Henderson-Hesselbach equation

Explain how the Henderson-Hesselbach equation can be used, in conjunction with a titration curve, to determine a pKa. 

When the progression of an acid-base titration is graphed as a function of pH vs the volume of acid or base added, the curve will appear as shown below. If we recall, from general chemistry coursework, that the steepest point on the curve represents the equivalence point of the titration (the point where the amount of acid and base are equal), we can locate the point on the curve that represents the midpoint of the titration. This point is found at half the concentration of base added to acid (or acid added to base) to reach the equivalence point. Once we have done this, we recall the Henderson-Hesselbach equation (Fig. 2.8)—specifically, the term dealing with the concentrations of the ionic and the neutral species. Realizing that at the midpoint of the titration, these concentrations are equal, the logarithmic term in the Henderson-Hesselbach equation reduces to log(l), which is equal to zero. Therefore, the equation reduces to pA a = pH at the midpoint of the titration. 

When acetic acid is titrated with j an equivalent of base, we realize that the term log [A-]/[HA] becomes log ( ) because one part out of four parts of acetic acid has been deprotonated. This leaves three parts acid to one part conjugate base. Filling in this value and that of the pK of acetic acid into the Henderson-Hesselbach equation (Fig. 2.8), solving for pH gives us a value of 4.27 as our answer. 

This behavior is described quantitatively by the Henderson-Hesselbach equation, which can be derived from the definition of Ka ... 

At physiological pH, the Henderson-Hesselbach equation shows that the second equilibrium is most important. Phosphoric acid and phosphate exist in vanishingly small quantities near neutrality. 

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