3}\right)_{2}}{162.12\: g}\right)\left(\dfrac{1\: mol\: C a^{2+}}{1\: mol\: Ca\left(H C O_{3}\right)_{2}}\right)= {8.66 \times 10^{-3}\: M\: C a^{2+}}$$
2. $$\left(8.66 \times 10^{-3}\: M\: Ca^{2+}\right)\left(\dfrac{40.08\: g\: Ca^{2+}}{mol\: Ca^{2+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{347\: mg\: Ca^{2+}}{L}$$

4. Magnesium
1. $$\left(\dfrac{1153\: {mg}\: {Mg}\left({HCO}_{3}\right)_{2}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {Mg}\left({HCO}_{3}\right)_{2}}{146.35\: {g}}\right)\left(\dfrac{1\: {mol}\: {Mg}^{2+}}{1\: {mol}\: {Mg}\left({HCO}_{3}\right)_{2}}\right)=7.88 \times 10^{-3}\: {M}\: {Mg}^{2+}$$
2. $$\left(7.88 \times 10^{-3}\: M\: Mg^{2+}\right)\left(\dfrac{24.31\: g\: Mg^{2+}}{mol\: Mg^{2+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{192\: mg\: Mg^{2+}}{L}$$
5. Chloride
1. $$\left(\dfrac{4515\: mg\: NaCl}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: NaCl}{58.44\: g}\right)\left(\dfrac{1\: mol\: Cl^{-}}{1\: mol\: NaCl}\right)=7.726 \times 10^{-2}\: M\: Cl^{-}$$
2. $$\left(7.726 \times 10^{-2}\: M\: Cl^{-}\right)\left(\dfrac{35.45\: g\: Cl^{-}}{mol\: Cl^{-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{2739\: mg\: Cl^{-}}{L}$$
6. Metaborate
1. $$\left(\dfrac{321.3\: {mg}\: NaBO_2}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {NaBO}_{2}}{65.80\: {g}}\right)\left(\dfrac{1\: {mol}\: {BO}_{2}^{-}}{1\: {mol\: NaBO}_{2}}\right)=4.883 \times 10^{-3}\: {M}\: {BO}_{2}^{-}$$
2. $$\left(4.883 \times 10^{-3}\: M\: BO_{2}^{-}\right)\left(\dfrac{42.81\: g\: BO_{2}^{-}}{mol\: BO_{2}^{-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{209\: mg\: BO_{2}^{-}}{L}$$
7. Sulfate
1. $$\left(\dfrac{3.895\: {mg}\: {Na}_{2} {SO}_{4}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {Na}_{2} {SO}_{4}}{142.04\: {g}}\right)\left(\dfrac{1\: {mol}\: {SO}_{4}^{2-}}{1\: {mol}\: {Na}_{2} {SO}_{4}}\right)=2.742 \times 10^{-5}\: {M}\: {SO}_{4}^{2-}$$
2. $$\left(2.742 \times 10^{-5}\: M\: SO_{4}^{2-}\right)\left(\dfrac{96.06\: g\: SO_{4}^{2-}}{mol\: SO_{4}^{2-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{2.63\: mg\: SO_{4}^{2-}}{L}$$
8. Silica (as silicate)
1. $$\left(\dfrac{94.9\: {mg}\: {SiO}_{2}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: {mol}\: {SiO}_{2}}{60.08\: g}\right)\left(\dfrac{1\: {mol}\: {SiO}_{3}^{2-}}{1\: {mol}\: SiO_2}\right)=1.58 \times 10^{-3}\: {M}\: {Si}O_{3}^{2-}$$
2. $$\left(1.58 \times 10^{-3}\: M\: SiO_{3}^{2-}\right)\left(\dfrac{76.08\: g\: SiO_{3}^{2-}}{mol\: SiO_{3}^{2-}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{120.2\: mg\: SiO_{3}^{2-}}{L}$$
9. Iron (assuming aluminum content is negligible)
1. $$\left(\dfrac{12.5\: {mg}\: {Fe}_{2}{O}_{3}}{L}\right)\left(\dfrac{1\: {g}}{1000\: {mg}}\right)\left(\dfrac{1\: {mol}\: {Fe}_{2} {O}_{3}}{159.69\: {g}}\right)\left(\dfrac{2\: {mol}\: {Fe}^{3+}}{1\: {mol}\: {Fe}_{2}{O}_{3}}\right)\left(\dfrac{1\: {mol}\: {Fe}^{2+}}{1\: {mol}\: {Fe}^{3+}}\right)=1.566\times10^{-4}\:M\:Fe^{2+}$$
2. $$\left(1.566 \times 10^{-4}\: M\: Fe^{2+}\right)\left(\dfrac{55.845\: g\: Fe^{2 +}}{mol\: Fe^{2+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{8.74\: mg\: Fe^{2+}}{L}$$

