Chemistry The Central Science 11 th edition Theodore

Chemistry, The Central Science, 11 th edition Theodore L. Brown, H. Eugene Le. May, Jr. , and Bruce E. Bursten Chapter 8 Acids and Bases Dr Nehdi Imededdine King Saud University Acids and Bases © 2009, Prentice-Hall, Inc.

Some Definitions • Arrhenius – An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. an Arrhenius acid HCl(g) HO H+(aq) + Cl-(aq) – A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions. +(aq) + OH-(aq) Acids Na. OH H O Na an Arrhenius base and Bases – 2 2 © 2009, Prentice-Hall, Inc.

The hydronium ion, H 3 O+(aq) An H+ ion is simply a proton with no surrounding valence electron. This small, positively charged particle interacts strongly with the nonbonding electron pairs of water molecules to form hydrated hydrogen ions. The interaction of a proton with one water molecule forms the hydronium ion, H 3 O+(aq): Acids and Bases © 2009, Prentice-Hall, Inc.

Some Definitions • Brønsted-Lowry – An acid is a proton donor ( HCl, H 2 O, …). – A base is a proton acceptor (NH 3, H 2 O, …). Acids and Bases © 2009, Prentice-Hall, Inc.

A Brønsted-Lowry acid… ( examp: HCl) …must have a removable (acidic) proton. A Brønsted-Lowry base…( examp: NH 3) …must have a pair of nonbonding electrons. Acids and Bases © 2009, Prentice-Hall, Inc.

Amphiprotic substance If a substance can be either acid or base it is amphiprotic HCO 3 HSO 4 H 2 O Base Acid HCO 3 - + H 2 O H 2 CO 3 + OH(CO 3 )2 - + H 3 O+ Acids and Bases © 2009, Prentice-Hall, Inc.

What Happens When an Acid Dissolves in Water? • Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid. • As a result, the conjugate base of the acid and a hydronium ion are formed. Acids and Bases © 2009, Prentice-Hall, Inc.

Conjugate Acids and Bases • The term conjugate comes from the Latin word “conjugare, ” meaning “to join together. ” • Reactions between acids and bases always yield their conjugate bases and acids. Acids and Bases © 2009, Prentice-Hall, Inc.

Acid and Base Strength • Strong acids are completely dissociated in water. – Their conjugate bases are quite weak. • Weak acids only dissociate partially in water. Figure 8. 1 Relative strengths of some conjugate acid–base pairs – Their conjugate bases are weak bases. Acids and Bases © 2009, Prentice-Hall, Inc.

Acid and Base Strength • Substances with negligible acidity do not dissociate in water. – Their conjugate bases are exceedingly strong. Acids and Bases © 2009, Prentice-Hall, Inc.

Acid and Base Strength • In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H 2 O (l) H 3 O+ (aq) + Cl- (aq) • H 2 O is a much stronger base than Cl-, so the equilibrium lies so far to the right that K is not measured (K >>1). Acids and Bases © 2009, Prentice-Hall, Inc.

Acid and Base Strength • In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. CH 3 CO 2 H (aq) + H 2 O (l) H 3 O+ (aq) + CH 3 CO 2 - (aq) • Acetate is a stronger base than H 2 O, so the equilibrium favors the left side (K< 1). Acids and Bases © 2009, Prentice-Hall, Inc.

Proton transfer reaction • A proton transfer reactions are governed by the relative abilities of two acids to transfer protons. For example, consider the proton transfer that occurs when an acid HX acid dissolves in water HX (aq) + H 2 O (l) H 3 O+ (aq) + X- (aq) In any acid-base reaction, the position of the equilibrium favors transfer of the proton from the stronger acid to the stronger base. Acids and Bases © 2009, Prentice-Hall, Inc.

Autoionization of Water • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. H 2 O (l) + H 2 O (l) H 3 O+ (aq) + OH- (aq) • This is referred to as autoionization. Acids and Bases © 2009, Prentice-Hall, Inc.

