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    consider the hydration of ions in water. which of the following interactions is involved in given process?

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    Hydrolysis

    Hydration and Hydrolysis of Metal Cations

    Hydration and Hydrolysis of Metal Cations Hydration

    When sodium chloride dissolves in water, the sodium and chloride ions and the polar water molecules are strongly attracted to one another by ion-dipole interactions. The solvent molecules (water in this case) surround the ions removing them from the crystal and forming the solution. As the dissolving process proceeds, the individual ions are removed from the solid surface becoming completely separate, hydrated species in the solution.

    Dissolution of NaCl in Water

    In order to dissolve the sodium chloride in the water, three processes have to occur:

    the water molecules must be separated (H-bonds must be broken)

    the forces of attraction between the ions in the NaCl lattice must be broken

    solute-solvent(ion-dipole)interations must be formed

    The overall enthalpy change in forming a solution is the sum of energy changes for each of these processes.

    Whether the dissolution process will be exothermic or endothermic depends on the relative magnitudes of the energy changes for the three steps. In general, a substance will be insoluble if the energy expended to break apart the solvent and solute particles is significantly greater than the energy given off when solute-solvent interactions are established.

    Hydration Energy

    If you were to take gaseous cations and put them into water, they would form hydrated ions and release large amounts of energy. This energy is called the hydration energy of the cation. Although this is an experiment that is impossible to perform, the hydration energies have been determined indirectly (you will see how this is done later in the term).

    PROBLEMS

    Using the data in the Table of Hydration Energies, explain the trends in hydration enthalpies versus ionic radii.

    Consider elements of similar charge and radius that have different electronegativity.

    compare Rb+1 and Tl+1

    compare Sr+2 and Pb+2

    compare Lu+3 and Tl+3

    Offer an explanation for the observed trend.

    As expected from Coulomb's Law , the hydration energy of a cation depends on the charge and radius of the cation.

    Hydration energy also depends on the electronegativity of the element.

    Latimer observed that when the electronegativity of the metal is not too great, a good approximation of the hydration energy of a metal ion is given by the following equation:

    Latimer's Equation

    In this equation Z is the charge on the cation and r is the cationic radius (in pm). The constant, 50, roughly equates to the radius of the oxygen atom in water.

    Latimer's equation does not include the effects of electronegativity. This equation is valid for cations that have electronegativities less than 1.5. Compare the hydration energy data for cations having electronegativities <1.5 and those having electronegativities >1.5. Cations with electronegativies >1.5 have substantially higher hydration energies than more electropositive cations of comparable radius and charge. These electronegative metals have some degree of covalent bonding in their interaction with the oxygen atom of the water molecule, i.e., oxygen's unshared pair of electrons is shared with the metal atom.

    Hydrolysis

    Metal ions in aqueous solution behave as Lewis acids. The positive charge on the metal ion draws electron density from the O-H bond in the water. This increases the bond's polarity making it easier to break. When the O-H bond breaks, an aqueous proton is released producing an acidic solution.

    The equilibrium constant for this reaction can be measured.

    [M(H2O)n]z+ + H2O [M(H2O)n-1(OH)](z-1)+ + H3O+

    Note the similarity of the equation for the hydrolysis of a hydrated cation with the equation for the ionization of a weak acid in aqueous solution.

    The Keq for the hydrolysis of a hydrated cation is analogous to the Ka for the ionization of a weak acid. Keq is an acid ionization constant. Generally, hydrolysis constants for cations are tabulated as -log Ka. These tabulated hydrolysis constants are averages of different experimental measurements that sometimes differ by more than one pKa unit. For this reason, small differences are not considered significant when comparing the values of these constants.

    PROBLEMS

    Use the data in the Table of Hydrolysis Constants for Metal Cations for these questions.

    Consider the ions of Group 2A.

    What trend exists between ionic radius and the hydrolysis constant?

    What trend exists between electronegativity and the hydrolysis constant?

    Consider Na+1, Ca+2, and Pu+3.

    What trend exists between ionic charge and hydrolysis?

    स्रोत : people.wou.edu

    Hydration Enthalpy

    Hydration Enthalpy also referred to as hydration energy is defined as the amount of energy released on dilution of one mole of gaseous ions.

    JEEIIT JEE Study MaterialHydration Enthalpy

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    Hydration Enthalpy

    Hydration Enthalpy What is Hydration Enthalpy?

    Hydration enthalpy (ꕔHHyd) is the change in enthalpy when one mole of gaseous ion under a standard condition of 1 bar pressure dissolves in a sufficient amount of water to form an infinitely dilute solution (infinite dilution means a further addition of solute will not cause any heat change).

