Acid-dissociation constant solving

Table 8.3 lists the base dissociation constants for several weak bases at 25°C. Nitrogen-containing compounds are Bronsted-Lowry bases, because the lone pair of electrons on a nitrogen atom can bond with H+ from water. The steps for solving problems that involve weak bases are similar to the steps you learned for solving problems that involve weak acids. 

When Ng is solved as a function oih = [H] (with Nj= ), we obtain the titration curve for dicarboxylic acid in terms of the dissociation constants ... 

The base-dissociation constant expression can be used to calculate the concentrations of ions just like the acid-dissociation expression. The problem-solving strategy is identical. 

In the calculation of protolytic equilibria, the ionic product of water, equations for dissociation constants of acids and bases, equations of analytical concentrations and equations of electroneutrality or proton balance are taken for the starting point. Due to the difficulty of the numerical calculation of pH these systems are generally solved by graphical methods. 

Let s consider a mixture of two weak acids HAi and HA2 of dissociation constants Kai and K 2 and of analytical concentrations Ci and C2. Taking into account all the relations that are necessarily satisfied gives a fourth-order equation in [H3O+] that is difficult to solve. However, neglecting the concentration [OH"] and formulating the hypothesis that both acids are weakly dissociated gives the resulting formula ... 

Only cases where Kof, / AT, ate useful for our purpose since Cab = ab if = JQb. Therefore, we consider cases where the dissociation constants of species A and B are different. For given pH, and X) , one can now solve for the ratio ( ab/ ab)i since is assumed known, a B can be determined. Figure 5.2.5 illustrates how ( ab/ ab) varies with -pXA - pH for different positive values of pKf, - pKs (Robinson and Cha, 1985). Here B is a stronger base than A (X, > K f it is also more water soluble. Thus, when the solution becomes more acidic, species B ionizes more and its water solubility increases further. Therefore, will decrease much more rapidly, leading to a high separation factor species A will be extracted much more into the solvent compared to species B, leading to a higher purity of the extract in species A. Of course, the value of Kao " iU be decreased due to an increase in pH. Consequendy, an optimum between the purity of A and its extent of recovery in the solvent has to be struck. The case where pKg > pKf, (Xja > X[Pg.292]

The traditional approximation made to solve for the instrinsic constants is to assume that the effect of the (unionized) carboxylic group on the dissociation of the NHj group is not much changed upon esterification of the carboxyl group. If we denote by the binding (or association) constant to the NHj group of the esterified amino acid, then the approximation... 

Weak acids are weak electrolytes and do not dissociate completely. An equilibrium exists between the reactants and the products, and the equilibrium constant must be taken into account to solve for the pH value. When a weak acid (HA) is dissolved in water, the conjugate base (A ) and conjugate acid (H+) are... 

The thermochemical cycle in Scheme 3 can be used to determine M-H bond dissociation energies (BDEs) of metal hydrides." The required quantities are the thermodynamic acidity (p-ATa) of the metal hydride MH and the oxidation potential for the corresponding base M , °(M /M), both in the same solvent (solv). The thermochemical cycle and its underlying parameters are based on Bordwell s method for determination of C-H BDEs." The constant term C in Equation (7), Scheme 3 contains contributions from the electrode potential in the solvent under consideration,... 

We recall that all sodium salts are soluble and that all soluble ionic salts are completely dissociated in H2O. We recognize that both NaCHsCOO and NaCN are salts of strong bases (which provide the cation) and weak acids (which provide the anion). The anions in such salts hydrolyze to give basic solutions. In the preceding text we determined that for CH3COO = 5.6 X 10" , and in Example 18-19 we determined that K, for CN" = 2.5 X 10 . As we have done before, we first write the appropriate chemical equation and equilibrium constant expression. Then we complete the reaction siunmary, substitute the algebraic representations of equilibriiun concentrations into the equilibrium constant expression, and solve for the desired concentration(s). 

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