What is ion pairing?

Short Answer

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Ion pairing is a phenomenon in which oppositely charged ions in a solution are attracted to each other due to their electrostatic forces, forming pairs that act as single entities. This process is influenced by the dielectric constant of the solvent, ionic strength of the solution, and specific ions involved. Ion pairing affects the physical and chemical properties of a solution, such as solubility, conductivity, and reactivity.

Step by step solution

01

Introduction to Ion Pairing

Ion pairing is a phenomenon that occurs when oppositely charged ions in a solution come together due to their electrostatic attraction. This behavior is most common among ionic species that have high charge numbers and are found in solutions with low ionic strength. The formed pairs of ions are considered a single entity, and they can significantly influence the physical and chemical properties of a solution, such as solubility, conductivity, and reactivity.
02

Electrostatic Interactions

Electrostatic interactions are the fundamental force behind the ion pairing phenomenon. In a solution, ions with opposite charges will naturally experience an attractive force, given by Coulomb's Law: \(F = k * \frac{q_1 * q_2}{r^2}\), where F is the force, k is the electrostatic constant, q_1 and q_2 are the charges of the ions, and r is the distance between them. This attraction drives the ions to come close together and form a stable ion pair.
03

Example of Ion Pairing

Let's consider a simple example of ion pairing: a mixture of sodium chloride (NaCl) and magnesium sulfate (MgSO4) in water. When both salts dissolve in water, they dissociate into their constituent ions: Na+, Cl-, Mg2+, and SO4(2-). In this solution, Na+ and Cl- can form ion pairs, as well as Mg2+ and SO4(2-). The shared electrostatic attraction between these oppositely charged ions creates a stronger interaction between them than with the surrounding water molecules, leading to ion pairing. The stability and occurrence of ion pairs depend on a few factors, including the dielectric constant of the solvent, the ionic strength of the solution, and the specific ions involved. In general, ion pairing is more likely to occur in solutions with low dielectric constants and high ionic strengths, as these conditions promote stronger electrostatic interactions between ions. In summary, ion pairing is the interaction between oppositely charged ions in a solution due to their electrostatic attraction, leading to the formation of pairs that act as single entities. This concept is essential in understanding the behavior of ions in solutions and their impact on various physical and chemical properties.

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Most popular questions from this chapter

In some regions of the southwest United States, the water is very hard. For example, in Las Cruces, New Mexico, the tap water contains about \(560 \mu \mathrm{g}\) of dissolved solids per milliliter. Reverse osmosis units are marketed in this area to soften water. A typical unit exerts a pressure of \(8.0 \mathrm{~atm}\) and can produce \(45 \mathrm{~L}\) water per day. a. Assuming all of the dissolved solids are \(\mathrm{MgCO}_{3}\) and assuming a temperature of \(27^{\circ} \mathrm{C}\), what total volume of water must be processed to produce \(45 \mathrm{~L}\) pure water? b. Would the same system work for purifying seawater? (Assume seawater is \(0.60 \mathrm{M} \mathrm{NaCl}\).)

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \mathrm{X}\) \(10^{-3} \mathrm{~mol} / \mathrm{L} \cdot\) atm for Henry's law in the form \(C=k P\), where \(C\) is the gas concentration \((\mathrm{mol} / \mathrm{L})\).

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In a coffee-cup calorimeter, \(1.60 \mathrm{~g} \mathrm{NH}_{4} \mathrm{NO}_{3}\) was mixed with \(75.0 \mathrm{~g}\) water at an initial temperature \(25.00^{\circ} \mathrm{C}\). After dissolution the salt, the final temperature of the calorimeter contents was \(23.34^{\circ} \mathrm{C}\). a. Assuming the solution has a heat capacity of \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), and assuming no heat loss to the calorimeter, calculate the enthalpy of solution \(\left(\Delta H_{\text {soln }}\right)\) for the dissolution of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) in units of \(\mathrm{kJ} / \mathrm{mol}\). b. If the enthalpy of hydration for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is \(-630 . \mathrm{kJ} / \mathrm{mol}\), calculate the lattice energy of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\).

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