Chemistry The Central Science 10 th edition Theodore

Chemistry, The Central Science, 10 th edition Theodore L. Brown; H. Eugene Le. May, Jr. ; and Bruce E. Bursten Chapter 17 Additional Aspects of Aqueous Equilibria John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall, Inc. Aqueous Equilibria

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. ” EXAMPLE: acetic acid and sodium acetate Consider a solution of acetic acid: HC 2 H 3 O 2(aq) + H 2 O(l) H 3 O+(aq) + C 2 H 3 O 2−(aq) What will happen if sodium acetate is added to this solution? Aqueous Equilibria

The Common-Ion Effect • If acetate ion, C 2 H 3 O 2−(aq), is added to the solution, Le Châtelier says the equilibrium will shift to the left. • So if sodium acetate Na. C 2 H 3 O 2 (a strong electrolyte containing acetate ion), is added to to solution, the acetic acid (a weak electrolyte) will be less ionized. Aqueous Equilibria

The Common-Ion Effect Calculate the fluoride ion concentration and p. H of a solution that is 0. 20 M in HF Ka for HF is 6. 8 10− 4. [H 3 O+] [F−] = 6. 8 10 -4 Ka = [HF] Aqueous Equilibria

The Common-Ion Effect HF(aq) + H 2 O(l) H 3 O+(aq) + F−(aq) [HF], M [H 3 O+], M [F−], M Initially Change At Equilibrium Aqueous Equilibria

The Common-Ion Effect HF(aq) + H 2 O(l) H 3 O+(aq) + F−(aq) [HF], M [H 3 O+], M [F−], M Initially 0. 20 0. 00 0 Change −x +x +x 0. 20 − x 0. 20 x x At Equilibrium Aqueous Equilibria

The Common-Ion Effect 2) (x 6. 8 10− 4 = (0. 20) (6. 8 10− 4) = x 2 1. 2 10− 2 = x Aqueous Equilibria

The Common-Ion Effect • Therefore, [F−] = x = 1. 2 10− 2 M [H 3 O+] = x = 1. 2 10− 2 M • So, p. H = −log (1. 2 10− 2) p. H = 1. 93 Aqueous Equilibria

The Common-Ion Effect Now compare the [F−] and p. H in this solution to the p. H of a solution that has the same initial [HF] but also has another source of one of the ions present. Calculate the fluoride ion concentration and p. H of a solution that is 0. 20 M in HF and 0. 10 M in HCl. [H 3 O+] [F−] = 6. 8 10 -4 Ka = [HF] Aqueous Equilibria

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 Change At Equilibrium Aqueous Equilibria

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 Aqueous Equilibria

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 Aqueous Equilibria

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 How do the [F−] and p. H of the two solutions. Aqueous Equilibria compare?

Buffers Are solutions of a weak conjugate acid-base pair. (e. g. HF and F−). What are two other examples? • Unlike other solutions (or pure water), they resist p. H changes when strong acid or base is added. • If a small amount of hydroxide (base) is added to a solution of HF and Na. F what happens? • If on the other hand a small amount of hydronium ion (acid) is added to a solution of HF in Na. F, the F - reacts with the H O+ to make HF and water. 3 Aqueous Equilibria

Buffers • If a small amount of hydroxide (base) is added to a solution of HF and Na. F, for example, the HF reacts with the OH− to make F− and water. HF + OH− F − + H 2 O • If on the other hand a small amount of hydronium ion (acid) is added to a solution of HF in Na. F, the F - reacts with the H O+ to make HF and water. 3 F - + H 3 O + HF + H 2 O Aqueous Equilibria

Making Buffers There are two basic ways to make a buffer. 1. Dissolve a approximately equal amounts in moles of a weak acid (or weak base) AND it’s conjugate base or (conjugate acid) together in a solution. EXAMPLE #1: Dissolve 20 m. L of 2. 0 M HF and 1. 7 g of Na. F in 100 m. L water. How many moles of HF and F- are in the solution? Is it a buffer? EXAMPLE #2: How much (NH 4)2 SO 4 should you add to a solution of 10 g NH 3 in 90 m. L water to make a Aqueous good buffer? Equilibria

