Equilibrium problems calculating concentrations

Solve different types of equilibrium problems calculate K given unknown quantities (concentrations or pressures), or unknown quantities given K, set up and use a reaction table, apply the quadratic equation, and make an assumption to simplify the calculations ( 17.5) (SPs 17.4-17.8) (EPs 17.30-17.45)... 

As mentioned above, in solving equilibrium problems, both concentrations and activity coefficients of the substances participating in the equilibrium must be evaluated. Implicit in Equation 3-3 is the basis for a very important approach to the incorporation of activity coefficients in all equilibrium calculations. As will be discussed below, it will be possible to arrive at a sufficiently good value of the activity coefficient quotient, Q. in terms of a single experimental parameter, the ionic strength. Hence, a useful strategy for taking activities into account in equilibrium calculations consists of ... 

Do the problem in two parts. First, you assume that the H30 ion from the strong acid and the conjugate base from the buffer react completely. This is a stoichiometric calculation. Actually, the HgO ion and the base from the buffer reach equilibrium just before complete reaction. So you now solve the equilibrium problem using concentrations from the stoichiometric calculation. Because these concentrations are not far from equilibrium, you can use the usual simplifying assumption about x. 

A quantitative solution to an equilibrium problem may give an answer that does not agree with the value measured experimentally. This result occurs when the equilibrium constant based on concentrations is matrix-dependent. The true, thermodynamic equilibrium constant is based on the activities, a, of the reactants and products. A species activity is related to its molar concentration by an activity coefficient, where a = Yi[ ] Activity coefficients often can be calculated, making possible a more rigorous treatment of equilibria. 

At the equivalence point, the moles of Fe + initially present and the moles of Ce + added are equal. Because the equilibrium constant for reaction 9.16 is large, the concentrations of Fe and Ce + are exceedingly small and difficult to calculate without resorting to a complex equilibrium problem. Consequently, we cannot calculate the potential at the equivalence point, E q, using just the Nernst equation for the analyte s half-reaction or the titrant s half-reaction. We can, however, calculate... 

The quantitative treatment of a reaction equilibrium usually involves one of two things. Either the equilibrium constant must be computed from a knowledge of concentrations, or equilibrium concentrations must be determined from a knowledge of initial conditions and Kgq. In this section, we describe the basic reasoning and techniques needed to solve equilibrium problems. Stoichiometry plays a major role in equilibrium calculations, so you may want to review the techniques described in Chapter 4, particularly Section 4- on limiting reactants. 

For this example, we summarize the first four steps of the method The problem asks for the concentration of ions. Sodium hydroxide is a strong base that dissolves in water to generate Na cations and OH- anions quantitatively. The concentration of hydroxide ion equals the concentration of the base. The water equilibrium links the concentrations of OH" and H3 O" ", so an equilibrium calculation is required to determine the concentration of hydronium ion. What remains is to organize the data, carry out the calculations, and check for reasonableness. 

Sol id Sol utions. The aqueous concentrations of trace elements in natural waters are frequently much lower than would be expected on the basis of equilibrium solubility calculations or of supply to the water from various sources. It is often assumed that adsorption of the element on mineral surfaces is the cause for the depleted aqueous concentration of the trace element (97). However, Sposito (Chapter 11) shows that the methods commonly used to distinguish between solubility or adsorption controls are conceptually flawed. One of the important problems illustrated in Chapter 11 is the evaluation of the state of saturation of natural waters with respect to solid phases. Generally, the conclusion that a trace element is undersaturated is based on a comparison of ion activity products with known pure solid phases that contain the trace element. If a solid phase is pure, then its activity is equal to one by thermodynamic convention. However, when a trace cation is coprecipitated with another cation, the activity of the solid phase end member containing the trace cation in the coprecipitate wil 1 be less than one. If the aqueous phase is at equil ibrium with the coprecipitate, then the ion activity product wi 1 1 be 1 ess than the sol ubi 1 ity constant of the pure sol id phase containing the trace element. This condition could then lead to the conclusion that a natural water was undersaturated with respect to the pure solid phase and that the aqueous concentration of the trace cation was controlled by adsorption on mineral surfaces. While this might be true, Sposito points out that the ion activity product comparison with the solubility product does not provide any conclusive evidence as to whether an adsorption or coprecipitation process controls the aqueous concentration. 