Q4. How does your approach to calculating ion concentrations represented by multiple salts differ from expressing ion concentrations represented by a single salt? Complete these calculations, and express the solute concentration in milligrams per liter (mg L-1) and in moles per liter (M).

Multiple sources of a common ion need to be converted from mg L-1 to mol L-1 and then added together. Sodium ion and bicarbonate ion have a common salt so the molarity of sodium bicarbonate may first be calculated. After this calculation, the molarity of the common ions can be calculated using stoichiometric relationships and the concentrations from the previous question.

1. Sodium bicarbonate
1. $$\left(\dfrac{2456\: mg\: NaHCO_{3}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: NaHCO_{3}}{84.01\: g}\right)\left(\dfrac{1\: mol\: Na^{+}}{1\: mol\: NaHCO_{3}}\right)=2.923 \times 10^{-2}\: M\: Na^{+}$$

2. $$\left(2.923 \times 10^{-2}\: M\: Na^{+}\right)\left(\dfrac{1\: mol\: HCO_{3}^{-}}{1\: mol\: Na^{+}}\right)=2.923 \times 10^{-2}\: M\: HCO_{3}^{-}$$

2. All sources of sodium ion
1. $$\left(2.923 \times 10^{-2}\: M\: Na^{+}\right)+\left(7.726 \times 10^{-2}\: M\: Cl^{-}\left(\dfrac{1\: mol\: Na^{+}}{1\: mol\: Cl^{-}}\right)\right)+ \left(4.883 \times 10^{-3}\: M\: BO_{2}^{-}\left(\dfrac{1\: mol\: Na^{+}}{1\: mol\: BO_{2}^{-}}\right)\right)+\left(2.742 \times 10^{-5}\: M\: SO_{4}^{2-}\left(\dfrac{2\: mol\: Na^{+}}{1\: mol\: SO^{2-}_{4}}\right)\right)= {0.1114\: M\: Na^{+}}$$
2. $$\left(0.1114\: M\: Na^{+}\right)\left(\dfrac{22.99\: g\: Na^{+}}{mol\: Na^{+}}\right)\left(\dfrac{1000\: mg}{g}\right)=\dfrac{2562\: mg\: Na^{+}}{L}$$

3. All sources of bicarbonate ion
1. $$(2.923 \times 10^{-2}\: M\: HCO_{3}^{-})+\left(2.792 \times 10^{-3}\: M\: K^{+}\left(\dfrac{1\: mol\: HCO_{3}^{-}}{1\: mol\: K^{+}}\right)\right)+\left(2.263 \times 10^{-3}\: M\: Li^{+}\left(\dfrac{1\: mol\: HCO_{3}^{-}}{1\: mol\: Li^+}\right)\right)+\left(8.66 \times 10^{-3}\: M\: Ca^{2+}\left(\dfrac{2\: mol\: HCO_{3}^{-}}{1\: mol\: Ca^{2+}}\right)\right)+\left(7.88 \times 10^{-3}\: M\: Mg^{2+}\left(\dfrac{2\: mol\: HCO_{3}^-}{1\: mol\: Mg^{2+}}\right)\right)=0.06736\: M\: HCO_{3}^{-}$$

