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The total concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in a sample of hard water was determined by titrating a \(0.100\) - L sample of the water with a solution of EDTA \({ }^{4-}\). The EDTA \({ }^{4-}\) chelates the two cations: $$ \begin{aligned} \mathrm{Mg}^{2+}+[\mathrm{EDTA}]^{4-} & \longrightarrow[\mathrm{Mg}(\mathrm{EDTA})]^{2-} \\ \mathrm{Ca}^{2+}+\left[\mathrm{EDTA}^{4-}\right.& \longrightarrow[\mathrm{Ca}(\mathrm{EDTA})]^{2-} \end{aligned} $$ It requires \(31.5 \mathrm{~mL}\) of \(0.0104 \mathrm{M}[\mathrm{EDTA}]^{4-}\) solution to reach the end point in the titration. A second \(0.100-L\) sample was then treated with sulfate ion to precipitate \(\mathrm{Ca}^{2+}\) as calcium sulfate. The \(\mathrm{Mg}^{2+}\) was then titrated with \(18.7 \mathrm{~mL}\) of \(0.0104 \mathrm{M}\) [EDTA] ]- Calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water in mg/I.

Short Answer

Expert verified
The concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water are 47.26 mg/L and 53.33 mg/L, respectively.

Step by step solution

01

Determine the moles of [EDTA]4- used in the titrations.

In the first titration, we have the volume of the \(\mathrm{EDTA}\) solution required to titrate both \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions: 31.5 mL with a concentration of 0.0104 M. For the second titration, only \(\mathrm{Mg}^{2+}\) was titrated. We have the volume of the \(\mathrm{EDTA}\) solution required for this: 18.7 mL with a concentration of 0.0104 M. Convert mL to L: titration 1 volume = 0.0315 L titration 2 volume = 0.0187 L Calculate the moles of \(\mathrm{EDTA}\) used in each titration: moles of \(\mathrm{EDTA}\) in titration 1 = 0.0315 L × 0.0104 mol/L = 0.0003276 mol moles of \(\mathrm{EDTA}\) in titration 2 = 0.0187 L × 0.0104 mol/L = 0.00019448 mol
02

Determine the moles of Mg2+ and Ca2+ ions in the titrations.

From the balanced equation, the stoichiometry of the reaction between \(\mathrm{Ca}^{2+}\) and \(\mathrm{EDTA}^{4-}\), as well as \(\mathrm{Mg}^{2+}\) and \(\mathrm{EDTA}^{4-}\), is 1:1. This means that the moles of \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) ions will be equal to the moles of \(\mathrm{EDTA}\) used in each titration. moles of \(\mathrm{Mg}^{2+}\) ions (titration 2) = moles of \(\mathrm{EDTA}\) in titration 2 = 0.00019448 mol Now, we can determine the moles of \(\mathrm{Ca}^{2+}\) ions by subtracting the moles of \(\mathrm{Mg}^{2+}\) ions from the total moles of \(\mathrm{EDTA}\) in titration 1. moles of \(\mathrm{Ca}^{2+}\) ions (titration 1) = moles of \(\mathrm{EDTA}\) in titration 1 - moles of \(\mathrm{Mg}^{2+}\) ions (titration 2) = 0.0003276 mol - 0.00019448 mol = 0.00013312 mol
03

Calculate the concentration of Mg2+ and Ca2+ ions in mg/L.

Now, we will use the given sample volume (0.100 L) and the molar masses (24.305 g/mol for \(\mathrm{Mg}\) and 40.078 g/mol for \(\mathrm{Ca}\)) to calculate the concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in mg/L: concentration of \(\mathrm{Mg}^{2+}\) in mg/L = (0.00019448 mol / 0.100 L) × (24.305 g/mol) × (1000 mg/g) = 47.26 mg/L concentration of \(\mathrm{Ca}^{2+}\) in mg/L = (0.00013312 mol / 0.100 L) × (40.078 g/mol) × (1000 mg/g) = 53.33 mg/L The concentrations of \(\mathrm{Mg}^{2+}\) and \(\mathrm{Ca}^{2+}\) in the hard water are 47.26 mg/L and 53.33 mg/L, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Titration
Titration is a fundamental analytical method in chemistry used to determine the concentration of a solute in a solution. It involves the careful addition of a solution of known concentration (called the titrant) to a solution of unknown concentration (the analyte) until the chemical reaction between the two is complete. This point of completion is known as the equivalence point and can be detected using different indicators, such as a color change.