    In simple terms, enthalpy of hydration is described as the amount of energy released on dilution of one mole of gaseous ions. It can be considered as enthalpy of solvation with the solvent being water. Hydration enthalpy is also called hydration energy and its values are always negative.

    For a chemical reaction,

    M+(g) + aq → M+(aq) Enthalpy change = ∆HHyd

    Water is considered to be a polar solvent because it has a positive (H atom) and negative (O atom) poles. When an ionic compound (any salt, say NaCl) is dissolved in water the solid-state structure of the compound is destroyed and the Na+ and Cl– are separated.

    In the solution, they exist as Na+ atoms surrounded by the negative ends of water molecules and similarly the Cl– is surrounded by the positive end of water molecules as shown in figure (i). Since additional bonds are formed between the atoms the process releases some energy.

    This energy is expressed as the hydration enthalpy or enthalpy of hydration between M+(g) and M+(aq) is that in M+(aq) the ion is surrounded by water molecules forming a weak bond.

    Hydration Enthalpy of Elements

    Hydration enthalpy values of various elements are tabulated in the table given below.

    Ion ꕔHHyd

    Li+ -520 Na+ -405 K+ -321 Rb+ -300 Cs+ -277 F- -506 Cl- -364 Br- -337 I- -296

    The magnitude of hydration enthalpy depends on the charge density of the ions. The charge density is more for smaller ions and hence the smaller ions have higher values of hydration enthalpy. The higher the charge density the higher will be the force of attraction between the ion and the water polar end. This makes the value of hydration enthalpy higher in smaller ions. The alkali metals are highly hydrated, and the extent of hydration decreases down the group.

    Hydration Enthalpy and Solubility

    The ions in a solute are bound together by coulombic force of attraction, to dissolve this solute into the solvent (here water) the water molecule should overcome this strong force of attraction. The energy required to break this string force of attraction is called lattice enthalpy.

    Most of the ionic compounds are insoluble in non-aqueous solutions but they show high solubility in water. The factor that determines the solubility of a salt is the interactions of the ions with the solvent. As explained earlier the water is a polar molecule with a partial positive charge on hydrogen and partial negative charge on oxygen, interacts with the ions and forms a strong bond releasing energy.

    The process of dissolution can be considered as a combination of two processes.

    The first one is,

    M+(s) → M+(g) ∆ = ∆HLatt Lattice enthalpy

    The second process is the hydration,

    M+(g) + aq → M+(aq) ∆ = ∆HHyd Hydration enthalpy

    The first process involves the breaking of bonds in the solid solute; therefore, it is an endothermic process. Lattice enthalpy can be defined as the energy released when one mole ionic solid is converted gaseous ions. Greater the lattice enthalpy greater energy is required to overcome the force of attraction. Some compounds are insoluble in water because of their higher lattice enthalpy value.

    Also Read: Entropy

    The enthalpy of the solution. The dissolution process is dependent on both lattice enthalpy and hydration enthalpy. Enthalpy of the solution is the difference between hydration enthalpy and lattice enthalpy.

    ie,

    ∆HSolution= ∆HHyd – ∆HLatt

    ∆HHyd = ∆HHyd(cation) + ∆HHyd

    The enthalpy of the solution involves two processes, i.e., lattice energy and enthalpy of hydration.

    The lattice energy of NaCl is the energy released when Na+ and Cl− ions come close to each other to form a lattice.

    NaCl(s) → Na+(g) +Cl−(g) (I)

    The enthalpy of hydration takes place when there is a dispersal of gaseous solute in water. It is the energy released when the solute transforms from a gaseous state to aqueous.

    Na+(g) + Cl−(g) → Na+(aq) + Cl−(aq) (II)

    On combining equations (I) and (II), the dissociation reaction of NaCl(s) into Na+(aq) and Cl−(aq) is obtained as shown below

    NaCl(s) → Na+(aq) + Cl−(aq)

    Therefore, the enthalpy of solution is calculated as:

    ΔHsolution = Enthalpy of hydration – Lattice energy­­

    Where,

    ΔHsolution is enthalpy of the solution.

    Substitute the values in the above expression.

    ΔHsolution = Enthalpy of hydration – Lattice energy = −783kJ mol-1 + 786kJ mol-1 = 3kJmol-1

    The favourable conditions for the formation of the solution involve a negative value for ∆HSolution, ie, when the heat released on hydration is more than the heat required to overcome the force of attraction ie the lattice enthalpy.

    Application of Hydration Enthalpy

    One application of enthalpy of hydration is the reaction of cement with water. The reaction being exothermic releases a large amount of heat. This heat released becomes significant in mass constructions like building dams and big structures. For the construction of massive concrete blocks, large quantities of cement are used.

    स्रोत : byjus.com

    11.4: Hydration of Ions

    The process of dissolving is more complicated than it might first appear. This section describes the process of dissolving for ionic compounds, which can be referred to as hydration.

