Simple Henderson-Hasselbalch Equations

The basic relationships between solubility and pH can be derived for any given equilibrium model. The model refers to a set of equilibrium equations and the associated equilibrium quotients. In a saturated solution, three additional equations need to be considered, along with the ionization Eqs. (2a)-(2d), which describe the equilibria between the dissolved acid, base or ampholyte in solutions containing a suspension of the (usually crystaUine) solid form of the compounds  

The concentrations of species in the solid phase, [HA(s)j, [B(s)j and [XH(s)], by convention are taken as unity. Hence, the quotients in Eqs. (10) reduce to the concentrations of the neutral species in the saturated solution, each called the intrinsic solubility of the compound, Sq. 

In a saturated solution, solubility, S, at a particular pH is defined as the sum of the concentrations of all of the species dissolved in the aqueous solution  

In Eqs. (11), [HA], [B] and [XH] are constant (intrinsic solubility), but the other concentrations are variable. The next step involves conversions of all variables into expressions containing only constants and [H j (as the independent variable). Substitution of Eqs. (2) and (10) into (11) produces the desired equations. 

Although Fig. 3.2 properly conveys the shapes of solubility-pH curves in saturated solutions of uncharged species, according to the Henderson-Hasselbalch equation, the indefinite ascendancy of the dashed curves in the plots can be misleading. When pH changes elevate the solubihty, at some value of pH, the solubihty product of the salt wiU be reached, causing the shape of the solubihty-pH curve to level off, as indicated in Fig. 3.2(a) for pH 8.38. 

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Figure 3.2(a) shows a plot of log S versus pH for naproxen, based on re-analysis (unpublished) of the shake-flask [49, 77] and microtiter plate [20] data reported in the literature.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. 

As in Example 6.13, the Henderson-Hasselbalch equation provides a simple way to calculate the pH of a buffer and to determine the change in pH upon adding a strong acid or strong base. 

A. pH = 4.5. This problem is a simple application of the Henderson-Hasselbalch equation. Remember to take the negative logcirithm of K to get p. Plugging in known values yields... 

Type 3 problems reduce to a very simple form because the value of H depends only on Ka and the initial ratio of A7HA. Thus, unlike type 1 and type 2 problems, the value of H+ does not depend on the actual concentrations of A- or HA (provided both A- and HA are large enough so that the H+ can be ignored). Since this type of problem is the one most frequently encountered, the Henderson-Hasselbalch equation is commonly employed ... 

The potentiometric titration of simple acids characterised by an acid dissociation constant Ka is described by the Henderson—Hasselbalch equation (Stumm and Morgan, 1970) ... 

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