Equilibrium problems finding concentrations

The second main type of equilibrium problem asks for values of equilibrium concentrations. We also use concentration tables for this type of problem, with one additional feature. In such problems, we need to assign a variable x to one unknown concentration, and then we use the equilibrium constant to find the value of x by standard algebraic techniques. Examples 16-11 and 16-12 illustrate this use and manipulation of unknowns. 

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. 

A typical equilibrium problem involves finding the equilibrium concentrations (or pressures) of reactants and products, given the value of the equilibrium constant and the initial concentrations (or pressures). 

So far, we have learned that if we know the chemical equilibria involved in a system, we can write a corresponding system of equations that allows us to solve for the concentrations of all species in the system. Although the systematic method gives us the means to solve equilibrium problems of great complexity, it is sometimes tedious and time consuming, particularly when a system must be solved for various sets of experimental conditions. For example, if we wish to find the solubility of silver chloride as a function of the concentration of added chloride, the system of five equations and five unknowns must be solved repetitively for each different concentration of chloride (see Example 11-9). 

In most equilibrium problems, we use quantities (concentrations or pressures) of reactants and products to find K, or we use K to find quantities. We use a reaction table to summarize the initial quantities, how they change, and the equilibrium quantities. When K is small and the initial quantity of reactant is large, we assume the unknown change in the quantity (x) is so much smaller than the initial quantity that it can be neglected. If this assumption is not justified (that is, if the error is greater than 5%), we use the quadratic formula to find x. To determine reaction direction, we compare the values of Q and K. 

Two common types of weak-acid equilibrium problems involve finding Kg from a concentration and finding a concentration from Kg. We summarize the information in a reaction table, and we simplify the arithmetic by assuming (1) is so... 

In Chapters 17 and 18, we described two types of equilibrium problems. In one type, we use concentrations to find K, and in the other, we use K to find concentrations. Here we encounter the same two types. 

A typical equilibrium problem Involves finding the equilibrium concentrations (or pressures) of reactants and products, given the value of the equilibrium constant and the initial concentrations (or pressures). However, since such problems sometimes become complicated mathematically, we will develop useful strategies for solving them by considering cases for which we know one or more of the equilibrium concentrations (or pressures). 

As chemical equilibrium problems are normally posed, we are given the total (analytical) concentrations of all components, the stoichiometry and stability constants of all species, and are asked to find the free equilibrium concentration of all components, from which we can easily compute the free concentrations of all species. This problem is solved as follows (see Tables I and II for description of symbols) from an initial guess for the concentration of components, the approximate concentration of the species can be computed ... 

In equilibrium problems, we typically use quantities (concentrations or pressures) of reactants and products to find K, or we use K to find quantities. 

The second type of equilibrium problem involving weak acids gives some concentration data and and asks for the equilibrium concentration of some component. Such problems are very similar to those we solved in Chapter 17 in which a substance with a given initial concentration reacted to an unknown extent (see Sample Problems 17.8 to 17.10). We will use a reaction table in these problems to find the values, and since we now know that [H3O IfromUiO is so small relative to [HjO ] we will neglect it and enter the initial [H30 ] in aU reaction tables as zero. 

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. 

PROBLEM STRATEGY This problem is the reverse of the preceding ones instead of finding K p from the solubility, here you calculate solubility from the K p. You follow the three steps for equilibrium problems, but since the molar solubility is not hnmediately known, you assign it the value x. For Step 1, you obtain the concentration of each ion by multiplying x by the coefficient of the ion in the chemical equation. In Step 2, you obtain AT as a cubic in x. In Step 3, you solve the equilibrium-constant equation... 

In all equilibrium constant problems, find the values of the concentrations first and then substitute as the K expression requires. Keep the steps separate. 

To find [CHO2 ] we must solve an equilibrium problem. However, the initial concentration of H30 in this case is not negligible (as it has been in all the other weak acid equilibrium problems that we have worked so far) because HCl has formed a significant amount of H30. The concentration of H3O+ formed by HCl becomes the initial concentration of H30 in the ICE table for HCHO2 as shown here ... 