    11.4: Hydration of Ions

    Last updated Oct 7, 2022

    11.3: Precipitation Reactions

    11.5: Hydrogen and Hydroxide Ions

    Ed Vitz, John W. Moore, Justin Shorb, Xavier Prat-Resina, Tim Wendorff, & Adam Hahn

    Chemical Education Digital Library (ChemEd DL)

    We have already stated several times that solubility in water is a characteristic property of many ionic compounds, and Table 1 from Precipitation Reactions provides further confirmation of this fact. We have also presented experimental evidence that ions in solution are nearly independent of one another. This raises an important question, though, because we have also stated that attractive forces between oppositely charged ions in a crystal lattice are large. The high melting and boiling points of ionic compounds provide confirmation of the expected difficulty of separating oppositely charged ions. How, then, can ionic compounds dissolve at room temperature? Surely far more energy would be required for an ion to escape from the crystal lattice into solution than even the most energetic ions would possess.

    Figure 11.4.1 11.4.1

    : The hydration of (a) a positive ion; (b) a negative ion. When ions are dissolved in water, they attract and hold several water dipoles around them shown in the circular area in the circular area in the center of each part of the diagram.

    The resolution of this apparent paradox lies in the interactions between ions and the molecules of water or other polar solvents. The negative (oxygen) side of a dipolar water molecule attracts and is attracted by any positive ion in solution. Because of this ion-dipole force, water molecules cluster around positive ions, as shown in Figure

    11.4.1 11.4.1

    . Similarly, the positive (hydrogen) ends of water molecules are attracted to negative ions. This process, in which either a positive or a negative ion attracts water molecules to its immediate vicinity, is called hydration.

    When water molecules move closer to ions under the influence of their mutual attraction, there is a net lowering of the potential energy of the microscopic particles. This counteracts the increase in potential energy which occurs when ions are separated from a crystal lattice against their attractions for other ions.

    Thus the process of dissolving an ionic solid may be divided into the two hypothetical steps shown in Figure

    11.4.2 11.4.2

    . First, the crystalline salt is separated into gaseous ions. The heat energy absorbed when the ions are separated this way is called the lattice enthalpy (or sometimes the lattice energy). Next, the separate ions are placed in solution; that is, water molecules are permitted to surround the ions. The enthalpy change for this process is called the hydration enthalpy.

    Figure 11.4.2 11.4.2

    : Enthalpy changes and solution. There is usually very little energy or enthalpy change when ionic solids like NaCl dissolve in H2O since the energy needed to separate the ions from each other is not very different from the energy liberated when the ions become hydrated by attracting H2O dipoles around them.

    Since there is a lowering of the potential energy of the ions and water molecules, heat energy is given off and hydration enthalpies are invariably negative. The heat energy absorbed when a solute dissolves (at a pressure of 1.00 atm) is called the enthalpy of solution. It can be calculated using Hess' law, provided the lattice enthalpy and hydration enthalpy are known.

    Example 11.4.1 11.4.1 : Enthalpy

    Using data given in Figure

    11.4.2 11.4.2

    , calculate the enthalpy of solution for NaCl().

    Solution

    According to the figure, the lattice enthalpy is 773 kJ mol–1. The hydration enthalpy is – 769 kJ mol–1. Thus we can write the thermo-chemical equations

    NaCl(s)→ Na + (g)+ Cl − (g)

    NaCl(s)→Na+(g)+Cl−(g)

    △ H f =773 kJ mol △Hf=773kJmol Na + (g)+ Cl − (g)→ Na + (aq)+ Cl − (aq) – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – –

    Na+(g)+Cl−(g)→Na+(aq)+Cl−(aq)_

    △ H h =−769 kJ mol △Hh=−769kJmol NaCl(s)→ Na + (aq)+ Cl − (aq)

    NaCl(s)→Na+(aq)+Cl−(aq)

    △ H s =△ H f +△ H h △Hs=△Hf+△Hh △ H s =(773−769) kJ mol =+4 kJ mol

    △Hs=(773−769)kJmol=+4kJmol

    When NaCl() dissolves, 773 kJ is required to pull apart a mole of Na+ ions from a mole of Cl– ions, but almost all of this requirement is provided by the 769 kJ released when the mole of Na+ and the mole of Cl– becomes surrounded by water dipoles. Only 4 kJ of heat energy is absorbed from the surroundings when a mole of NaCl() dissolves. You can verify the small size of this enthalpy change by putting a few grains of salt on your moist tongue. The quantity of heat energy absorbed as the salt dissolves is so small that you will feel no cooling, even though your tongue is quite a sensitive indicator of temperature changes.

    स्रोत : chem.libretexts.org

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