Titration curves points, calculation

Each leg of the titration curve is calculated separately. I he first leg, from pH 1 to 6, corresponds to the dissociation of protonated alanine, HjA" " the secund leg, from pH 6 to 11, corresponds to the dissociation of zwitter-ionic alanine, HA. Exactly halfway between the two legs ia the isooloctric point at 6.01. In essence, it s os if we started with HaA at low pH and then titrated with NaOIL When 0.5 equiv of NaOH is added, the de proto nation of H.A " is 50%. done when 1.0 equiv of NaOH is added, the deprotonaiion of HgA " is f mplete and HA predominates (the isoelectric poinU when... 

The extra adsorption function (i.e., TH — TOH , or the adsorption density in micromoles/g) is calculated numerically with a Pc [2]. The concentration (Ca, Cb) and strength (pKa, pKb) of surface sites from titration curves are calculated from the difference at the equivalent points, in the presence and absence of the sample powder, by considering that at halfway to each equivalent point, pH = pKa or pH = pKw — pKb for acidic and basic sites, respectively. 

See Problem 21 for a spreadsheet calculation of the Ca-EDTA titration curve in Figure 9.3 at pH 10. As with calculated acid-base titration curves, the calculations here break down very near the equivalence point due to simplifying assumptions we have made. 

At the various points in your titration curve, list the major species present after the strong base (NaOH, for example) reacts to completion with the weak acid, HA. What equilibrium problem would you solve at the various points in your titration curve to calculate the pH Why is pH > 7.0 at the equivalence point of a weak acid-strong base titration Does the pH at the halfway point to equivalence have to be less than 7.0 What does the pH at the halfway point equal Compare and contrast the titration curves for a strong acid-strong base titration and a weak acid-strong base titration. 

Selected points along the titration curve are calculated in a manner that closely resembles those we used in the last two chapters. With oxidation-reduction titrations, the method is even simpler when we recognize that on either side of the equivalence point there is an excess of one of the two redox couples, allowing us to calculate the concentration ratio of oxidized to reduced forms of the substance being titrated or of the titrant. Knowing... 

The first task in constructing the titration curve is to calculate the volume of NaOIi needed to reach the equivalence point. At the equivalence point we know from reaction 9.1 that... 

Sketching an Acid—Base Titration Curve To evaluate the relationship between an equivalence point and an end point, we only need to construct a reasonable approximation to the titration curve. In this section we demonstrate a simple method for sketching any acid-base titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. 

Where Is the Equivalence Point We have already learned how to calculate the equivalence point for the titration of a strong acid with a strong base, and for the titration of a weak acid with a strong base. We also have learned to sketch a titration curve with a minimum of calculations. Can we also locate the equivalence point without performing any calculations The answer, as you may have guessed, is often yes ... 

Where Is the Equivalence Point In discussing acid-base titrations and com-plexometric titrations, we noted that the equivalence point is almost identical with the inflection point located in the sharply rising part of the titration curve. If you look back at Figures 9.8 and 9.28, you will see that for acid-base and com-plexometric titrations the inflection point is also in the middle of the titration curve s sharp rise (we call this a symmetrical equivalence point). This makes it relatively easy to find the equivalence point when you sketch these titration curves. When the stoichiometry of a redox titration is symmetrical (one mole analyte per mole of titrant), then the equivalence point also is symmetrical. If the stoichiometry is not symmetrical, then the equivalence point will lie closer to the top or bottom of the titration curve s sharp rise. In this case the equivalence point is said to be asymmetrical. Example 9.12 shows how to calculate the equivalence point potential in this situation. 

As the titration begins, mostly HAc is present, plus some H and Ac in amounts that can be calculated (see the Example on page 45). Addition of a solution of NaOH allows hydroxide ions to neutralize any H present. Note that reaction (2) as written is strongly favored its apparent equilibrium constant is greater than lO As H is neutralized, more HAc dissociates to H and Ac. As further NaOH is added, the pH gradually increases as Ac accumulates at the expense of diminishing HAc and the neutralization of H. At the point where half of the HAc has been neutralized, that is, where 0.5 equivalent of OH has been added, the concentrations of HAc and Ac are equal and pH = pV, for HAc. Thus, we have an experimental method for determining the pV, values of weak electrolytes. These p V, values lie at the midpoint of their respective titration curves. After all of the acid has been neutralized (that is, when one equivalent of base has been added), the pH rises exponentially. 