Ion-Product Constant • The equilibrium expression for this process is Kc = [H 3 O+] [OH-] • This special equilibrium constant is referred to as the ion-product constant for water, Kw. • At 25 C, Kw = 1. 0 10 -14 Acids and Bases © 2009, Prentice-Hall, Inc.

p. H is defined as the negative base-10 logarithm of the concentration of hydronium ion. p. H = -log [H 3 O+] Acids and Bases © 2009, Prentice-Hall, Inc.

p. H • In pure water, Kw = [H 3 O+] [OH-] = 1. 0 10 -14 • Since in pure water [H 3 O+] = [OH-], [H 3 O+] = 1. 0 10 -14 = 1. 0 10 -7 Acids and Bases © 2009, Prentice-Hall, Inc.

p. H • Therefore, in pure water, p. H = -log (1. 0 10 -7) = 7. 00 • An acid has a higher [H 3 O+] than pure water, so its p. H is <7. • A base has a lower [H 3 O+] than pure water, so its p. H is >7. Acids and Bases © 2009, Prentice-Hall, Inc.

p. H These are the p. H values for several common substances. Acids and Bases © 2009, Prentice-Hall, Inc.

Other “p” Scales • The “p” in p. H tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). • Some similar examples are – p. OH: -log [OH-] – p. Kw: -log Kw Acids and Bases © 2009, Prentice-Hall, Inc.

Watch This! Because [H 3 O+] [OH-] = Kw = 1. 0 10 -14, we know that -log [H 3 O+] + -log [OH-] = -log Kw = 14. 00 or, in other words, p. H + p. OH = p. Kw = 14. 00 Acids and Bases © 2009, Prentice-Hall, Inc.

How Do We Measure p. H? • For less accurate measurements, one can use – Litmus paper • “Red” paper turns blue above ~p. H = 8 • “Blue” paper turns red below ~p. H = 5 – Or an indicator. Acids and Bases © 2009, Prentice-Hall, Inc.

How Do We Measure p. H? For more accurate measurements, one uses a p. H meter, which measures the voltage in the solution. Acids and Bases © 2009, Prentice-Hall, Inc.

Strong Acids • You will recall that the seven strong acids are HCl, HBr, HI, HNO 3, H 2 SO 4, HCl. O 3, and HCl. O 4. • These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. • For the monoprotic strong acids, [H 3 O+] = [acid]. p. H = -log [acid] Molarity of strong acid solution Acids and Bases © 2009, Prentice-Hall, Inc.

Strong Bases • Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca 2+, Sr 2+, and Ba 2+). • Again, these substances dissociate completely in aqueous solution. For the monoprotic strong bases, [OH-] = [base]. p. OH = -log [base] p. H = p. Kw - p. OH Molarity of strong base solution Acids and Bases © 2009, Prentice-Hall, Inc.

Dissociation Constants • For a generalized acid dissociation, HA (aq) + H 2 O (l) A- (aq) + H 3 O+ (aq) the equilibrium expression would be [H 3 O+] [A-] Ka = [HA] • This equilibrium constant is called the acid-dissociation constant, Ka. Acids and Bases © 2009, Prentice-Hall, Inc.

Dissociation Constants The greater the value of Ka, the stronger is the acid. Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating Ka from the p. H The p. H of a 0. 10 M solution of formic acid, HCOOH, at 25 C is 2. 38. Calculate Ka formic acid at this temperature. We know that [H 3 O+] [COO-] Ka = [HCOOH] HCOOH (aq) + H 2 O (l) HCOO- (aq) + H 3 Acids O+ and(aq) Bases © 2009, Prentice-Hall, Inc.

