Strong acids equilibrium problems with

Sketch the titration curve for a weak base titrated by a strong acid. Weak base-strong acid titration problems also follow a two-step procedure. 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 acid (HNO3, for example) reacts to completion with the weak base, B. 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 base-strong acid titration If pH = 6.0 at the halfway point to... 

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. 

Note that we solved this problem by first performing a stoichiometric (limiting reactant) calculation and then an equilibrium calculation. A similar strategy works if a strong base such as OH is added instead of a strong acid. The base reacts with formic acid to produce formate ions. Adding 0.10 mol of OH to the HCOOH/HCOO buffer of Example 15.7 increases the pH only to 3.58. In the absence of the buffer system, the same base would raise the pH to 13.00. 

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. 

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. 

If ethanol is the solvent, there is a problem. Ethanol has a pKg of about 15.9 and it is clearly much more acidic than 2-butanone. Once formed, the enolate anion (also a strong base) will react with ethanol to give 2-butanone as the conjugate acid. In other words, in the protic solvent, 34 will react with ethanol to regenerate 32, and this reaction shifts the equilibrium back to the left (Kgi is small), which favors the thermodynamic process. Therefore, an aprotic solvent will favor a large and kinetic control whereas a protic solvent will favor a small and thermodynamic control. 

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... 

In order to gain an insight into the mechanism on the basis of the slope of a Type A correlation requires a more complicated procedure. Consider the Hammett equation. The usual statement that electrophilic reactions exhibit negative slopes and nucleophilic ones positive slopes may not be true, especially when the values of the slopes are low. The correct interpretation has to take the reference process into account, for example, the dissociation equilibrium of substituted benzoic acids at 25°C in water for which the slope was taken, by definition, as unity (p = 1). The precise characterization of the process under study is therefore that it is more or less nucleophilic than the reference process. However, one also must consider the possible influence of temperature on the value of the slope when the catalytic reaction has been studied under elevated temperatures there is disagreement in the literature over the extent of this influence (cf. 20,39). The sign and value of the slope also depend on the solvent. The situation is similar or a little more complex with the Taft equation, in which the separation of the molecule into the substituent, link, and reaction center may be arbitrary and may strongly influence the values of the slopes obtained. This problem has been discussed by Criado (33) with respect to catalytic reactions. 

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