2. $$\left(0.06736\: M\: HCO_{3}^{-}\right)\left(\dfrac{61.02\: g\: {HCO}_{3}^{-}}{mol\: {HCO}_{3}^{-}}\right)\left(\dfrac{1000\: {mg}}{g}\right)=\dfrac{4110\: {mg}\: {HCO}_{3}^{-}}{L}$$

Q5. Determine the number of milligrams of each analyte present in 25.00 mL of Lithia water from the 1915 analysis, assuming that the composition of Lithia water has not changed since 1915. Using these analyte quantities, determine appropriate analytical techniques based on the classification scheme in Table 1.

Students should recognize that multiplying the analyte concentration by a given volume will result in the mass (or moles) of analyte in that given volume. Each analyte concentration was calculated in the previous question in mg L-1, and those concentrations can be used for the determination of milligrams of each analyte in a 25.00 mL sample of Lithia water. The analytical technique(s) listed are suggested techniques based on analyte classification. It is up to the students, in consultation with the instructor, to determine which methods are most appropriate for a given laboratory project.

Table 1. Suggested analyte classification table for Lithia water analysis. See Analytical Chemistry 2.0 (Chapter 3.4, Scale of Operation) on the Analytical Sciences Digital Library website for additional information.

Analyte

Calculation

Analyte Classification (Potential Technique)

Sodium

$$\left(\dfrac{2562\: mg\: Na^{+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=64.04\: mg$$

Minor (ISE)

Potassium

$$\left(\dfrac{109.1\: {mg}\: {K}^{+}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL})=2.7\: {mg}$$

Minor (ISE or Spectroscopy)

Lithium

$$\left(\dfrac{15.7\: {mg}\: Li^{+}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL})=0.4\: {mg}$$

Trace (ISE or Spectroscopy)

Calcium

$$\left(\dfrac{347\: mg\: Ca^{2+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=8.7\: mg$$

Minor (Titrimetry, ISE or Spectroscopy)

Magnesium

$$\left(\dfrac{192\: mg\: Mg^{2+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=4.8\: mg$$

Minor (Titrimetry, ISE or Spectroscopy)

Iron

$$\left(\dfrac{8.74\: mg\: Fe^{2+}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=0.2\: mg$$

Trace (Spectroscopy)

Chloride

$$\left(\dfrac{2739\: mg\: Cl^{-}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=68.5\: mg$$

Minor (ISE or Titrimetry)

Metaborate

$$\left(\dfrac{209\: mg\: BO_{2}^{-}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=5.2\: mg$$

Minor (Titrimetry, ISE or Spectroscopy)

Sulfate

$$\left(\dfrac{2.63\: {mg}\: {SO}_{4}^{2-}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL})=0.07\: {mg}$$

Trace (Spectroscopy)

Bicarbonate

$$\left(\dfrac{4110\: {mg}\: {HCO}_{3}^{-}}{L}\right)\left(\dfrac{1\: {L}}{1000\: {mL}}\right)(25.00\: {mL}) =102.8\: {mg}$$

Minor (Titrimetry)

Silicate

$$\left(\dfrac{120.2\: mg\: SiO_{3}^{2-}}{L}\right)\left(\dfrac{1\: L}{1000\: mL}\right)(25.00\: mL)=3.0\: mg$$

Minor (Spectroscopy)

Q6. Once the appropriate analytical technique(s) for each analyte have been identified based on the classification scheme in Table 1, consider, discuss, and identify additional criteria that would be important in the selection of an appropriate method for each analyte in Lithia water. An appropriate starting point for this discussion would be to read Chapter 3.4 in Analytical Chemistry 2.0.