Calculating Ka from the p. H The p. H of a 0. 10 M solution of formic acid, HCOOH, at 25 C is 2. 38. Calculate Ka formic acid at this temperature. To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H 3 O+], which is the same as [HCOO-], from the p. H. Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating Ka from the p. H = -log [H 3 O+] 2. 38 = -log [H 3 O+] -2. 38 = log [H 3 O+] 10 -2. 38 = 10 log [H 3 O+] = [H 3 O+] 4. 2 10 -3 = [H 3 O+] = [HCOO-] Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating Ka from p. H HCOOH (aq) + H 2 O (l) HCOO- (aq) + H 3 O+ (aq) Now we can set up a table… 0. 10 – 4. 2 X 10 -3 M ≈ 0. 10 M Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating Ka from p. H [4. 2 10 -3] Ka = [0. 10] = 1. 8 10 -4 Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating Percent Ionization [H 3 O+]eq • Percent Ionization = [HA] 100 initial • In this example [H 3 O+]eq = 4. 2 10 -3 M [HCOOH]initial = 0. 10 M 4. 2 10 -3 Percent Ionization = 100 0. 10 = 4. 2% Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating Percent Ionization • If the percent Ionization < 5% We can make the simplifying approximation that x, the amount of acid that dissociates, is small compared with the initial concentration of acid; so [HA] eq ≈ [HA]initial – x ≈ [HA]initial Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H from Ka Calculate the p. H of a 0. 30 M solution of acetic acid, CH 3 COOH, at 25 C. CH 3 COOH (aq) + H 2 O (l) H 3 O+ (aq) + CH 3 COO - (aq) Ka for acetic acid at 25 C is 1. 8 10 -5. Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H from Ka The equilibrium constant expression is [H 3 O+] [CH 3 COO-] Ka = [CH 3 COOH] Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H from Ka We next set up a table… [CH 3 COOH], M [H 3 O+], M [CH 3 COO-], M Initially 0. 30 0 0 Change -x +x +x 0. 30 - x 0. 30 x x At Equilibrium We are assuming that x will be very small compared to 0. 30 and can, therefore, be ignored. Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H from Ka Now, 2 (x) 1. 8 10 -5 = (0. 30) (1. 8 10 -5) (0. 30) = x 2 5. 4 10 -6 = x 2 2. 3 10 -3 = x Our approximation (2. 3 10 -3/0. 3)*100 = 0. 76 % < 5 % is appropriate Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H from Ka p. H = -log [H 3 O+] p. H = -log (2. 3 10 -3) p. H = 2. 64 Acids and Bases © 2009, Prentice-Hall, Inc.

Polyprotic Acids… …have more than one acidic proton If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the p. H generally depends only on the first dissociation. Acids and Bases © 2009, Prentice-Hall, Inc.

Weak Bases react with water to produce hydroxide ion. Acids and Bases © 2009, Prentice-Hall, Inc.

Weak Bases The equilibrium constant expression for this reaction is [HB] [OH-] Kb = [B-] where Kb is the base-dissociation constant. Acids and Bases © 2009, Prentice-Hall, Inc.

Weak Bases Kb can be used to find [OH-] and, through it, p. H. Acids and Bases © 2009, Prentice-Hall, Inc.

p. H of Basic Solutions What is the p. H of a 0. 15 M solution of NH 3? NH 3 (aq) + H 2 O (l) NH 4+ (aq) + OH- (aq) [NH 4+] [OH-] Kb = = 1. 8 10 -5 [NH 3] Acids and Bases © 2009, Prentice-Hall, Inc.

p. H of Basic Solutions Tabulate the data. Initially At Equilibrium [NH 3], M [NH 4+], M [OH-], M 0. 15 - x 0. 15 0 x Approximation : 0. 15 - x ≈ 0. 15 Acids and Bases © 2009, Prentice-Hall, Inc.

p. H of Basic Solutions 2 (x) 1. 8 10 -5 = (0. 15) (1. 8 10 -5) (0. 15) = x 2 2. 7 10 -6 = x 2 1. 6 10 -3 = x 2 Our approximation : (x/0. 15 ) x 100% = 1. 06 % < 5% , is appropriate Acids and Bases © 2009, Prentice-Hall, Inc.

p. H of Basic Solutions Therefore, [OH-] = 1. 6 10 -3 M p. OH = -log (1. 6 10 -3) p. OH = 2. 80 p. H = 14. 00 - 2. 80 p. H = 11. 20 Acids and Bases © 2009, Prentice-Hall, Inc.

Ka and Kb are related in this way: Ka Kb = Kw p. Ka +p. Kb = p. Kw Therefore, if you know one of them, you can calculate the other. Acids and Bases © 2009, Prentice-Hall, Inc.

Reactions of Anions with Water • Anions are bases. • As such, they can react with water in a hydrolysis reaction to form OH- and the conjugate acid: X- (aq) + H 2 O (l) HX (aq) + OH- (aq) Example : Hypochlorite anion Cl. O- (aq) + H 2 O (l) HCl. O (aq) + OH- (aq) Acids and Bases © 2009, Prentice-Hall, Inc.

Reactions of Cations with Water • Cations with acidic protons (like NH 4+) will lower the p. H of a solution. • Most metal cations that are hydrated in solution also lower the p. H of the solution. Acids and Bases © 2009, Prentice-Hall, Inc.

