Equilibria Equilibrium solubility

Composition of initial Equilibrium solubility Equilibrium composition... 

In our first example, the stress to the equilibrium was applied directly. It is also possible to apply a concentration stress indirectly. Consider, for example, the following solubility equilibrium involving AgCl... 

In this experiment the equilibrium constant for the dissociation of bromocresol green is measured at several ionic strengths. Results are extrapolated to zero ionic strength to find the thermodynamic equilibrium constant. Equilibrium Constants for Calcium lodate Solubility and Iodic Acid Dissociation. In J. A. Bell, ed. Chemical Principles in Practice. Addison-Wesley Reading, MA, 1967. 

The acid dissociation constants (p/iT 2.7), equilibrium solubilities in several solvents (71JPS503) and lipophilicity properties (77JPS1675) for various penicillins have been summarized. 

Crystal Formation There are obviously two steps involved in the preparation of ciystal matter from a solution. The ciystals must first Form and then grow. The formation of a new sohd phase either on an inert particle in the solution or in the solution itself is called nucle-ation. The increase in size of this nucleus with a layer-by-layer addition of solute is called growth. Both nucleation and ciystal growth have supersaturation as a common driving force. Unless a solution is supersaturated, ciystals can neither form nor grow. Supersaturation refers to the quantity of solute present in solution compared with the quantity which would be present if the solution were kept for a veiy long period of time with solid phase in contac t with the solution. The latter value is the equilibrium solubility at the temperature and pressure under consideration. The supersaturation coefficient can be expressed... 

Fig. 10.5. TTT diagram for the precipitation of CuAh from the Al + 4 wt% Cu solid solution. Note that the equilibrium solubility of Cu in Al at room temperature is only 0.1 wt% (see Fig. 10.3). The quenched solution is therefore carrying 4/0.1 = 40 times as much Cu as it wants to.

The alloy aluminium-4 wt% copper forms the basis of the 2000 series (Duralumin, or Dural for short). It melts at about 650°C. At 500°C, solid A1 dissolves as much as 4 wt% of Cu completely. At 20°C its equilibrium solubility is only 0.1 wt% Cu. If the material is slowly cooled from 500°C to 20°C, 4 wt% - 0.1 wt% = 3.9 wt% copper separates out from the aluminium as large lumps of a new phase not pure copper, but of the compound CuAlj. If, instead, the material is quenched (cooled very rapidly, often by dropping it into cold water) from 500°C to 20°C, there is not time for the dissolved copper atoms to move together, by diffusion, to form CuAlj, and the alloy remains a solid solution. 

The equilibrium solubility of common inorganic gases in petroleum liquids can now be estimated by nomograph. The relationship is based on an earlier correlation established by the U.S. National Aeronautics and Space Administration and the subject of a standard method approved by the American Society of Testing and Materials. 

A third method, or phenomenon, capable of generating a pseudo reaction order is exemplified by a first-order solution reaction of a substance in the presence of its solid phase. Then if the dissolution rate of the solid is greater than the reaction rate of the dissolved solute, the solute concentration is maintained constant by the solubility equilibrium and the first-order reaction becomes a pseudo-zero-order reaction. 

Now interpret phase X as pure solute then Cs and co become the equilibrium solubilities of the solute in solvents S and 0, respectively, and we can apply Eq. (8-58). Again the concentrations should be in the dilute range, but nonideality is not a great problem for nonelectrolytes. For volatile solutes vapor pressure measurements are suitable for this type of determination, and for electrolytes electrode potentials can be used. 

Henry s Law. This is an empirical formulation that describes equilibrium solubilities of noncondensable gases in a liquid when Raoult s law fails. It states that the mole fraction of a gas (solute i) dissolved in a liquid (solvent) is proportional to the partial pressure of the gas above the liquid surface at given temperature. That is,... 

Common ion effect The tube at the left contains a saturated solution of silver acetate (AgC2H302). Originally the tube at the right also contained a saturated solution of silver acetate. With the addition of a solution of silver nitrate (AgNOs), the solubility equilibrium of the silver acetate is shifted by the common ion Ag+ and additional silver acetate precipitates. 

