Weak acids equilibrium problems with

As is often the case with equilibrium problems, we arrive at a quadratic equation in Jc, which we can solve using the quadratic formula (see Appendix IC). However, in many cases we can apply the x is small approximation (first discussed in Section 14.8). In Examples 15.5 and 15.6, we examine the general procedure for solving weak acid equilibrium problems. In both of these examples, the x is small approximation works well. In Example 15.7, we solve a problem in which the x is small approximation does not work. In such cases, we can solve the quadratic equation explicitly, or apply the method of successive approximations (also discussed in Section 14.8). Einally, in Example 15.8, we work a problem in which we find the equilibrium constant of a weak acid from its pH. 

To find the pH, you must find the equilibrium concentration of H30. Treat the problem as a weak acid pH problem with a single ionizable proton. The second proton contributes a negligible amount to the concentration of H30 and can be ignored. Follow the procedure from Example 15.6, shown in condensed form here. Use for ascorbic acid from Table 15.10. 

Plan Although we are dealing specifically with the ionization of a weak acid, this problem is very similar to the equilibrium problems we encountered in Chapter 15. We can solve this problem using the method first outlined in Sample Exercise 15.9, starting with the chemical reaction and a tabulation of initial and equilibrium concentrations. 

Analyze We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because tire neutralization of a weak acid produces the corresponding conjugate base, we ejqsect the pH to be basic at the equivalence point. Plan We should first determine how many moles of acetic acid there are initially. This win teU us how many moles of acetate ion there will be in solution at the equivalence point. We then must determine the final volume of the resulting solution, and the concentration of acetate ion. From this point this is simply a weak-base equilibrium problem like those in Section 16.7. 

Determine equilibrium constants for the reaction of amines with strong or weak acids. (Example 22.8 Problems 27,28) 27... 

As an example of the problem of species in solution, consider the case of a solution made by dissolving some potassium chrome alum, KCrfSO s-12H20, in water. On testing, the solution is distinctly acidic. A currently accepted explanation of the observed acidity is based upon the assumption that, in water solution, chromic ion is associated with six H20 molecules in the complex ion, Cr(H20) a. This complex ion can act as a weak acid, dissociating to give a proton (or hydronium ion). Schematically, the dissociation can be represented as the transfer of a proton from one water molecule in the Cr(H20) 3 complex to a neighboring H20 to form a hydronium ion, H30+. Note that removal of a proton from an H20 bound to a Cr+3 leaves an OH- group at that position. The reaction is reversible and comes to equilibrium ... 

Unfortunately, it is not easy to measure acid strengths of very weak acids like the conjugate acids of simple unsubstituted carbanions. There is little doubt that these carbanions are very unstable in solution, and in contrast to the situation with carbocations, efforts to prepare solutions in which carbanions such as ethyl or isopropyl exist in a relatively free state have not yet been successful. Nor has it been possible to form these carbanions in the gas phase. Indeed, there is evidence that simple carbanions such as ethyl and isopropyl are unstable toward loss of an electron, which converts them to radicals. Nevertheless, there have been several approaches to the problem. Applequist and O Brien studied the position of equilibrium for the reaction... 

We can ignore ions such as Sr2+, which come from strong acids or strong bases in this type of problem. Ions, such as C2H3O2", from a weak acid or a base, weak acid in this case, will undergo hydrolysis, a reaction with water. The acetate ion is the conjugate base of acetic acid (Ka = 1.74 x 10 5). Since acetate is a weak base, this will be a Kb problem, and OH will form. The equilibrium is ... 

In this section, you compared strong and weak acids and bases using your understanding of chemical equilibrium, and you solved problems involving their concentrations and pH. Then you considered the effect on pH of buffer solutions solutions that contain a mixture of acid ions and base ions. In the next section, you will compare pH changes that occur when solutions of acids and bases with different strengths react together. 

As an example of a weak acid-strong base titration, let s consider the titration of 40.0 mL of 0.100 M acetic acid with 0.100 M NaOH. Calculation of the pH at selected points along the titration curve is straightforward because we ve already met all the equilibrium problems that arise. 

Having a conceptual understanding of the effect is a good starting point, but we still need to be able to understand the quantitative relationships between the different components in the equilibrium mixture. In this section, we will see how to deal with the common-ion effect in acid-base equilibrium problems. You will find that these problems are very similar to the weak acid problems earlier in the chapter. 

There are a few things that you need to pay attention to as you set up this problem. First, the sodium need not be written in the equation because it doesn t do anything (it s a spectator). Second, unlike the previous weak acid problems, these problems aren t starting out with no products. The common ion in this reaction is the acetate ion, which is a product of the acetic acid dissociation. When you set up your chart, you need to include all amounts of all substances present at the start of the reaction. We re going to omit water in our chart because it is not part of the equilibrium expression. 

C) pH = 4.0. This is a buffer solution containing a weak acid and its conjugate base. In this problem, you set up the equilibrium calculation table with amounts for HCH02 and CH02 in the start column ... 

Type 1. What is the pH of a 0.01 M solution of a weak acid (HA) with a pKa of 5.0 This is a type 1 problem because initially only the acid form (HA) is present. Hydrogen ions (H+) arise from the dissociation of both HA and H20. Water is a very weak acid, and most of the will come from the dissociation of HA thus, the H20 dissociation can be ignored (see Eq. Qb). Each mole of HA that dissociates gives 1 mol of H+ and 1 mol of A". The HA dissociates until equilibrium is met. 