Reactions of Cations with Water • Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. • This makes the O-H bond more polar and the water more acidic. • Greater charge and smaller size make a cation more acidic. Acids and Bases © 2009, Prentice-Hall, Inc.

Effect of Cations and Anions Six summary points 1. An anion that is the conjugate base of a strong acid will not affect the p. H (Cl-, NO 3 -, Cl. O 4…). These anions are called ‘’spectator’’ anions 2. An anion that is the conjugate base of a weak acid will increase the p. H (CH 3 COO-, Cl. O-, . . . ). 3. A cation that is the conjugate acid of a weak base will decrease the p. H (NH 4+, Acids and CH 3 NH 3+. . . ) Bases © 2009, Prentice-Hall, Inc.

Effect of Cations and Anions 4. Cations of the strong Arrhenius bases will not affect the p. H: the alkali metal of the group IA (Na, K, …) and heavier alkaline earth metal of the group IIA Mg, Ca, Ba, …). These cations are called ‘’spectator’’ cations 5. Other metal ions will cause a decrease in p. H. 6. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the affect on p. H depends on the Ka and Kb values. Sample 8 -18, 8 -19 Acids and Bases © 2009, Prentice-Hall, Inc.

One finds the percent yield by comparing the amount actually obtained (actual yield) to the amount it was possible to make (theoretical yield). Actual Yield Percent Yield = x 100 Theoretical Yield Acids and Bases © 2009, Prentice-Hall, Inc.

The Common-Ion Effect • Consider a solution of acetic acid: CH 3 COOH(aq) + H 2 O(l) H 3 O+(aq) + CH 3 COO−(aq) • If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. Acids and Bases © 2009, Prentice-Hall, Inc.

The Common-Ion Effect “The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte. ” Acids and Bases © 2009, Prentice-Hall, Inc.

The Common-Ion Effect Calculate the fluoride ion concentration and p. H of a solution that is 0. 20 M in HF and 0. 10 M in HCl. Ka for HF is 6. 8 10− 4. [H 3 O+] [F−] = 6. 8 10 -4 Ka = [HF] Acids and Bases © 2009, Prentice-Hall, Inc.

The Common-Ion Effect HF(aq) + H 2 O(l) H 3 O+(aq) + F−(aq) Because HCl, a strong acid, is also present, the initial [H 3 O+] is not 0, but rather 0. 10 M. [HF], M [H 3 O+], M [F−], M Initially 0. 20 0. 10 0 Change −x +x +x 0. 20 − x 0. 20 0. 10 + x 0. 10 x At Equilibrium Acids and Bases © 2009, Prentice-Hall, Inc.

The Common-Ion Effect 6. 8 10− 4 (0. 10) (x) = (0. 20) (6. 8 10− 4) =x (0. 10) 1. 4 10− 3 = x Acids and Bases © 2009, Prentice-Hall, Inc.

The Common-Ion Effect • Therefore, [F−] = x = 1. 4 10− 3 [H 3 O+] = 0. 10 + x = 0. 10 + 1. 4 10− 3 = 0. 10 M • So, p. H = −log (0. 10) p. H = 1. 00 Sample 8. 20, 8. 21 Acids and Bases © 2009, Prentice-Hall, Inc.

Buffers • Buffers are solutions of a weak conjugate acid-base pair. • They are particularly resistant to p. H changes, even when strong acid or base is added. Acids and Bases © 2009, Prentice-Hall, Inc.

Buffers If a small amount of hydroxide is added to an equimolar solution of HF in Na. F, for example, the HF Acids reacts with the OH− to make F− and water. and Bases © 2009, Prentice-Hall, Inc.

Buffers Similarly, if acid is added, the F− reacts with it to form Acids HF and water. and Bases © 2009, Prentice-Hall, Inc.

Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H 2 O H 3 O + + A − [H 3 O+] [A−] Ka = [HA] Acids and Bases © 2009, Prentice-Hall, Inc.

Buffer Calculations Rearranging slightly, this becomes −] [A Ka = [H 3 O+] [HA] Taking the negative log of both side, we get −] [A −log Ka = −log [H 3 O+] + −log [HA] p. Ka p. H base acid Acids and Bases © 2009, Prentice-Hall, Inc.