To use activity coefficients, first solve the equilibrium problem with all activity coefficients equal to unity. From the resulting concentrations, compute the ionic strength and use the Davies equation to find activity coefficients. With activity coefficients, calculate the effective equilibrium constant K for each chemical reaction. K is the equilibrium quotient of concentrations at a particular ionic strength. Solve the problem again with K values and find a new ionic strength. Repeat the cycle until the concentrations reach constant values. 

As each aliquot of Ce4+ is added, titration reaction 16-1 consumes the Ce4+ and creates an equal number of moles of Ce3+ and Fe3+. Prior to the equivalence point, excess unreacted Fe2+ remains in the solution. Therefore, we can find the concentrations of Fe2+ and Fe3+ without difficulty. On the other hand, we cannot find the concentration of Ce4+ without solving a fancy little equilibrium problem. Because the amounts of Fe2+ and Fe3+ are both known, it is convenient to calculate the cell voltage by using Reaction 16-2 instead of Reaction 16-3. 

Because K, depends on concentrations and the product KyKx is concentration independent, Kx must also depend on concentration. This shows that the simple equilibrium calculations usually carried out in first courses in chemistry are approximations. Actually such calculations are often rather poor approximations when applied to solutions of ionic species, where deviations from ideality are quite large. We shall see that calculations using Eq. (47) can present some computational difficulties. Concentrations are needed in order to obtain activity coefficients, but activity coefficients are needed before an equilibrium constant for calculating concentrations can be obtained. Such problems are usually handled by the method of successive approximations, whereby concentrations are initially calculated assuming ideal behavior and these concentrations are used for a first estimate of activity coefficients, which are then used for a better estimate of concentrations, and so forth. A G is calculated with the standard state used to define the activity. If molality-based activity coefficients are used, the relevant equation is... 

Determine the concentration of Z at equilibrium by repeating each of the steps in the prior problem. Calculate the equilibrium concentration of R if 1.0 mol of W and 2.5 mol of Q are placed in a 1.0-L vessel and allowed to attain equilibrium. 

We have seen that fairly complicated calculations are often necessary to solve equilibrium problems. However, under certain conditions, we can make simplifications that greatly reduce the mathematical difficulties. For example, consider gaseous NOCl, which decomposes to form the gases NO and Cl2. At 35°C the equilibrium constant is 1.6 X 10 5 mol/L. In an experiment in which 1.0 mole of NOCl is placed in a 2.0-liter flask, what are the equilibrium concentrations ... 

For more practice calculating concentrations from the equilibrium constant expression, go to Supplemental Practice Problems in Appendix A. 

Solution of a multiple-equilibrium problem requires us to develop as many independent equations as there are participants in the system being studied. For example, if we wish to compute the solubility of barium sulfate in a solution of acid, we need to be able to calculate the concentration of all the species present in the solution. There are five species [Ba ], [SOj"l, [HSOj], [H30" ], and [OH ]. To calculate the solubility of barium sulfate in this solution rigorously, it is then necessary to develop five independent algebraic equations that can be solved simultaneously to give the five concentrations. 

Although the theory of activity coefficients, especially in concentrated solutions, is still rather incomplete, one must remember that in the quantitative treatment of all chemical equilibrium problems, it is the activity and not the concentration of the participating components which must be used in calculations. 

The pH is minus the logarithm of the activity of protons rather than of their concentration. For ionic strengths less than 0.001 M, the difference in numerical value between the activity and the concentration is immaterial, but with concentrations of a 1-1 electrolyte greater than 0.01 M, the difference is in excess of 0.1. This is especially important when the recorded pH values are used to calculate the chemical speciation (solution of the chemical equilibrium problem), for which the measured activity has to be properly converted into concentration in the mass balance calculation. 

