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CHAPTER 8 TITRATING POLYFUNCTIONAL ACIDS AND BASES Introduction to Analytical Chemistry
Copyright © 2011 Cengage Learning 8-2 8A Polyfunctional Acids With this acid, as with other polyprotic acids, K a1 ＞ K a2 ＞ K a3.
Copyright © 2011 Cengage Learning 8-3 8B Describing Polyfunctional Bases 8-3
Copyright © 2011 Cengage Learning 8-4 8B Describing Polyfunctional Bases The overall basic dissociation reaction of sodium carbonate is described by the equations Solving the several simultaneous equations that are involved can be difficult and time consuming. Fortunately, simplifying assumptions can be invoked when the successive equilibrium constants for the acid (or base) differ by a factor of about 10³ (or more). 8-4
Copyright © 2011 Cengage Learning 8-5 8C Finding the pH of Solutions of Amphiprotic Salts The solution could be acidic because of or basic because of 8-5
Copyright © 2011 Cengage Learning 8-6 8C Finding the pH of Solutions of Amphiprotic Salts Whether a solution of NaHA is acidic or basic depends on the relative magnitude of the equilibrium constants for these processes: 8-6 (8-1) (8-2)
Copyright © 2011 Cengage Learning 8-7 8C Finding the pH of Solutions of Amphiprotic Salts As we can see in Feature 8-2, solution of these equations yields an approximate value of [H₃O⁺] that is given by the equation (8-3)
Copyright © 2011 Cengage Learning 8-8 8C Finding the pH of Solutions of Amphiprotic Salts Frequently, the ratio c NaHA / K a1 is much larger than unity in the denominator of Equation 8-3 and K a 2 c NaHA is considerably greater than K w in the numerator. (8-4)
Copyright © 2011 Cengage Learning 8-9 Example 8-1 Calculate the hydronium ion concentration of a 0.100 M NaHCO₃ solution. We first examine the assumptions leading to Equation 8-4. The dissociation constants for H₂CO₃ are K a1 = 1.5 × 10⁻⁴ and K a2 = 4.69 × 10⁻¹¹. Clearly, c NaHA / K a1 in the denominator is much larger than unity; in addition, K a2 c NaHA has a value of 4.69 × 10¯¹², which is substantially greater than K w. Thus, Equation 8-4 applies and
Copyright © 2011 Cengage Learning 8-10 Example 8-2 Calculate the hydronium ion concentration of a 1.00 × 10¯³ M Na₂HPO₄ solution. The pertinent dissociation constants are K a2 and K a3, which both contain [HPO₄ 2– ]. Their values are K a2 = 6.32 × 10¯⁸ and K a3 = 4.5 × 10⁻¹³. Considering again the assumptions that led to Equation 8-4, we find that the ratio (1.00 × 10¯³)/(6.32 × 10¯⁸) is much larger than 1, so the denominator can be simplified. The product K a3 c Na₂HPO₄ is by no means much larger than K w, however. We therefore use a partially simplified version of Equation 8-3: Use of Equation 8-4 yields a value of 1.7 × 10¯¹⁰M.
Copyright © 2011 Cengage Learning 8-11 Figure 8-1 Figure 8-1 Titration of 20.00 mL of 0.1000 M H 2 A with 0.1000 M NaOH. For H 2 A, K a1 = 1.00 10 –3 and K a2 = 1.00 10 –7. The method of pH calculation is shown for several points and regions on the titration curve.
Copyright © 2011 Cengage Learning 8-12 Example 8-4 Construct a curve for the titration of 25.00 mL of 0.1000 M maleic acid, with 0.1000 M NaOH. Consider the initial pH, the first buffer region, the first equivalence point, the second buffer region, the second equivalence point, and the region beyond the second equivalence point. The calculations are done manually here.
Copyright © 2011 Cengage Learning 8-13 Example 8-4 Because the ratio K a1 /K a2 is large (2 × 10⁴), we proceed as just described for constructing Figure 8-1.
