PKa equalization

Now consider the overall shape of the pH curve. The slow change in pH about halfway to the stoichiometric point indicates that the solution acts as a buffer in that region (see Fig. 11.3). At the halfwayr point of the titration, [HA] = [A ] and pH = pfCa. In fact, one way to prepare a buffer is to neutralize half the amount of weak acid present with strong base. The flatness of the curve near pH = pKa illustrates very clearly the ability of a buffer solution to stabilize the pH of the solution. Moreover, we can now see how to determine pKa plot the pH curve during a titration, identify the pH halfway to the stoichiometric point, and set pKa equal to that pH (Fig. 11.8). To obtain the pfCh of a weak base, we find pK3 in the same way but go on to use pKa -1- pfq, = pKw. The values recorded in Tables 10.1 and 10.2 were obtained in this way. 

This very simple result makes it easy to make an initial choice of a buffer we just select an acid that has a pKa equal to the pH we require and pre-... 

For convenience we often express equilibrium constants as the negative logarithm, or pK value. Thus the relative proportion of the neutral and charged species, will be a function of the pKa and solution pH. When the pH is equal to pKa, equal concentrations of the neutral and ionized forms will be present. When pH is less than the pKa, the neutral species will be predominant when pH is greater than pKa, the ionized species will be in excess. The exact equilibrium distribution can be calculated from the equilibrium expression above and the law of mass conservation. 

Any sharp-eyed readers may notice an Inconsistency in the statement that the pKa equals the pH when we have equal concentrations of acid and conjugate base. If. when [A ] = [AH], thepKa = pH, then the pKg for water equals the pH when [H20] - [H30+], We assume that [H20] Is constant at 55.50 mol dm"3 and so [H30+] must also equal 55.56 mol drrf3 and hence pH = pKg --log(55.56). This assumption cannot be valid here rathar [H20] + [H30+] should equal approximately 55.56 mol dm 3. 

The infrared spectrum of cycloserine presented in Fig. 1 was taken in a KBr pellet. A spectrum of the same sample taken in a Nujol Mull is essentially identical to the one presented. Hidy- -, Kuehl, and Stammer showed that the solid state spectrum of cycloserine has two ionizable groups with pKa, equal to. A -. 5 and pKa2 equal to 7 Spectral bands typical of an amino acid zwitterion (2200 cm assigned to the -NH +) and a resonance stabilized hydroxamafe anion (1600 to 1500 cm-1) are in agreement with the peaks represented in Fig. 1. 

When the concentrations of HA and A are equal, equation 9.9 reduces to = [HaO ]) ot pH = pKa. Thus, the piweak acid can be determined by measuring the pH for a solution in which half of the weak acid has been neutralized. On a titration curve, the point of half-neutralization is approximated by the volume of titrant that is half of that needed to reach the equivalence point. As shown in Figure 9.20, an estimate of the weak acid s piQ can be obtained directly from the titration curve. 

Equations (7-66) and (7-67), or related versions, have been used by Hupes and Jencks and by Castro and co-workers to account for curvature. The quantity p/(S defines the center of eurvature of the plot and is expected to occur when the Y>Ka of the nucleophile is equal to the p/( of the leaving group." For weaker nucleophiles pKa < p/(S), breakdown of the tetrahedral intermediate will be rate determining, because the leaving group X is a stronger nucleophile than is N , so 2 < -li if, however, p/( > p/(S, the nucleophilic attack is rate determining. 

This equation says that the logarithm of the concentration of dissociated acid [A-] divided by the concentration of undissociated acid [HA] is equal to the pH of the solution minus the pKa of the acid. Thus, if we know both the pH of the solution and the pKa of the acid, we can calculate the ratio of [A-] to [HA]. Furthermore, when pH = pKai the two forms HA and A- are present in equal amounts because log 1 = 0. 

These equations say that the Kh of an amine multiplied by the of the corresponding ammonium ion is equal to Kw, the ion-product constant for water (1.00 x 10 14). Thus, if we know Ka for an ammonium ion, we also know for the corresponding amine base because /stronger base has an ammonium ion with a larger p... 

Confining attention to the case in which the concentrations of the acid and its salt are equal, i.e. of a half-neutralised acid then pH = pKa. Thus the pH of a half-neutralised solution of a weak acid is equal to the negative logarithm of the dissociation constant of the acid. For acetic (ethanoic) acid, Ka = 1.75 x 10 5 mol L 1, pKfl = 4.76 a half-neutralised solution of, say 0.1M acetic acid will have a pH of 4.76. If we add a small concentration of H + ions to such a solution, the former will combine with acetate ions to form undissociated acetic acid ... 

The pKa value of the equilibrium was found to be equal to 10.2. Meissner and coworkers36 studied also the reaction of OH radicals with DMSO, however, the product of this reaction has no optical absorption in the range 270-800 nm and they measured only the rate of this reaction by a competition method and obtained k = 4.2 x 109m-1s-1. 

