CHAPTER 15 Titration Curves for Complex Acid/Base Systems

Two acids or bases of different strengths In this chapter, we describe methods for calculating curves for complex acid/ base systems. Two acids or bases of different strengths An acid or base that has two or more acidic or basic functional groups An amphiprotic substance Equations for more than one equilibrium are required to describe the characteristics of any of these system 395

§ 15A Mixtures of strong and weak acids or strong and weak bases It is possible to determine each of the components in a mixture containing a strong acid and weak acid (or a strong base and weak base) provided that the concentrations of the two are of the same order of magnitude and that the dissociation constant for the weak acid or base is < 10-4

Ex 15-1 Calculate the pH of a 25mL mixture acid that is 0 Ex 15-1 Calculate the pH of a 25mL mixture acid that is 0.1200 M in hydrochloride acid and 0.0800 M in the weak acid HA (Ka = 1.00 x 10-4) during its titration with 0.1000 M KOH. Compute results for additions of the following volumes of base; (a) 0.00 mL and (b) 5.00 mL (CASE I ) 25mL 的混合酸液(0.1200 M in HCl 與 0.0800 M 的弱酸 HA )，以0.1000M的KOH滴定，試求pH值

Ex 15-2 Calculate the pH of the solution that results when 29 Ex 15-2 Calculate the pH of the solution that results when 29.00 mL of 0.1000 M NaOH is added to 25.00 mL of the solution described in Ex 15-1 (CASE II) 延續Ex15-1，25 mL 的混合酸液(0.1200 M in HCl 與 0.0800 M 的弱酸 HA )，以29 mL 0.1000M的KOH滴定，試求pH值

CASE III : the amount of base added = the amount of HCl ～～～ appropriate quantities of the weak acid and sodium chloride in a suitable volume of water 強鹼與強酸到達當量點後，滴定曲線可視為 弱酸強鹼的滴定， pH值受到弱酸解離的影響

Figure 15-1 Curves for the titration of strong acid/weak acid mixtures with 0.1000 M NaOH. Each titration is on 25.00 mL of a solution that is 0.1200 M in HCl and 0.0800 M in HA 399

§ 15B Polyfunctional acids and bases Poly:多，polyfunctional acids：多質子酸 ~~ generally, the two groups differ in strength and exhibit two or more end points in a neutralization titration

The phosphoric acid system Ka1 > Ka2 > Ka3

The carbon dioxide carbonic acid system

Ex 15-3 Calculate the pH of a solution that is 0.02500M CO2. 求0.02500M CO2 溶液的 pH值

§ 15C Buffer Solutions Involving Polyprotic Acids Free acid H2A + its conjugate base salt NaHA The acic NaHA + its conjugate base salt Na2A The pH of 2) is higher than 1) 401

Ex 15-4 Calculate the hydronium ion concentration for a buffer solution that is 2.00M in phosphoric acid and 1.50M in potassium dihydrogen phosphate. 求下列緩衝溶液的[H3O+]， 2.00 M H3PO4 磷酸與1.50 M KH2PO4 磷酸二氫鉀 402

Ex 15-5 Calculate the hydronium ion concentration of a buffer that is 0.050M in potassium hydrogen phthalate (KHP) and 0.150M in potassium phthalate (K2P). 求下列緩衝溶液的[H3O+]， 0.050 M KHP 與0.150 M K2P 403

§ 15D Calculation of the pH of Solution of NaHA Calculate the pH of solutions of salts that have both acidic and basic properties ~~ amphiprotic salts The pH of the solution is determine by 404

CNaHA = [HA-] + [H2A] + [A2-] [Na+] + [H3O+] = [HA-] + 2[A2-] + [OH-] ……charge-balance CNaOH + [H3O+] = [HA-] + 2[A2-] + [OH-] CNaOH = [H2A] +[HA-] + [A2-] ……..mass balance [H3O+] = [A2-] + [OH-] – [H2A]

Ex 15-6 Calculate the hydronium concentration of a 1 Ex 15-6 Calculate the hydronium concentration of a 1.00X 10-3M Na2HPO4 solution. The pertinent dissociation constant are Ka2 and Ka3, which both contain [HPO42-]. Their value are Ka2 = 6.32X10-8 and Ka3 = 4.5X10-13. 求下列溶液的[H3O+]， 1.00x10-3M Na2HPO4

求下列溶液的[H3O+]， 0.00100 M NaH2PO4 Ex 15-7 Find the hydronium ion concentration of a 0.0100M NaH2PO4 solution.

求下列溶液的[H3O+]， 0.100 M NaHCO3 Ex 15-8 Calculate the hydronium ion concentration of a 0.100M NaHCO3 solution.

§ 15E Titration Curves For Polyfunctional Acids Compounds with two or more acid functional groups yield multiple end points in a titration curve. 407

Figure 15-2 Titration of 20.00 mL of 0.1000 M H2A with 0.1000 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. 407

HOOC-CH=CH-COOH, with 0.1000M NaOH. Ex 15-9 Construct a curve for the titration of 25.00mL of 0.1000M maleic acid, HOOC-CH=CH-COOH, with 0.1000M NaOH. 建構下列酸鹼滴定的曲線：以 0.1000 M NaOH 滴定25.00mL 0.1000M 的丁烯二酸 408

Figure 15-3 Titration curve for 25.00 mL of 0.1000 M maleic acid, H2M, with 0.1000 M NaOH. 414

Figure 15-4 Curves for the titration of polyprotic acids. A 0.1000 M NaOH solution is used to titrate 25.00 ML of 0.1000 M H3PO4(curve A), 0.1000 M oxalic acid (curve B), and 0.1000 M H2SO4 (curve C). 414

The dissociation of sulfuric acid H2SO4 ~~ one proton is a strong acid ~~ one proton is a weak acid Ka2 = 1.02 x 10-2 Example: How the [H3O+] is computed in 0.0400M solution. Ans: [H3O+] is computed in 0.0471M

§ 15F Titration Curves For Polyfunctional Bases Consider the titration of sodium carbonate solution with standard HCl

Figure 15-5 Curve for the titration of 25.00 mL of 0.1000 M Na2CO3 with 0.1000 M HCl. 417

§ 15G Titration Curves For Amphiprotic Species An amphiprotic substance, when dissolved in a suited solvent, behaves both as a weak acid and as a weak base.

Example: sodium dihydrogen phosphate solution NaH2PO4 Na+ + H2PO4- acid base

§ 15H The Composition of Solutions of a Polyprotic Acid as a Function of pH

Composition of H2M solutions as a function of pH. Figure 15-6 Composition of H2M solutions as a function of pH. 421

Figure 15-7 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 a plot of pH as a function of volume. 421