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**Introduction to Analytical Chemistry**

CHAPTER 8 TITRATING POLYFUNCTIONAL ACIDS AND BASES

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**8A Polyfunctional Acids**

With this acid, as with other polyprotic acids, Ka1＞Ka2＞Ka3.

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**8B Describing Polyfunctional Bases**

8-3

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**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

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**8C Finding the pH of Solutions of Amphiprotic Salts**

The solution could be acidic because of or basic because of 8-5

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**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: (8-1) (8-2) 8-6

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**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)

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**8C Finding the pH of Solutions of Amphiprotic Salts**

Frequently, the ratio cNaHA / Ka1 is much larger than unity in the denominator of Equation 8-3 and Ka2cNaHA is considerably greater than Kw in the numerator. (8-4)

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Example 8-1 Calculate the hydronium ion concentration of a M NaHCO₃ solution. We first examine the assumptions leading to Equation 8-4. The dissociation constants for H₂CO₃ are Ka1 = 1.5 × 10⁻⁴ and Ka2 = 4.69 × 10⁻¹¹. Clearly, cNaHA / Ka1 in the denominator is much larger than unity; in addition, Ka2cNaHA has a value of 4.69 × 10¯¹², which is substantially greater than Kw . Thus, Equation 8-4 applies and

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Example 8-2 Calculate the hydronium ion concentration of a 1.00×10¯³ M Na₂HPO₄ solution. The pertinent dissociation constants are Ka2 and Ka3 , which both contain [HPO₄2–]. Their values are Ka2 = × 10¯⁸ and Ka3 = 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 Ka3cNa₂HPO₄ is by no means much larger than Kw , 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.

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Figure 8-1 Figure 8-1 Titration of mL of M H2A with M NaOH. For H2A, Ka1 = 1.00 10–3 and Ka2 = 1.00 10–7. The method of pH calculation is shown for several points and regions on the titration curve.

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Example 8-4 Construct a curve for the titration of mL of M maleic acid, with 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.

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Example 8-4 Because the ratio Ka1/Ka2 is large (2 × 10⁴), we proceed as just described for constructing Figure 8-1.

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Example 8-4 Initial pH Substituting these relationships into the expression for Ka1 gives

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Example 8-4

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**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,

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Example 8-4 Substitution of these values into the equilibrium-constant expression for Ka1 yields a tentative value of 5.2 × 10¯² M for [H₃O⁺]. It is clear, however, that the approximation [H₃O⁺] << cH₂M or cHM¯ is not valid; therefore, Equations 7-6 and 7-7 must be used, and

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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

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Example 8-4 First Equivalence Point

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**Example 8-4 Second Buffer Region**

Further additions of base to the solution create a new buffer system consisting of HM¯ and M²¯.

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**Example 8-4 Second Equivalence Point**

After the addition of mL of M sodium hydroxide, the solution is M in Na2M (2.5 mmol/75.00 mL). Reaction of the base M²¯ with water is the predominant equilibrium in the system

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Example 8-4

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**Example 8-4 pH Beyond the Second Equivalence Point**

When mL of NaOH have been added,

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**8D Constructing Titration Curves for Polyfunctional Acids**

Figure 8-3 shows titration curves for three other polyprotic acids. The ratio Ka1/Ka2 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.

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**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 Ka1/Ka2 is approximately 10⁵, as is Ka2/Ka3. 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 (Ka2 = × 10¯²).

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Figure 8-3 Figure 8-3 Curves for the titration of polyprotic acids. A M NaOH solution is used to titrate mL of M H3PO4 (curve A), M oxalic acid (curve B), and M H2SO4 (curve C).

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**8E Drawing Titration Curves for Polyfunctional Bases**

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Figure 8-4 Figure 8-4 Curve for the titration of mL of M Na2CO3 with M HCl.

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**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.

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**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

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**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,

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**Feature 8-4 Titration Curves for Amino Acids**

At the isoelectric point,

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**Feature 8-4 Titration Curves for Amino Acids**

If we substitute Kw/[H₃O⁺] for [OH¯] and rearrange, we get The isoelectric point for glycine occurs at a pH of 6.0. That is,

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**8F The Composition of Polyprotic Acid Solutions as A Function of pH**

(8-11) 8-34

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**8F The Composition of Polyprotic Acid Solutions as A Function of pH**

(8-12) (8-13) (8-14)

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Figure 8-6 Figure 8-6 Titration of mL of M maleic acid with 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. 8-36

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THE END

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