FIGURE 10.17 As shown here for five conjugate acid-base pairs, the sum of the pKa of an acid (pink) and the p/Ch of its conjugate base (blue) is constant and equal to pKw, which is 14.00 at 25°C. 

This very simple result makes it easy to make an initial choice of a buffer we just select an acid that has a pfC, close to the pH that we require and prepare an equimolar solution with its conjugate base. When we prepare a buffer for pH > 7 we have to remember that the acid is supplied by the salt, that the conjugate base is the base itself, and that the pKa is that of the conjugate acid of the base (and hence related to the pKh of the base by pR l + pKh = p/conjugate acid and base have unequal concentrations—such as those considered in Examples 11.1 and 11.2—are buffers, but they may be less effective than those in which the concentrations are nearly equal (see Section 11.3). Table 11.1 lists some typical buffer systems. 

Figure 11.8 summarizes the changes in pH of a solution during a titration of a weak acid by a strong base. Halfway to the stoichiometric point, the pH is equal to the pKa of the acid. The pH is greater than 7 at the stoichiometric point of the titration of a weak acid and strong base. The pH is less than 7 at the stoichiometric point of the titration of a weak base and strong acid. 

Figure 3.2(a) shows a plot of log S versus pH for naproxen, based on re-analysis (unpublished) of the shake-flask [49, 77] and microtiter plate [20] data reported in the literature. The dashed curves in Fig. 3.2 were calculated with the simple Henderson-Hasselbalch equations. For pH pKa, the function reduces to the horizontal line log S = log Sq. For pH pXi, log S is a straight line as a function of pH, exhibiting a slope of 1 (and an intercept of log So-pKj). Where the slope is 0.5, the pH equals to the pKj. 

It is convenient to summarize the various reactions in a box diagram, such as Fig. 4.1 [17,275,280], illustrated with the equilibria of the weak base, propranolol. In Fig. 4.1 is an equation labeled pA °et. This constant refers to the octanol pKa, a term first used by Scherrer [280]. When the concentrations of the uncharged and the charged species in octanol are equal, the aqueous pH at that point defines p which is indicated for a weak acid as... 

Characteristic of a box diagram, the difference between the partition coefficients is equal to the difference between the two pKa values [229,275,280,362] ... 

It may surprise some that for a diprotic molecule with overlapping pKa values the region of maximum log I) (0.76 in Fig. 4.4a) does not equal log P a displaced horizontal line in Fig. 4.4a indicates the logP to be 0.89 for morphine [161,162],... 

Figure 4.7 Octanol-water Bjerrum plots for a monoprotic (a) acid and (b) base. The volumes of octanol and water are equal, so that the difference between the apparent pKa and the true pKa is about equal to the partition coefficient. [Avdeef, A., Curr. Topics Med. Chem., 1, 277-351 (2001). Reproduced with permission from Bentham Science Publishers, Ltd.]...

Figure 5.1 shows a tetrad of equilibrium reactions related to the partitioning of a drug between an aqueous environment and that of the bilayer formed from phospholipids. (Only half of the bilayer is shown in Fig. 5.1.) By now, these reaction types might be quite familiar to the reader. The subscript mem designates the partitioning medium to be that of a vesicle formed from a phospholipid bilayer. Equations (4.1)-(4.4) apply. The pAi m in Fig. 5.1 refers to the membrane pKa. Its meaning is similar to that of pAi when the concentrations of the uncharged and the charged species in the membrane phase are equal, the aqueous pH at that point defines pAi em, which is described for a weak base as...

As can be seen in Fig. 6.8, the presence of precipitate causes the apparent pKa, pA jjPP, to shift to higher values for acids and to lower values for bases, and in opposite but equal directions for ampholytes, just as with octanol (Chapter 4) and liposomes (Chapter 5). The intrinsic solubility can be deduced by inspection of the curves and applying the relationship [472]... 

Thus, the pKa value equals pH if the concentrations of protonated and deprotonated states are identical. For this reason, pKa is also called p/fi/2- Equation (10-4) can be rearranged and written in a generalized form in the presence of multiple titration sites,... 

Membrane uptake of nonionized solute is favored over that of ionized solute by the membrane/water partition coefficient (Kp). If Kp = 1 for a nonionized solute, membrane permeability should mirror the solute ionization curve (i.e., membrane permeability should be half the maximum value when mucosal pH equals solute pKa). When the Kp is high, membrane uptake of nonionized solute shifts the ionization equilibrium in the mucosal microclimate to replace nonionized solute removed by the membrane. As a result, solute membrane permeability (absorption rate) versus pH curves are shifted toward the right for weak acids and toward the left for weak bases (Fig. 7). 

The widely used term log P refers to the logarithm of the partition coefficient P of the unionized species, which is P° for a base with no acidic functions, and P1 for a monoprotic acid or an ordinary ampholyte. Log P is equal to log D at any pH remote from the pKa where the molecule exists entirely in its unionized form. Log D at any pH is equal to log P for a non-ionizable molecule. 

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