I/i ge/ieral, considering the process to be a combined chemical and phase equilibrium problem will result in more complicated calculations, but will yield somewhat more information—here the concentration of ammonia in the liquid phase. 

Atmospheric solubility equilibrium. The dissolved concentrations of the noble gases in equilibrium with the atmosphere can easily be calculated from Henry s law (Eqn. 2), using the noble gas partial pressures in moist air given by Equation (3). The practical problem is to calculate the Henry coefficients Hi(T,S) in appropriate units. Several different expressions for gas solubilities are common in the literature and a variety of units for the concentrations in the two phases are used. In Equation (2), the units atm for Pi and cm STP g for Ci are widely used in noble gas studies. For some calculations it is convenient to use the same concentration units (e.g., mol m ) for both phases, resulting in the dimensionless Henry coefficient Hj, as in Equation (1). The conversion from Hj in volumetric units (e.g., (mol/lgas)(mol/lwater) ) to Hi in units atm (cm STP g ) is a... 

These last three equations are used to define the total concentrations of the three electrostatic components (see also Table IV). When the equilibrium problem is solved, the charge calculated by summation of charged species in the 0 and p planes must be equal to the electrostatically calculated charge given previously (Equations 9-11). 

Because acid-base reactions in solution generally are so rapid, we can concern ourselves primarily with the determination of species concentrations at equilibrium. Usually, we desire to know [H+], [OH ], and the concentration of the acid and its conjugate base that result when an acid or a base is added to water. As we shall see later in this text, acid-base equilibrium calculations are of central importance in the chemistry of natural waters and in water and wastewater treatment processes. The purpose of this section is to develop a general approach to the solution of acid-base equilibrium problems and to apply this approach to a variety of situations involving strong and weak acids and bases. 

Knowledge Required (1) Algebraic form of the equilibrium expression. (2) Substitutions into the equilibrium expression to calculate equilibrium concentrations. (3) Alternative approaches to solving equilibrium problems. 

Comment The same method can be applied to gaseous equilibrium problems to calculate Kp, in which case partial pressures are used as table entries in place of molar concentrations. Your instructor may refer to this kind of table as an ICE chart, where ICE stands for Initial - ange - Equilibrium. 

Problem Calculating equilibrium concentrations from rate constants... 

Our comments on the stability of metal complexes have evolved from the study of stability constant data. The experimental determination of stability constants is an important but often difiBcult task. Perhaps the greatest problem in equilibrium measurements is to determine which species are actually present in solution. EquiUbriinn constants have been measured by many different methods. Usually a solution of the metal ion and ligand is prepared, sufficient time is allowed for the system to come to equilibrium, and the concentrations of the species in solution are then measured. From these equilibrium concentrations one can calculate the equilibrium constant using an expression such as equation (18). 

Third, calculate the initial and equilibrium concentrations of each of fhe species fhaf parficipafes in fhe equilibrium. We can fabulafe fhe concenfrafions much as we have done in solving ofher equilibrium problems. (Section 15.5)... 

Eor a typical equilibrium problem, the value of K and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equflihrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equUih-riurn concentrations ... 

Once you have determined the equilibrium constant for a reaction, you can use it to calculate the concentrations of substances in an equilibrium mixture. The next example illustrates a simple type of equilibrium problem. 

Consider, for example, how you could answer the following questions What is the hydronium-ion concentration of 0.10 M niacin (nicotinic acid) What is the hydronium-ion concentration of the solution obtained by dissolving one 5.00-grain tablet of aspirin (acetylsalicylic acid) in 0.500 L of water If these were solutions of strong acids, the calculations would be simple 0.10 M monoprotic acid would yield 0.10 M HsO ion. However, because niacin is a weak monoprotic acid, the HsO concentration is less than 0.10 M. To find the concentration, you need the equilibrium constant for the reaction involved, and you need to solve an equilibrium problem. 

As in the previous example, you can do a quick check of your work by essentially reversing the problem, calculating Ka from the equilibrium concentrations of acid, acid ion, and hydronium ion. You can obtain this from the last line of the equilibrium table, now that you have the value of x. 

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