Daniel L. Reger Scott R. Goode David W. Ball www.cengage.com/chemistry/reger Chapter 16 Reactions Between Acids and Bases Titration of Strong Acids and Bases • Titration: a method used to determine the concentration of a substance known as the analyte by adding another substance, the titrant, which reacts in a known manner with the analyte. • analyte + titrant → products Laboratory Titrations (a) A known volume of acid is measured into a flask. (b) Standard base is added from a buret. (c) The endpoint is indicated by a color change. (d) The volume of base is recorded. Titration: Strong Acid and Base • Titration Curve: a graph of pH of a solution as titrant is added. • For a titration of a strong acid with a strong base, the pH will start at a very low value and stay low as long as strong acid is still present. Titration: Strong Acid and Base • The pH will rise sharply to 7 at the equivalence point, where the acid and base are present in stoichiometrically equivalent amounts. • After excess strong base has been added, the pH levels off at a high value. Titration Curve for a Strong Acid with a Strong Base The Equivalence Point in a Titration • Calculate the equivalence point in the titration of 20.00 mL of 0.1252 M HCl with 0.1008 M NaOH. HCl + NaOH H2O + NaCl Test Your Skill • Calculate the equivalence point in the titration of 40.00 mL of 0.2387 M HNO3 with 0.3255 M NaOH. Units of Millimoles • Millimole: one thousandth of a mole. • If molarity is expressed in moles/liter (M) and volume in milliliters (mL), n will be in millimoles (mmol). moles • n liter milliliters millimoles Liters cancel but the milli- multiplier remains. . Calculating a Titration Curve • Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 0, 2.00, 10.00, and 20.00 mL base are added. Test Your Skill • Calculate the pH in the titration of 20.0 mL of 0.125 M HCl with 0.250 M NaOH after 5.00 mL and 12.00 mL base are added. Calculating a Titration Curve Strong Base + Strong Acid Curve • Titration curve of 50.00 mL 0.500 M KOH with 1.00 M HCl. Stoichiometry and Titration Curves • 10 mL of two different 0.100 M acids titrated with 0.100 M NaOH. Estimating the pH of Mixtures • Fill in first 3 three lines of sRf table. • Look at the final solution (f-line). • If a strong acid is present, the solution will be strongly acidic. • If a strong base is present the solution will be strongly basic. Solution Estimate of pH • If only water is Strongly acidic 1 present, the Neutral 7 solution will be Strongly basic 13 neutral. Buffers • Buffer: a solution that resists changes in pH. • A buffer is a mixture of a weak acid or base and its conjugate partner. • HA + OH- → H2O + AWeak acid reacts with any added OH-. • A- + H3O → HA + H2O Weak base reacts with any added H3O+. The pH of a Buffer System • For the chemical reaction HA + H2O → H3O+ + A - K a [HA] [H3 O ][A ] or [H3 O ] Ka [HA] [A ] take the - log (p - function) to obtain - [A ] • pH pK a log [HA] The pH of a Buffer System • Calculate the pH of a solution of 0.50 M -10 HCN and 0.20 M NaCN, Ka = 4.9 x 10 . The pH of a Buffer System • Calculate the pH of a solution of 0.40 M NH3 and 0.10 M NH4Cl. For NH3 Kb = -5 1.8 x 10 . Test Your Skill • Calculate the pH of a buffer that is 0.25 M HCN and 0.15 M NaCN, -10 Ka = 4.9 x 10 . The Composition of a pH Buffer • Calculate the amount of sodium acetate that must be added to 250 mL of 0.16 M acetic acid in order to prepare a pH 4.68 -5 buffer. Ka = 1.8 x 10 Test Your Skill • How many moles of NaCN should be added to 100 mL of 0.25 M HCN to prepare a buffer with pH = 9.40? -10 Ka = 4.9 x 10 . Determining the Response of a Buffer to Added Acid or Base • Calculate the initial and final pH when 10 mL of 0.100 M HCl is added to (a) 100 mL of water, and (b) 100 mL of a buffer which is 1.50 M CH3COOH and 1.20 M CH3COONa. Test Your Skill • Calculate the final pH when 10 mL of 0.100 M NaOH is added to 100 mL of a buffer which is 1.50 M CH3COOH and 1.25 M CH3COONa. Qualitative Aspects: Titration: Weak Acid + Strong Base Before any base added. (a) Part way to equivalence point. (b) Equivalence point. (c) Beyond equivalence point. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (a) Before any base is added the solution is a weak acid has a low pH. • Estimated pH = 3 pH 2-4 is typical –Depends on the concentration of the acid. –Depends on the value of Ka. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (b) After some base is added, but before the equivalence point is reached. • The solution is a mixture of the weak acid HA and its conjugate base A ; therefore, the solution is a buffer. • The estimated pH is equal to pKa. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (c) At the equivalence point the solution is salt of A , all the HA having been consumed by the stoichiometric amount of OH-. • A- is the weak conjugate base of HA. • The estimated pH is 10. Titration: Weak Acid + Strong Base • HA + OH- A- + H2O • (d) After excess strong base is added OH- is in excess. • The estimated pH is 13. pH Estimates Solution Strongly acidic Weakly acidic Neutral Weakly basic Strongly basic Buffer (acidic to basic) Estimate of pH 1 3 7 11 13 pKa 4-10 typical Titration: Weak Acid + Strong Base Titration Curves for Acids of Different Strengths Calculating the Titration Curve for a Weak Acid • Calculate the pH in the titration of 20.00 mL of 0.500 M formic acid (HCOOH Ka=1.8 x 10-4) with 0.500 M NaOH after 0, 10.00, 20.00, and 30.00 mL of base have been added. • The titration reaction is HCOOH + OH- HCOO- + H2O Titration of 25.00 mL of 0.500 M Formic Acid with 0.500 M NaOH Test Your Skill • Calculate the pH in the titration of 12.00 mL of 0.100 M HOCl with 0.200 M NaOH after 0, 3.00, 6.00, and 9.00 mL of base have been added. Titration of 20.00 mL of 0.500 M Methylamine with 0.500 M HCl pH Indicators • Indicator: a substance that changes color at the endpoint of a titration. • pH indicators are weak acids or bases whose conjugate species are a different color. pH Indicators • HIn + H2O H3O+ + In • KIn [H3O ][In ] [HIn] • pKIn = -log(KIn) pH Indicators • When pH is lower than pKIn, the indicator will be in the acid form. • When pH is greater than pKIn, the indicator will be in the base form. • An indicator should be chosen which changes at or just beyond the equivalence point. Properties of Indicators Name Acid Color Base Color pH Range pKIn Thymol blue* Red Yellow 1.2–2.8 1.6 Methyl orange Red Yellow 3.1–4.4 3.5 Methyl red Red Yellow 4.2–6.3 5.0 Bromthymol blue Yellow Blue 6.2–7.6 7.3 Phenolphthalein Clear Pink 8.3–10.0 8.7 Thymol blue* Yellow Blue 8.0–9.6 9.2 * Thymol blue is polyprotic and has three color forms. Titration Curves for Strong and Weak Acids Polyprotic Acids • Polyprotic acids provide more than one proton when they ionize. • Polyprotic acids ionize in a stepwise manner. H2A + H2O ⇌ H3O+ + HA- Step 1 HA- + H2O ⇌ H3O+ + A2- Step 2 Polyprotic Acids • There is a separate acid ionization constant for each step H2A + H2O ⇌ H3O+ + HA- Step 1 [H3 O ][HA ] K a1 [H2 A] HA- + H2O ⇌ H3O+ + A2K a2 2- [H3 O ][A ] [HA - ] Step 2 Polyprotic Acids - • HA is the conjugate base of H2A, so it is a weaker acid than H2A. • Ka1 is always larger than Ka2. • For triprotic acids (such as H3PO4), Ka2 is always larger than Ka3. Calculating Concentrations of Species in Polyprotic Acid Solutions • When successive Ka values differ by a factor of 1000 or more, each step can be assumed to be essentially unaffected by the occurrence of the subsequent step. Concentrations of Species in Polyprotic Acid Solutions • Calculate the concentrations of all species in 0.250 M malonic acid, -2 Ka = 1.6 x 10 . • Consider the first ionization and solve by usual approach. H2C3H2O4 + H2O ⇌ HC3H2O4- + H3O+ Concentrations of Species in Polyprotic Acid Solutions • The second step is needed only to calculate the concentration of C3H2O42because the concentration of H3O+ is determined by the first step. • You can ignore the effect of the second step on the pH because the Ka1 is so much larger than Ka2. Test Your Skill • Calculate the pH of a 0.040 M solution of ascorbic acid. (Ka1 = 8.0 x 10-5, Ka2 = 1.6 x 10-12) Amphoteric Species • Amphoteric: having both acidic and basic properties. • Conjugate bases of weak polyprotic acids are amphoteric. • The hydrogen oxalate ion, HC2O4-, is a -4 weak acid (Ka2 = 1.6 ×10 ). • HC2O4- + H2O ⇌ C2O42- + H3O+ Amphoteric Species • Weak Acid • The hydrogen oxalate ion, HC2O4-, is a weak acid. • Weak Base • The hydrogen oxalate ion, HC2O4-, can also act as a weak base. HC2O4- + H2O ⇌ C2O42- + H3O+ HC2O4- + H2O → H2C2O4 + OH- • Ka2 = 1.6 x 10-4 • Kb = Kw/Ka1 = 1.0 x 10-14 / 5.6 x 10-2 • Kb = 1.9 x 10-13 • Since Ka > Kb, the ion will act as a weak acid in water. • When comparing Ka to Kb note that Ka is Ka2 and Kb is Kw/Ka1. Test Your Skill • Ka for the hydrogen malonate ion, -6 HC3H2O4 , is 2.1 x 10 . Is a solution of sodium hydrogen malonate acidic or basic? Factors That Influence Solubility • The pH affects the solubility of salts of weak acids. • Complex ion formation affects the solubility of salts of transition metal cations. Salts of Anions of Weak Acids • The solubility of salts of anions of weak acids is enhanced by lowering the pH. • Cd(CN)2(s) ⇌ Cd2+(aq) + 2CN-(aq) -8 Ksp = 1.0 x 10 • Adding acid reduces [CN-] in solution, by the reaction • H3O+(aq) + CN-(aq) ⇌ HCN(aq) + H2O(l) Salts of Transition Metal Cations • Transition metal cations form complexes with Lewis bases such as H2O, NH3, or OH . • Formation of a complex reduces the concentration of metal ion and increases the solubility of the salt. Solubility of Amphoteric Species • Amphoteric species, such as Be(OH)2, Al(OH)3, Sn(OH)2, Pb(OH)2, Cr(OH)3, Ni(OH)2, Cu(OH)2, Zn(OH)2, and Cd(OH)2 react with acid or base to form the soluble metal ion or complex ions + x+ M(OH)x + xH M + xH2O x = 2,3 y+ M(OH)x + yOH M(OH)x x = 2,3, y = 1,2