Type 2. What is the pH of a 0.01 M solution of NaA (pXa of HA = 5.0) The salt NaA completely dissociates in H20 to give Na+ and A". This is a type 2 problem because only the base form (A-) is initially present. HA is a weak acid, the A" will tend to combine with any available H+ to form HA. The only H+ available, however, comes from H20 dissociation, and H20 is such a weak acid that only a limited amount of H+ will become available. The two reactions below will proceed simultaneously until equilibrium is obtained. 

In Chapter 7 we were concerned with calculating the equilibrium concentrations of species (particularly H+ ions) in solutions containing an acid or a base. In this section we discuss solutions that contain not only the weak acid HA but also its salt NaA. Although this case appears to be a new type of problem, it can be handled rather easily by using the procedures developed in Chapter 7. 

The problems involved in the measurement of acidities or relative acidities of weak acids are illustrated by the widely different estimates which have been given for the acidity of substituted acetylenes. Two different approaches have been used for measuring the equilibrium acidity of carbon acids which do not ionize in the pH range. In one approach, the ionization of a carbon acid is studied in mixed solvents containing base. Some of these solutions are more basic than aqueous solutions and by varying the solvent mixture the ionization of acids with pK values in the range 12—25 can be studied. Values at the low end of the pK range are directly compared with aqueous p/iC values. It is assumed that ratios of the activity coefficients (f) for the ionized (S-) and unionized acids (SH) are the same for all the acids studied and an acidity function (86)... 

Very many problems in solution chemistry are solved with use of the acid and base equilibrium equations. The uses of these equations in discussing the titration of weak acids and bases, the hydrolysis of salts, and the properties of buffered solutions are illustrated in the following sections of this chapter. 

The analogous problem of determining changes in the standard thermodynamic functions associated with the ionization of weak acids from the relation between the equilibrium constant (K) and the temperature has received further attention within the last two years. Ives and Mardsen (1965) have expressed In A as a fonction of several orthogonal polynomials of the temperature. Their method has several advantages but... 

The concept of eqnilibrinm constants is extremely important in chemistry. As you will soon see, equilibrinm constants are the key to solving a wide variety of stoichiometry problems involving eqnilibrium systems. For example, an industrial chemist who wants to maximize the yield of sulfuric acid, say, must have a clear understanding of the equilibrium constants for all the steps in the process, starting from the oxidation of sulfur and ending with the formation of the final product. A physician specializing in clinical cases of acid-base imbalance needs to know the equilibrium constants of weak acids and bases. And a knowledge of equilibrium constants of pertinent gas-phase reactions will help an atmospheric chemist better understand the process of ozone destraction in the stratosphere. 

Generally, we can calculate the hydrogen ion concentration or pH of an acid solution at equilibrium, given the initial concentration of the acid and its value. Alternatively, if we know the pH of a weak acid solution and its initial concentration, we can determine its K. The basic approach for solving these problems, which deal with equilibrium concentrations, is the same one outlined in Chapter 14. However, because acid ionization represents a major category of chemical equilibrium in aqueous solution, we will develop a systematic procedure for solving this type of problem that will also help us to understand the chemistry involved. 

These can be illustrated by the use of emf measurements to find the equilibrium constants for weak acids and bases, for the self ionisation of water, for the formation of a complex or ion pair and for the solubility of sparingly soluble salts. This, taken with the situations described in the previous worked problems, illustrates the extreme versatility of emf studies. 

Such a mixture of two weak acids seems as a complex problem. As nevertheless HNO2 from the Ka values is a far stronger acid than HCN (these may be directly compared as the Ka s have a similar unit), we will assume that this acid is the dominating contributor to H ions in the solution. We will thereby focus only on this equilibrium with corresponding equilibrium expression ... 

We have seen earlier how calculations of pH in solutions with strong acid and strong base are relatively simple because strong acids and strong bases are completely dissociated. On the contrary, pH calculations in cases where the titrated acid is weak is not as simple. In order to be able to calculate the concentration of HsO ions after the addition of a given amount of strong base it is necessary to look at the weak acids dissociation equilibrium. Calculations of pH curves for titration of a weak acid with a strong base involve a series of buffer-related problems. 

This problem is one we have met earlier. We are to determine pH in a solution of a weak acid. We are to write the equation of equilibrium with corresponding expression of equilibrium. 

However, the fact that HOCl is a weak acid, while HCIO3 and HCl are strong ones (see Table 6.3) means that, in the presence of hydrogen ions, [OCl] is protonated and this affects the position of equilibrium 16.81 HOCl is more stable with respect to disproportionation than [OCl]. On the other hand, the disproportionation of chlorate into perchlorate and chloride is realistically represented by equilibrium 16.82. From the data in Figure 16.9, this reaction is easily shown to be thermodynamically favourable (see problem 16.18b at the end of the chapter). Nevertheless, the reaction does not occur in aqueous solution owing to some undetermined kinetic factor. 

The second type of equilibrium problem involving weak acids gives some concentration data and the value 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 17.6 to 17.8). 

At the various points in your titration curve, list the major species present after the strong base (NaOH, for example) reacts to completion with the weak acid, HA. What equilibrium problem would you solve at the various points in your titration curve to calculate the pH Why is pH > 7.0 at the equivalence point of a weak acid-strong base titration Does the pH at the halfway point to equivalence have to be less than 7.0 What does the pH at the halfway point equal Compare and contrast the titration curves for a strong acid-strong base titration and a weak acid-strong base titration. 

The compounds at the beginning of Table 4.2 are very strong acids. Their equilibrium constants are very large and cannot be measured accurately because the concentrations of the reactants are extremely small. The equilibrium constants for these compounds are determined by some indirect method and only approximate values can be obtained. Because the pKg values cannot be determined very precisely, they are listed without any figures right of the decimal place. A similar problem occurs with the extremely weak acids at the end of the table. 

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. 

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