Acid-base equilibria problems weak acids

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. 

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. 

Outline the procedure for the exact treatment of acid-base equilibrium and use it to find the pH of a very dilute solution of a weak acid or base (Section 15.8, Problems 69-70). 

Besides equilibrium constant equations, two other types of equations are used in the systematic approach to solving equilibrium problems. The first of these is a mass balance equation, which is simply a statement of the conservation of matter. In a solution of a monoprotic weak acid, for example, the combined concentrations of the conjugate weak acid, HA, and the conjugate weak base, A , must equal the weak acid s initial concentration, Cha- ... 

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 ... 

C17-0037. Outline the procedure for working an equilibrium problem for a weak acid-base system. 

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 ... 

The reasoning above allows us to find good qualitative answers, but in order to be able to do quantitative problems (how much is present, etc.), the extent of the dissociation of the weak acids and bases must be known. That is where a modification of the equilibrium constant is useful. 

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. 

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 ... 

All acid-base problems involving aqueous solutions of weak acids (HA) and/or their corresponding base forms (A-) fall into three distinct types type 1, Solutions of acid form only (e.g., HA) type 2, solutions of base form only (e.gM A-) type 3, solutions of both acid and base forms (e.g., HA and A"). For all three types of problems, two equilibrium conditions must be satisfied ... 

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 aqueous solutions of weak acids or weak bases, two or more equilibrium systems exist at the same time, including the ionization of water. In such solutions, all equilibrium constant expressions must be satisfied in order for the system to be at equilibrium. Therefore, we can use any or all of the equilibrium constant expressions in these problems. 

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. 

Formulate the equilibrium expression for the ionization of a weak acid or base, and use it to determine the pH and fraction ionized (Section 15.4, Problems 27-36). 

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. 

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. 

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. 

A1 acetic acid is now brought to acetate form (CH3COO ). The problem is now to determine pH in a solution of a weak base with a concentration of 0.05 M (half of the initial concentration). The base equilibrium and the corresponding base equilibrium constant Kb (5.6 10 ° M) are to be written ... 

Just as you saw in Chapter 17 for equilibrium problems in general, there are two types of equilibrium problems involving weak acids and their conjugate bases ... 

A buffer generally contains a weak acid and its weak conjugate base, or a weak base and its weak conjugate acid, in water. You can solve for the pH by setting up the equilibrium problem using the reaction of the weak acid or the K, reaction of the conjugate base. Both reactions give the same answer for the pH of the solution. Explain. 

Sketch the titration curve for a weak acid titrated by a strong base. When performing calculations concerning weak acid-strong base titrations, the general two-step procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH. What reaction takes place in the stoichiometry part of the problem What is assumed about this reaction ... 

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. 

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. 

Solve Because HF is a weak acid and HCl is a strong add, the major species in solution are HF, H, and Cl. The Cl, which is the conjugate base of a strong acid, is merely a spectator ion in any acid-base chemistry. The problem asks for [F ], which is formed by ionization of HF. Thus, the important equilibrium is HF(aq)... 

The final area that can be discussed involves the kinetics and mechanism of reactions of Grignard reagents with ketones, nitriles, Schiff bases, and weak acids. This area is intimately tied to the problem of the structure of the organomagnesium reagent itself and this in turn centers around the values of the equilibrium constants for the processes... 

Before any base is added, the solution contains just HA in water. This is a weak-acid problem in which the pH is determined by the equilibrium... 

---