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Titration curves millimole

We can calculate pH titration curves using the principles of aqueous solution equilibria. To understand why titration curves have certain characteristic shapes, let s calculate these curves for four important types of titration (1) strong acid-strong base, (2) weak acid-strong base, (3) weak base-strong acid, and (4) polyprotic acid-strong base. For convenience, we ll express amounts of solute in millimoles (mmol) and solution volumes in milliliters (mL). Molar concentration can thus be expressed in mmol/mL, a unit that is equivalent to mol/L ... [Pg.679]

FIGURE 6-2 Titration curves of mixed carbonates 0.15 to 1.S7 millimoles of Na2C03 and 0.05 to 0.21 millimoles of NaHCOs per liter. (From Cooper. )... [Pg.114]

Figure 1 9-4 Spreadsheet and plot for titration of 50.00 mL of 0.0500 M Fe " with 0.1000 M Ce. Prior to the equivalence point, the system potential is calculated from the and Fe + concentrations. After the equivalence point, the Ce and Ce + concentrations are used in the Nernst equation. The Fe concentration in cell B7 is calculated from the number of millimoles of Ce added, divided by the total volume of solution. The formula used for the first volume is shown in documentation cell A21. In cell Cl, [Fe- ] is calculated as the initial number of millimoles of Fe present, minus the number of millimoles of Fe formed, divided by the total solution volume. Documentation cell A22 gives the formula for the 5.00-mL volume. The system potential prior to the equivalence point is calculated in cells F7 F12 by using the Nernst equation, expressed for the first volume by the formula shown in documentation cell A23. In cell F13, the equivalence-point potential is found from the average of the two formal potentials, as shown in documentation cell A24. After the equivalence point, the Ce(lll) concentration (cell D14) is found from the number of millimoles of Fe- initially present divided by the total solution volume, as shown for the 25.10-mL volume by the formula in documentation cell D21. The Ce(IV) concentration (El 4) is found from the total number of millimoles of Ce(lV) added, minus the number of millimoles of Fe + initially present, divided by the total solution volume, as shown in documentation cell D22. The system potential in cell FI4 is found from the Nernst equation as shown in documentation cell D23. The chart is then the resulting titration curve. Figure 1 9-4 Spreadsheet and plot for titration of 50.00 mL of 0.0500 M Fe " with 0.1000 M Ce. Prior to the equivalence point, the system potential is calculated from the and Fe + concentrations. After the equivalence point, the Ce and Ce + concentrations are used in the Nernst equation. The Fe concentration in cell B7 is calculated from the number of millimoles of Ce added, divided by the total volume of solution. The formula used for the first volume is shown in documentation cell A21. In cell Cl, [Fe- ] is calculated as the initial number of millimoles of Fe present, minus the number of millimoles of Fe formed, divided by the total solution volume. Documentation cell A22 gives the formula for the 5.00-mL volume. The system potential prior to the equivalence point is calculated in cells F7 F12 by using the Nernst equation, expressed for the first volume by the formula shown in documentation cell A23. In cell F13, the equivalence-point potential is found from the average of the two formal potentials, as shown in documentation cell A24. After the equivalence point, the Ce(lll) concentration (cell D14) is found from the number of millimoles of Fe- initially present divided by the total solution volume, as shown for the 25.10-mL volume by the formula in documentation cell D21. The Ce(IV) concentration (El 4) is found from the total number of millimoles of Ce(lV) added, minus the number of millimoles of Fe + initially present, divided by the total solution volume, as shown in documentation cell D22. The system potential in cell FI4 is found from the Nernst equation as shown in documentation cell D23. The chart is then the resulting titration curve.
Consider the titration of 100 mL of 0.1M Fe + with 0.1M Ce + in 1M HNO3. Each millimole Ce " will oxidize one millimole Fe ", and so the end point will occur at 100 mL. The titration curve is shown in Figure 14.1. This is actually a plot of the potential of the titration solution relative to the NHE, whose potential by definition is zero. [Pg.418]


See other pages where Titration curves millimole is mentioned: [Pg.1184]    [Pg.748]    [Pg.1184]    [Pg.748]    [Pg.113]    [Pg.819]    [Pg.316]    [Pg.398]   
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