After students have read Chapter 3.4 in Analytical Chemistry 2.0 and calculated the analyte concentrations in Q5, each group should develop a set of criteria for the selection of each analytical method. Appropriate criteria include, but are not limited to:

1. Method Sensitivity. In this project, students are guided in their selection of an analytical technique based on the analyte classification scheme in Table 1.
2. Cost of materials & equipment. In this analytical chemistry course, the selection of an analytical technique that minimizes the cost of reagents and equipment is preferable.
3. Timeframe of analysis. An analyte determination that can be completed within one laboratory period is preferable. However, it may also be acceptable if one or more group members agree to complete a determination by remaining beyond the scheduled lab period. Such arrangements should be made ahead of time, and in such cases, solution and standard preparation should be completed ahead of time if possible.
4. Is the proper equipment for performing the determination and handling any toxic or noxious reagents available in the department?

The most appropriate choice of a proper technique and method for the determination of a chosen analyte depends on the equipment, time and resources available. A student may discover that there are multiple techniques available for the determination of an analyte. If this is the case, the criteria are designed to guide each group toward the least expensive, and presumably, the least complicated analytical technique that can be used for a given determination. However, it is up to the instructor to decide how the criteria are prioritized based on course goals.

In the past, the analytical chemistry laboratory prescribed the following analytical techniques for Lithia water analytes:

• Sodium, potassium: Ion-selective electrodes
• Calcium, magnesium: Complexometric titrations
• Chloride: Precipitation titration
• Bicarbonate: Acid-base titration
• Boron, Iron, Silicate, Sulfate: Molecular spectroscopy
• Lithium: Atomic emission spectroscopy, molecular spectroscopy

Q7. Lithium carbonate is typically prescribed to control manic depression. A typical dosage is 900 mg day-1 of lithium carbonate. How many liters of Lithia water would someone have to drink to ingest an equivalent amount of lithium that is present in a prescribed dose of lithium carbonate? Based on your calculated volume of Lithia water in the previous question, are there health related issues that could arise due to the presence of other ions that are listed on the Lithia water analysis plaque? To help answer this question, perform an internet search on the health effects and the dietary intake of any of the other cations or anions in Lithia water. Cite any sources you use to support your answer.

If a typical daily dosage of lithium carbonate to treat manic depression is 900 mg per day, the volume of Lithia water needed to supply the appropriate dosage of lithium ion is:

$\left(\dfrac{0.900\: g\: Li_{2}CO_{3}}{day}\right)\left(\dfrac{1\: {mol}\: Li_{2}CO_{3}}{73.892\: g\: Li_{2}CO_{3}}\right)\left(\dfrac{2\: mol\: Li^{+}}{1\: mol\: Li_{2}CO_{3}}\right)\left(\dfrac{1\: L\: Lithia\: water}{0.002263\: {mol}\: Li^{+}}\right)=\left(\dfrac{10.8\: L\: Lithia\: water}{day}\right)\nonumber$

Instructors may want to point out to their students that one of the inorganic ions in Lithia water that serves an important role in human health is sodium. Although sodium is necessary for proper electrolyte balance, which controls such functions as nerve impulse conduction, only small amounts of dietary sodium are necessary to maintain such functions. Excess sodium in the diet leads to health problems such as high blood pressure. According to the Harvard School of Public Health website on Salt and Health, U.S. government recommendations are 2300 milligrams of sodium as a maximum daily dietary intake. The sodium concentration in Lithia water, calculated in the previous question in this section, is 2573 milligrams per liter. Therefore, the volume of Lithia water that must be consumed in order to ingest the maximum daily intake of sodium ions is:

$\left(\dfrac{2.300\: g\: Na^{+}}{day}\right)\left(\dfrac{L\: Lithia\: water}{2.573\: g\: Na^{+}}\right)=\left(\dfrac{0.894\: L\: Lithia\: water}{day}\right)\nonumber$

This volume is twelve times less than the volume needed to supply the therapeutic amount of lithium to treat manic depression. Drinking the Lithia water from the Ashland, OR source would not be a healthy or effective way of administering a therapeutic dose of lithium.