The second equivalence point

At the second equivalence point all of the stoichiometric amount of H2A(aq) initially present will have been neutralised, i.e. the H2A(aq) has been converted to an equal stoichiometric amount of Na2A(aq) which is fully ionised to Na+(aq) and A (aq). This occurs when a stoichiometric amount of base has been added equal to twice the stoichiometric amount of H2A(aq) originally present. This can be handled in the same way as the first equivalence point. 

Depending on the values of Ki and K2 either the more approximate or the more rigorous methods will have to be used. 

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Figure 9.8b shows a titration curve for a mixture consisting of two weak acids HA and HB. Again, there are two equivalence points. In this case, however, the equivalence points do not require the same volume of titrant because the concentration of HA is greater than that for HB. Since HA is the stronger of the two weak acids, it reacts first thus, the pH before the first equivalence point is controlled by the HA/A buffer. Between the two equivalence points the pH reflects the titration of HB and is determined by the HB/B buffer. Finally, after the second equivalence point, the excess strong base titrant is responsible for the pH. 

The end point for this titration is improved by titrating to the second equivalence point, boiling the solution to expel CO2, and retitrating to the second equivalence point. In this case the reaction is Na2C03 + 2H3O+ -> CO2 + 2Na+ + 3H2O TRIS stands for tr/s-(hydroxymethyl)aminomethane. 

There are two ways in which the sensitivity can be increased. The first, and most obvious, is to decrease the concentration of the titrant, since it is inversely proportional to the sensitivity, k. The second method, which only applies if the analyte is multiprotic, is to titrate to a later equivalence point. When H2SO3 is titrated to the second equivalence point, for instance, equation 9.10 becomes... 

To the second equivalence point NaOH + NaH2P04------> Na2HP04 + H20... 

To reach the second equivalence point means titrating 0.550 mmol H2P04, which requires an additional volume of titrant given by... 

Why does the titration curve of sodium carbonate have two inflection points Why does this titration require that the solution be boiled as you approach the second equivalence point Why can bromcresol green be used as the indicator and not phenolphthalein ... 

Describe the chemistry that occurs in each of the following regions in curve (a) in Figure 7-8 (i) before the first equivalence point (ii) at the first equivalence point (iii) between the first and second equivalence points (iv) at the second equivalence point and (v) past the second equivalence point. For each region except (ii), write the equation that you would use to calculate [Ag4 ]. 

The volume at the second equivalence point must be 2Ve, because the second reaction requires the same number of moles of HCI as the first reaction. 

At the second equivalence point, we have made BH which can be treated as a monoprotic weak acid. 

Point E is the second equivalence point, at which the solution is formally the same as one prepared by dissolving BH2C12 in water. The formal concentration of BHj4 is... 

There is no perceptible break at the second equivalence point (J), because BH 4 is too strong an acid (or, equivalently, BH is too weak a base). As the titration approaches low pH (S3), ... 

Alkalinity is defined as the capacity of natural water to react with H+ to reach pH 4.5, which is the second equivalence point in the titration of carbonate (CO5 ) with H. To a good approximation, alkalinity is determined by OH-, CO j, and HCOf ... 

Acidity of natural waters refers to the total acid content that can be titrated to pH 8.3 with NaOH. This pH is the second equivalence point for titration of carbonic acid (H2C03) with OH-. Almost all weak acids in the water also will be titrated in this procedure. Acidity is expressed as millimoles of OH- needed to bring 1 L of water to pH 8.3. 

Table 11-6 gives useful equations derived by writing a charge balance and substituting fractional compositions for various concentrations. For titration of the diprotic acid, H2A, ct> is the fraction of the way to the first equivalence point. When = 2, we are at the second equivalence point. It should not surprise you that, when cj> = 0.5, pH = pAj and, when = 1.5, pH pAT2. When = 1, we have the intermediate HA " and pH j(pAj -I- pA j). 

At the Second Equivalence Point At this point, we have added enough NaOH to convert all the HA to A-, and we have a 1.00 M solution of a basic salt (Section 15.14). The principal reaction is... 

Since the initial solution of H2A+ contained 1.00 mol of H2A+, the amount of NaOH required to reach the second equivalence point is 2.00 mol. Beyond the second equivalence point, the pH is determined by [OH-] from the excess NaOH. 

A 25.0 mL sample of a diprotic acid is titrated with 0.240 M KOH. If 60.0 mL of base is required to reach the second equivalence point, what is the concentration of the acid ... 

A 40.0 mL sample of a mixture of HC1 and H3PO4 was titrated with 0.100 M NaOH. The first equivalence point was reached after 88.0 mL of base, and the second equivalence point was reached after 126.4 mL of base. 

At point E on the titration curve, 2.5 of base per mole of phosphoric acid have been added. Again, it is erroneous to assume that the P04- ion concentration equals the HP04- ion concentration at this half-equivalence point because a significant fraction of the base added beyond the second equivalence point (D in Figure 2.4) has not reacted with the HP04- ion. Thus, the proper expression for the third dissociation constant for phosphoric acid is given by the relations... 

Assay (Note Use a combination pH-electrode for all titrations.) Accurately weigh between 0.100 and 0.150 g of sample, and dissolve it in 50 mL of ethanol. Perform the titration under a flow of nitrogen. Titrate with standardized 0.1 A tetrabutylammonium hydroxide in methanol or 2-propanol. Determine the volume of titrant needed to reach the first equivalence point (Vi mL) and the second equivalence point (V2 mL). 

Example 8.10 shows how the pH is calculated at the first equivalence point of the titration of phosphoric acid with sodium hydroxide, where the solution contains H2PO4-. Note that the pH of the solution does not depend on the concentration of H2P04 in the solution. Thus the pH is 4.67 at the first equivalence point in any typical titration of H3PO4. Likewise, at the second equivalence point, where the major species are HP042 and H20, the pH is calculated from the expression... 

For the titration of 0.01 M V(V) with relatively concentrated V(II) at pH = 0, neglecting dilution and assuming activity coeflScients of unity, calculate the potential at the following points (a) the point where half the V(V) has been titrated to V(FV) (b) the V(IV) equivalence point (Hint calculate the equi-hbrium constant of the reaction 2VO + 2H2O V(OH)4 + (c)the point where half the V(IV) has been titrated to V(III) (d) the V(III) equivalence point (e) a point beyond the second equivalence point corresponding to the volume required to reach the first equivalence point. 

We can also see that aspartic acid carries the maximum total number of charges at the second equivalence point. 

We assumed that at the second equivalence point the a-amino group is completely uncharged and the e-amino group is completely ionized. These assumptions are valid for calculating the net charge on lysine for the same reasons described earlier concerning aspartic acid. 

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