Buffer Calculations • So [base] p. Ka = p. H − log [acid] • Rearranging, this becomes [base] p. H = p. Ka + log [acid] • This is the Henderson–Hasselbalch equation. Acids and Bases © 2009, Prentice-Hall, Inc.

Henderson–Hasselbalch Equation What is the p. H of a buffer that is 0. 12 M in lactic acid, CH 3 CH(OH)COOH, and 0. 10 M in sodium lactate? Ka for lactic acid is 1. 4 10− 4. Acids and Bases © 2009, Prentice-Hall, Inc.

Henderson–Hasselbalch Equation [base] p. H = p. Ka + log [acid] p. H = −log (1. 4 10− 4) (0. 10) + log (0. 12) p. H = 3. 85 + (− 0. 08) p. H = 3. 77 Acids and Bases © 2009, Prentice-Hall, Inc.

p. H Range • The p. H range is the range of p. H values over which a buffer system works effectively. • It is best to choose an acid with a p. Ka close to the desired p. H. Acids and Bases © 2009, Prentice-Hall, Inc.

When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base is consumed in the reaction. Acids and Bases © 2009, Prentice-Hall, Inc.

Addition of Strong Acid or Base to a Buffer 1. Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. 2. Use the Henderson–Hasselbalch equation to determine the new p. H of the solution. Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H Changes in Buffers A buffer is made by adding 0. 300 mol CH 3 COOH and 0. 300 mol CH 3 COONa to enough water to make 1. 00 L of solution. The p. H of the buffer is 4. 74. Calculate the p. H of this solution after 0. 020 mol of Na. OH is added. Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H Changes in Buffers Before the reaction, since mol CH 3 COOH = mol CH 3 COO− p. H = p. Ka = −log (1. 8 10− 5) = 4. 74 Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H Changes in Buffers The 0. 020 mol Na. OH will react with 0. 020 mol of the acetic acid: CH 3 COOH(aq) + OH−(aq) CH 3 COO−(aq) + H 2 O(l) Acids and Bases © 2009, Prentice-Hall, Inc.

Calculating p. H Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new p. H: (0. 320) p. H = 4. 74 + log (0. 280) p. H = 4. 74 + 0. 06 p. H = 4. 80 Acids and Bases © 2009, Prentice-Hall, Inc.

Solubility Products Consider the equilibrium that exists in a saturated solution of Ba. SO 4 in water: Ba. SO 4(s) Ba 2+(aq) + SO 42−(aq) Acids and Bases © 2009, Prentice-Hall, Inc.

Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba 2+] [SO 42−] where the equilibrium constant, Ksp, is called the solubility product. Sample 8. 26 Acids and Bases © 2009, Prentice-Hall, Inc.

Solubility Products • Ksp is not the same as solubility. • Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 m. L (g/m. L) of solution, or in mol/L (M). Sample 8. 27 Acids and Bases © 2009, Prentice-Hall, Inc.

Will a Precipitate Form? • In a solution, – If Q = Ksp, the system is at equilibrium and the solution is saturated. – If Q < Ksp, more solid can dissolve until Q = Ksp. – If Q > Ksp, the salt will precipitate until Q = Ksp. Sample 8 -31, 8 -32 Acids and Bases © 2009, Prentice-Hall, Inc.

Factors Affecting Solubility • The Common-Ion Effect – If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease. Ba. SO 4(s) Sample 8 -28 Ba 2+(aq) + SO 42−(aq) Acids and Bases © 2009, Prentice-Hall, Inc.

Factors Affecting Solubility • p. H – If a substance has a basic anion, it will be more soluble in an acidic solution. – Substances with acidic cations are more soluble in basic solutions. Sample 8 -29 Acids and Bases © 2009, Prentice-Hall, Inc.

Factors Affecting Solubility • Complex Ions – Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. Acids and Bases © 2009, Prentice-Hall, Inc.

Factors Affecting Solubility • Complex Ions – The formation of these complex ions increases the solubility of these salts. Sample 8 -30 Acids and Bases © 2009, Prentice-Hall, Inc.

Factors Affecting Solubility • Amphoterism – Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. – Examples of such cations are Al 3+, Zn 2+, and Sn 2+. Acids and Bases © 2009, Prentice-Hall, Inc.

Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture. Acids and Bases © 2009, Prentice-Hall, Inc.