During a titration experiment, the volume of titrant needed to reach the equivalence point is measured. This volume, along with the known concentration of the titrant, allows us to calculate the unknown concentration of the analyte using stoichiometry principles. Titration is widely used in different fields, including environmental science, medicine, and industry, for quality control and quantification of substances.
Complexometric Titration
Complexometric titration is a type of titration which involves the formation of a complex between the analyte and the titrant. It is particularly useful for the determination of a mixture of different metal ions in solution. In this method, a chelating agent is used as the titrant that forms a very stable, water-soluble complex with the metal ions, causing a significant change that indicates the end point.

One commonly used chelating agent is EDTA (ethylenediaminetetraacetic acid), which binds with metal ions such as \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) in hard water. In hard water analysis, complexometric titration allows for the precise calculation of the total concentration of these ions, providing valuable information for water treatment processes.
EDTA
EDTA, an acronym for ethylenediaminetetraacetic acid, is a powerful chelating agent commonly used in complexometric titrations. It has the ability to 'grab' metal ions by forming four stable bonds, effectively sequestering them into a complex. EDTA is useful because it can form complexes with a variety of metal ions, and the resulting complexes are usually colorless, which simplifies the detection of the end point when suitable indicators are used.

When EDTA is added to a solution containing metal ions, it binds to those ions and forms a one-to-one complex. This property is key in calculating the concentration of metal ions, as the stoichiometry directly correlates with the amount of EDTA used in the titration process.
Concentration of Ions
The concentration of ions in a solution refers to the amount of ion species present in a given volume of solution. It is typically expressed in terms such as molarity (moles per liter). Ions play crucial roles in a myriad of chemical processes and are of particular importance in analyzing water hardness, where \(\mathrm{Ca}^{2+}\) and \(\mathrm{Mg}^{2+}\) are the primary components contributing to the hardness level.

Understanding the concentration of these ions is essential for water treatment, as excessive concentrations can lead to scale formation and interfere with soap's effectiveness. Analyzing and quantifying the concentration of ions allows for the management and treatment of water for both domestic and industrial purposes.
Stoichiometry
Stoichiometry is the area of chemistry that pertains to the quantitative relationships between the reactants and products in a chemical reaction. It is based on the conservation of mass where all atoms that enter a reaction must be accounted for among the products. For titrations, stoichiometry provides the calculations to determine the concentration of an unknown analyte in a solution.

In the context of the original exercise, the stoichiometry of the reaction between metal ions in hard water and EDTA is 1:1. This means that one mole of \(\mathrm{Ca}^{2+}\) or \(\mathrm{Mg}^{2+}\) reacts with one mole of EDTA. Consequently, the amount of EDTA used in the titration allows us to directly calculate the moles and subsequently the concentration of the metal ions in the water sample, based on the stoichiometric ratios of the reaction.

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Most popular questions from this chapter

Polydentate ligands can vary in the number of coordination positions they occupy. In each of the following, identify the polydentate ligand present and indicate the probable number of coordination positions it occupies: (a) \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4}(o-\mathrm{phen})\right] \mathrm{Cl}_{3}\) (b) $\left[\mathrm{Cr}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\right] \

The lobes of which \(d\) orbitals point directly between the ligands in (a) octahedral geometry, (b) tetrahedral geometry?

Consider an octahedral complex MA \(\mathrm{A}_{3}\). How many geometric isomers are expected for this compound? Will any of the isomers be optically active? If so, which ones?

Consider the tetrahedral anions \(\mathrm{VO}_{4}^{3-}\) (orthovanadate ion), \(\mathrm{CrO}_{4}^{2-}\) (chromate ion), and \(\mathrm{MnO}_{4}^{-}\)(permanganate ion). (a) These anions are isoelectronic. What does this statement mean? (b) Would you expect these anions to exhibit \(d-d\) transitions? Explain. (c) As mentioned in "A Closer Look" on charge- transfer color, the violet color of \(\mathrm{MnO}_{4}^{-}\)is due to a ligand- to-metal charge transfer (LMCT) transition. What is meant by this term? (d) The LMCT transition in \(\mathrm{MnO}_{4}^{-}\)occurs at a wavelength of \(565 \mathrm{~nm}\). The \(\mathrm{CrO}_{4}^{2}\) ion is yellow. Is the wavelength of the LMCT transition for chromate larger or smaller than that for \(\mathrm{MnO}_{4}^{-}\)? Explain. (e) The \(\mathrm{VO}_{4}^{3-}\) ion is colorless. Do you expect the light absorbed by the LMCT to fall in the UV or the IR region of the electromagnetic spectrum? Explain your reasoning.

A complex is written as \(\mathrm{NiBr}_{2} \cdot 6 \mathrm{NH}_{2}\), (a) What is the oxidation state of the Ni atom in this complex? (b) What is the likely coordination number for the complex? (c) If the complex is treated with excess \(\mathrm{AgNO}_{3}\left(\mathrm{aq}_{2}\right)\), how many moles of AgBr will precipitate per mole of complex?

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