In Examples 15.17 and 15.18, we illustrated how to find the concentration of H30 for such a solution, which is eqnal to the concentration of HX . What if instead we needed to find the concentration of X To find the concentration of X , we nse the concentration of HX and H30 (from the first ionization step) as the initial concentrations for the second ionization step. We then solve a second equilibrium problem using the second ionization eqnation and as shown in Example 15.19. 

To find the [CsHsOe ], use the concentrations of [HCsHsOs ] and H30 produced by the first ionization step (as calculated in Example 15.17) as the initial concentrations for the second step. Because of the 1 1 stoichiometry, [HCgHgOg ] = [H30 ]. Then solve an equilibrium problem for the second step similar to that of Example 15.6, shown in condensed form here. Use for ascorbic acid from Table 15.10. 

In this section, we will work through some additional examples, all of which involve the ionization of a weak acid or a weak base in water. Typically, we are required to find the equilibrium concentrations or determine the pH. To do so, we must solve an equilibrium problem. 

Example 4.2 used the method of false transients to solve a steady-state reactor design problem. The method can also be used to find the equilibrium concentrations resulting from a set of batch chemical reactions. To do this, formulate the ODEs for a batch reactor and integrate until the concentrations stop changing. This is illustrated in Problem 4.6(b). Section 11.1.1 shows how the method of false transients can be used to determine physical or chemical equilibria in multiphase systems. 

For the case of a three-phase problem, where the solute is accessible to the a, (3, and y phases, Whitaker [427] finds the overall average phase concentration for the case of local mass equilibrium given by... 

Construct a table of initial concentrations, changes in concentration, and equilibrium concentrations for each species that appears in the equilibrium constant expression. The equilibrium concentrations from the last row of the table are needed to find Kgq. Start by entering the data given in the problem. The initial concentration of benzoic acid is 0.125 M. Pure water contains no benzoate ions and a negligible concentration of hydronium ions. The problem also states the equilibrium concentration of hydronium ions, 0.0028 M. 

The problem asks for an equilibrium constant, which means we need to find equilibrium concentrations of the species involved in the solubility reaction. Use the seven-step strategy, which we present here without step numbers. 

The problem asks for mass of AgBr that will dissolve. To find this, we must calculate the equilibrium concentration of silver-containing ions in the solution. 

In this problem, we know the initial concentration of iodine and we want to find a new concentration at equilibrium. Looking for a new concentration is a very strong hint to create a reaction table. To begin a reaction table, you need to list each substance present in the reaction at the head of a column. Below each heading, you should enter the initial value for each substance. In this case, our table will begin as ... 

The only substance remaining in the solution that can influence the pH is the nitrite ion. This ion is the conjugate base of a weak acid. Since a base is present, the pH will be above 7. The presence of this weak base means this is a Kb problem. However, before we can attack the equilibrium portion of the problem, we must finish the stoichiometry part by finding the concentration of the nitrite ion. 

You can use the value of Kgp for a compound to determine the concentration of its ions in a saturated solution. The following Sample Problem shows you how to do this. You will use an approach that is similar to the approach you used in section 7.2 to find equilibrium amounts using Kc for homogeneous equilibria. 

However, in doing so one tests two theories the network formation theory and the rubber elasticity theory and there are at present deeper uncertainties in the latter than in the former. Many attempts to analyze the validity of the rubber elasticity theories were in the past based on the assumption of ideality of networks prepared usually by endllnklng. The ideal state can be approached but never reached experimentally and small deviations may have a considerable effect on the concentration of elastically active chains (EANC) and thus on the equilibrium modulus. The main issue of the rubber elasticity studies is to find which theory fits the experimental data best. This problem goes far beyond the network... 

The problem is to find the pH of a solution of the weak acid HA, given the formal concentration of HA and the value of Ka.4 Let s call the formal concentration F and use the systematic treatment of equilibrium ... 

E. feU (a) Using the ion-pair equilibrium constant in Appendix J, with activity coefficients = 1, find the concentrations of species in 0.025 M MgS()4. Hydrolysis of the cation and anion near neutral pH is negligible. Only consider ion-pair formation. You can solve this problem exactly with a quadratic equation. Alternatively, if you use SOLVER, set Precision to le-6 (not le-16) in the SOLVER Options. If Precision is much smaller. SOLVER does not find a satisfactory solution. The success of SOLVER in this problem depends on how close your initial guess is to the correct answer. 

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