The pH values at other points on the titration curve are similarly calculated. After the equivalence point has been passed, the solution contains excess of OH- ions which will repress the hydrolysis of the salt the pH may be assumed, with sufficient accuracy for our purpose, to be that due to the excess of base present, so that in this region the titration curve will almost coincide with that... 

Principle. By means of potentiometric titration (in nonaqueous media) of a blend of sulfonic and sulfuric acids, it is possible to split the neutralization points corresponding to the first proton of sulfuric acid plus that of sulfonic acid, and to the second proton of sulfuric acid. The first derivate of the titration curve allows identification of the second points the corresponding difference in the volume of titrating agent is used as a starting point in the calculation method (Fig. 4). 

When plotted on a graph of pH vs. volume of NaOH solution, these six points reveal the gross features of the titration curve. Adding additional calculated points helps define the pH curve. On the curve shown here, the red points A-D were calculated using the buffer equation with base/acid ratios of 1/3 and 3/1. Point E was generated from excess hydroxide ion concentration, 2.00 mL beyond the second stoichiometric point. You should verify these additional five calculations. 

The calculated curve shows the general features of the pH titration curve for a diprotic acid. The pH of the solution is acidic at the first stoichiometric point (major species = weak acid HA ) and basic at the second (major species =... 

C18-0020. Glycolic acid (HOCH2 CO2 H), a constituent of sugar cane juice, has a p Zg of 3.9. Sketch the titration curve for the titration of 60.0 mL of 0.010 M glycolic acid with 0.050 M KOH. Indicate the stoichiometric point, the buffer region, and the point of the titration where pH- p. S a. Sketch the curve qualitatively without doing any quantitative calculations. 

Subsequently, Bos and Dahmen used in m-cresol65 (e = 12.29 at 25° C) a potentiometric titration method combined with conductometry. Essential precautions were the preparation of water-free m-cresol (<0.01% of water), the use of a genuine Bronsted base B, e.g., tetramethylguanidine (TMG), and the application of a glass electrode combined with an Ag-AgCl reference electrode filled with a saturated solution of Me4NCl in m-cresol. The ion product of the self-dissociation of m-cresol, Ks, was determined from the part beyond the equivalence point of the potentiometric titration curve of HBr with TMG comparison with titration curves calculated with various Ka values showed the best fit for Ks = 2 10 19... 

To select an indicator for an acid-base titration it is necessary to know the pH of the end point before using equation (5.5) or standard indicator tables. The end point pH may be calculated using equations (3.27), (3.29) or (3.30). Alternatively, an experimentally determined titration curve may be used (see next section). As an example, consider the titration of acetic acid (0.1 mol dm 3), a weak acid, with sodium hydroxide (0.1 mol dm-3), a strong base. At the end point, a solution of sodium acetate (0.05 mol dm 3) is obtained. Equation (3.28) then yields... 

The next point in the titration curve is the equivalence point. At this point, both the material added and the material originally present are limiting. At this point, neither of the reactants will be present and therefore will not affect the pH. If the titration involves a strong acid and a strong base, the pH at the equivalence point is 7. If the titration involves a weak base, only the conjugate acid is present to affect the pH. This will require a Ka calculation. If the titration involves a weak acid, only the conjugate base is present to affect the pH. This will require a Kb calculation. The calculation of the conjugate acid or base will be the moles produced divided by the total volume of the solution. 

The acidity constants calculated from every point in the titration curve (Figure 2.2a and b) are microscopic acidity constants (Eqs. 2.5, 2.6). Each loss of a proton reduces the charge on the surface and thus affects the acidity of the neighboring... 

Surface charge of MnCC>3 (rhodochrosite) as a function of pH and pcOo as determined from surface titration curves. The values of pHpzc (point of zero charge as calculated from equilibrium (cf. Eq. 3.12)) are given by arrows. 

When the second derivative of (5.32) is calculated and set equal to zero, the inflection point of the titration curve is obtained [23, 24, 133, 134). It has been found that the theoretical titration error generally increases with decreasing sample concentration, with increasing value of the solubility product or of the dissociation constant, with increasing value of the dilution factor and with increasing concentration of the interferents. Larger errors are obtained with unsymmetrical titration reactions. The overall error is a combination of these factors the greatest effect is exerted by the sample concentration, a smaller one by the equilibrium constant and the interferents, and the smallest by dilution. To obtain errors below 1%, it must approximately hold that eg, > 10 2 i,K< 10 , < 10 to 10" and r < 0.3. 

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