Half-neutralization

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. 

It follows that at half-neutralization pH - pKa, while at A = 1/11 the pH = pjFira - land at A = 10/11 the pH = pKa + 1 in fact this means that the whole titration takes place within 2 pH units, which agrees with the maximum pH range of acid-base colour indicators. 

On the basis of the Henderson equation for titration of acid or base one can prove mathematically that the half-neutralization point represents a true inflection point and that as the titration end-point dpH/dA is maximal or minimal, respectively (the latter is only strictly true for titration of a weak acid with a weak base and vice versa). 

In practice, the volume of titrant is plotted instead of the titration parameter, a. At the titration end-points Ej and E2, the volumes of titrant consumed indicate the respective amounts of the acids, while their pH values or better the pH heights around the half-neutralization points, h.n.p and h.n.p2, are related to the identities of the acids. Therefore, in the two-dimensional figure the abscissa represents the quantitative aspects and the ordinate the qualitative aspects. 

While the acceleration afforded to the cyclization of 32 by La3 + in methanol is certainly spectacular, this is not a biologically relevant metal ion and its charge exceeds that of the natural metal ion Zn2+. Very recent investigations of Zn2+-catalysis of the methanolysis and ethanolysis of 32 indicated that there were indeed interesting catalytic effects, and that the situation in pure ethanol is quite different.85 Shown in Figs 15 and 16 are plots of the pseudo-first-order rate constant (kobs) for ethanolysis and methanolysis of HPNPP (32) as a function of [Zn2+]total when the [ OR]/[Zn2+] ratio is 0.5. This ratio was chosen to buffer the system at the half neutralization jjpH of 7 in ethanol8 and 9.5 in methanol at [Zn2+]to)ai = l-2 mM7... 

When the acid is half neutralized, half of the number of moles of HA originally present will be converted to the conjugate base A. So [HA] = [A ] and Ka will be equal to [H30+] since [HA] / [A ] = 1. Therefore, when the acid is half-neutralized, pH = pKa. The point at which the pH is equal to pKa can be seen in Figure 1. 

Refer to the volume of NaOH on your graph (from question 2). Calculate half this volume. On your graph, find the pH when the solution was half-neutralized. 

Calculate [HaO" ] when the CH3COOH was half-neutralized. How does this value compare with your value of for CH3COOH ... 

In collaboration with Jon Belisle, octanol pKa values were measured for a series of benzoic acids and phenols. A coupled electrode calibrated in aqueous buffers was used. The haIf-neutralization potential was measured since the Renderson-Hasselbalch equations would not apply. The titrant was 0.1 sodium hydroxide in isopropanol methanol 4 1. The titrant was only 6% of the total volume at half-neutralization, so the medium was essentially octanol-like. The results are listed in Table I and some benzoic acid values are plotted in Figure 6. 

Half neutralization of a phthalic anhydride solution forms potassium hydrogen phthalate. 

The acid dissociation constant, pKD, was determined for 4-methylimino-pentane-2-one as a function of mole fraction dioxane and the data conform (within 0.03) to the expression, pKD = 10.78 + 14.19 N2, in the range N2=0.08 to 0.173. The half-neutralized ligand was stable in 50 volume % dioxane for at least an hour. Formation constants could not be determined for the ligand again, hydrolysis was indicated. 

Before the half-neutralization point, A-, formed by the addition of titrant (strong base) reacts with the excess HA to form HA2, thus the value of the [A-]/ [HA] ratio is kept lower than in the absence of homoconjugation. On the other... 

At the half-neutralization point, however, this relation does not hold and / is expressed as follows, if Kf(HA )(Ca+Cs) l ... 

It is readily seen from this equation that the pKa of an acid or a base is the pH of half neutralization when the concentrations of B and BH+ are equal. 

The influence of substituents in mono- and disubstituted 2,3-dimethyl-quinoxalines and their conjugate acids on the UV and NMR spectra have been correlated with a,196 while stretching frequencies of the carbonyl group of 6-substituted quinazolinediones and their half-neutralization potentials have been related to ctp.197... 

The Henderson-Hasselbach equation is useful for calculating the molar ratio of base (proton acceptor) to acid (proton donor) for a given pH and pK or for calculating the pK, given the ratio of base (proton acceptor) to acid (proton donor). It can be seen that when the concentration of anion or base is equal to the concentration of undissociated acid (i.e., when the acid is half neutralized), the pH of the solu-tion is equal to the pK of the acid. 

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