Copyright © 2011 Cengage Learning 8-14 Example 8-4 Initial pH Substituting these relationships into the expression for K a1 gives
Copyright © 2011 Cengage Learning 8-15 Example 8-4
Copyright © 2011 Cengage Learning 8-16 Example 8-4 First Buffer Region A buffer consisting of the weak acid H₂M and its conjugate base HM¯. To the extent that dissociation of HM¯ to give M²¯ is negligible,
Copyright © 2011 Cengage Learning 8-17 Example 8-4 Substitution of these values into the equilibrium- constant expression for K a1 yields a tentative value of 5.2 × 10¯² M for [H₃O⁺]. It is clear, however, that the approximation [H₃O⁺] << c H₂M or c HM¯ is not valid; therefore, Equations 7-6 and 7-7 must be used, and
Copyright © 2011 Cengage Learning 8-18 Example 8-4 Because the solution is quite acidic, the approximation that [OH¯] is very small is surely justified. Substitution of these expressions into the dissociationconstant relationship gives
Copyright © 2011 Cengage Learning 8-19 Example 8-4 First Equivalence Point
Copyright © 2011 Cengage Learning 8-20 Example 8-4 Second Buffer Region Further additions of base to the solution create a new buffer system consisting of HM¯ and M²¯.
Copyright © 2011 Cengage Learning 8-21 Example 8-4 Second Equivalence Point After the addition of 50.00 mL of 0.1000 M sodium hydroxide, the solution is 0.0333 M in Na 2 M (2.5 mmol/75.00 mL). Reaction of the base M²¯ with water is the predominant equilibrium in the system
Copyright © 2011 Cengage Learning 8-22 Example 8-4
Copyright © 2011 Cengage Learning 8-23 Example 8-4 pH Beyond the Second Equivalence Point When 51.00 mL of NaOH have been added,
Copyright © 2011 Cengage Learning 8-24 8D Constructing Titration Curves for Polyfunctional Acids Figure 8-3 shows titration curves for three other polyprotic acids. The ratio K a1 /K a2 for oxalic acid (curve B) is approximately 1000. The magnitude of the pH change is too small to permit precise location of equivalence with an indicator.
Copyright © 2011 Cengage Learning 8-25 8D Constructing Titration Curves for Polyfunctional Acids Curve A in Figure 8-3 is the theoretical titration curve for triprotic phosphoric acid. Here, the ratio K a1 /K a2 is approximately 10⁵, as is K a2 /K a3. Curve C is the titration curve for sulfuric acid, a substance that has one fully dissociated proton and one that is dissociated to a relatively large extent (K a2 = 1.02 × 10¯²).
Copyright © 2011 Cengage Learning 8-26 Figure 8-3 Figure 8-3 Curves for the titration of polyprotic acids. A 0.1000 M NaOH solution is used to titrate 25.00 mL of 0.1000 M H 3 PO 4 (curve A), 0.1000 M oxalic acid (curve B), and 0.1000 M H 2 SO 4 (curve C).
Copyright © 2011 Cengage Learning 8-27 8E Drawing Titration Curves for Polyfunctional Bases
Copyright © 2011 Cengage Learning 8-28 Figure 8-4 Figure 8-4 Curve for the titration of 25.00 mL of 0.1000 M Na 2 CO 3 with 0.1000 M HCl.
Copyright © 2011 Cengage Learning 8-29 Feature 8-4 Titration Curves for Amino Acids The internal proton transfer from the carboxyl group to the amine group of an amino acid is very favorable in aqueous solution so that amino acids exist as zwitterions as shown below for glycine.
Copyright © 2011 Cengage Learning 8-30 Feature 8-4 Titration Curves for Amino Acids The conjugate acid of the zwitterion can be treated as a polyprotic acid. The zwitterion can act as an acid because of the second proton transfer or as a base because of
Copyright © 2011 Cengage Learning 8-31 Feature 8-4 Titration Curves for Amino Acids The pH at which no net migration occurs is called the isoelectric point; this point is an important physical constant for characterizing amino acids. The isoelectric point is readily related to the ionization constants for the species. Thus, for glycine,
Copyright © 2011 Cengage Learning 8-32 Feature 8-4 Titration Curves for Amino Acids At the isoelectric point,
Copyright © 2011 Cengage Learning 8-33 Feature 8-4 Titration Curves for Amino Acids If we substitute K w /[H₃O⁺] for [OH¯] and rearrange, we get The isoelectric point for glycine occurs at a pH of 6.0. That is,
Copyright © 2011 Cengage Learning 8-34 8F The Composition of Polyprotic Acid Solutions as A Function of pH 8-34 (8-11)
Copyright © 2011 Cengage Learning 8-35 8F The Composition of Polyprotic Acid Solutions as A Function of pH (8-13) (8-14) (8-12)
Copyright © 2011 Cengage Learning 8-36 Figure 8-6 8-36 Figure 8-6 Titration of 25.00 mL of 0.1000 M maleic acid with 0.1000 M NaOH. The solid curves are plots of alpha values as a function of volume. The broken curve is the titration curve of pH as a function of volume.
Copyright © 2011 Cengage Learning 8-37 THE END
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