The simple form of the equilibrium expression (4) follows directly from the dynamic nature of the solubility equilibrium. There must be a dynamic balance between the rate that iodine molecules leave the ciystal and the rate that iodine molecules return to the crystal. To understand this dynamic balance, we must consider the factors that determine these two rates. 

Thus we see that the equilibrium solubility of a gas again involves a balance between randomness and energy as it does for a solid, but the effects are opposite. For a gas, the tendency toward maximum randomness favors the gas phase, opposing dissolving. The tendency toward minimum energy favors the liquid state, hence favors dissolving. 

Expression (2) applies to a solubility equilibrium, provided we write the chemical reaction to show the important molecular species present. In Section 10-1 we considered the solubility of iodine in alcohol. Since iodine dissolves to give a solution containing molecules of iodine, the concentration of iodine itself fixed the solubility. The situation is quite different for substances that dissolve to form ions. When silver chloride dissolves in water, no molecules of silver chloride, AgCl, seem to be present. Instead, silver ions, Ag+, and chloride ions, Cl-, are found in the solution. The concentrations of these species, Ag+ and Cl-, are the ones which fix the equilibrium solubility. The counterpart of equation (7) will be... 

Solubility equilibrium constants, such as (20) and (22), are given a special name—the solubility product. It is symbolized K,p. A low value of K,p means the concentrations of ions are low at equilibrium. Hence the solubility must be low. Table 10-11 lists solubility products for some common compounds. 

The concentrations of free carbonate and bicarbonate ions determined at solubility equilibrium as a function of pH. Decrements of the concentration near pH = 10 suggest the formation of the Pu(0H)2C03 precipitate and hence lowering solubilities of Pu02 (cf. Figure 2). 

The equilibrium constant for the solubility equilibrium between an ionic solid and its dissolved ions is called the solubility product, Ksp, of the solute. For example, the solubility product for bismuth sulfide, Bi2S3, is defined as... 

When a precipitate has been formed during the qualitative analysis of the ions present in a solution, it may be necessary to dissolve the precipitate again to identify the cation or anion. One strategy is to remove one of the ions from the solubility equilibrium so that the precipitate will continue to dissolve in a fruitless chase for equilibrium. Suppose, for example, that a solid hydroxide such as iron(IIl) hydroxide is in equilibrium with its ions in solution ... 

Adsorption The first stage, applying when a polymer contacts a fluid, is that the latter adsorbs into the polymer surface, reaching equilibrium solubility here. 

C12-0055. Draw a molecular picture illustrating the solubility equilibrium between KCl(s) and KCl(aq). 

A molecular view of the solubility equilibrium for a solution of sodium chloride in water. At equilibrium, ions dissolve from the crystal surface at the same rate they are captured, so the concentration of ions in the solution remains constant. 

By convention, a solubility equilibrium is written in the direction of a solid dissolving to give aqueous ions, and the equilibrium constant for this reaction is called the solubility product ( sp). Here, for example, is the reaction... 

When equal volumes of 0.100 M solutions of sodium bromide and silver nitrate are mixed, a white solid precipitates from the solution. Identify the precipitate, write the net ionic reaction for the solubility equilibrium, and identify any spectator ions. 

All sodium salts are soluble, and so are all nitrate salts, so It makes sense that neither of these ions participates in a solubility equilibrium. Furthermore, nitrate and sodium cations are neither acidic nor basic, so it makes sense that neither participates in an acid-base equilibrium. 

Over the eons, the flow and evaporation of water inside a cavern creates a stunning array of rock sculptures. Stalagmites grow upward from the floor, sometimes joining stalactites to form massive columns. Limestone dams create beautiful pools of water. Limestone draperies fall like curtains from water flowing around overhanging rock. Delicate mineral flowers sprout from the walls. All these features result from the aqueous solubility equilibrium of calcium carbonate. 

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