EXAMPLE #2: How much (NH 4)2 SO 4 should you add to a solution of 10 g NH 3 in 90 m. L water to make a good buffer? 10 g NH 3/17 g/mol = 0. 59 mol NH 3 0. 59 mol NH 4+ (1 mol (NH 4)2 SO 4/ 2 mol NH 4+) = 0. 30 mol (NH 4)2 SO 4 *132 g/mol = 39. 6 g Aqueous Equilibria

Making Buffers There are two basic ways to make a buffer. 2. Dissolve a weak acid (or weak base) in water. Add half as many moles of hydroxide ion (or hydrogen ion) by adding strong base (or strong acid). This converts half of the weak acid or weak base to its conjugate, producing a buffer. Aqueous Equilibria

Making Buffer Example EXAMPLE : Dissolve 20 m. L of 2. 0 M HF in water. Add 3. 3 m. L of 6. 0 M Na. OH. The reaction that occurs is: H 2 O (aq) + F−(aq) HF(aq) + OH- (aq) HF, mol HO-, mol F−, mol Initially Change At Equilibrium Note that all of the strong acid or base added always reacts with the weak base or acid Aqueous Equilibria

Making Buffer Example EXAMPLE #1: Dissolve 20 m. L of 2. 0 M HF in water. Add 3. 3 m. L of 6. 0 M Na. OH. The reaction that occurs is: H 2 O (aq) + F−(aq) HF(aq) + OH- (aq) HF, mol HO-, mol F−, mol Initially 0. 020 L * 2. 0 M = 0. 0033 L * 6. 0 M = 0 mol 0. 040 mol 0. 020 mol Change At Equilibrium -0. 020 mol 0 mol + 0. 020 mol Note that all of the strong acid or base added always reacts with the weak base or acid Aqueous Equilibria

Buffer Calculations The equilibrium constant expression for a weak acid HA is: [H 3 O+] [A−] Ka = [HA] or −] [A Ka = [H 3 O+] [HA] Taking the negative log of both sides, we get −] [A −log Ka = −log [H 3 O+] + −log [HA] p. Ka p. H base acid Aqueous Equilibria

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 (H-H) equation. H-H can be used to solve for p. H, p. Ka or the ratio of acid to conjugate base if the other two terms Aqueous Equilibria are known.

Henderson–Hasselbalch Equation What is the p. H of a buffer that is 0. 12 M in lactic acid, HC 3 H 5 O 3, and 0. 10 M in sodium lactate? Ka for lactic acid is 1. 4 10− 4. [base] p. H = p. Ka + log [acid] Aqueous Equilibria

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 Aqueous Equilibria

p. H Range • Often a buffer is chosen to maintain a solution at a particular p. H. • The “buffer range” of a specific buffer pair is the range of p. H values over which a buffer system works effectively. It is about +/- 1 p. H unit from the p. Ka of the acid. • So when making a buffer, it is best to choose an acid with a p. Ka close to the desired p. H. EXAMPLE: Human blood needs to be at a p. H of about 7. 4. So HCO 3 -/H 2 CO 3 is a good buffer for blood because the p. Ka of H 2 CO 3 is 6. 4. HC 2 H 3 O 2/C 2 H 3 O 2 would not be a good blood buffer because the p. KAqueous a of Equilibria HC 2 H 3 O 2 is 4. 8.

When Strong Acids or Bases Are Added to a Buffer… …it is safe to assume that all of the strong acid or base added reacts with the weak base or the strong acid, respectively, that is present in the buffer. Aqueous Equilibria

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. Aqueous Equilibria

Calculating p. H Changes in Buffers A buffer is made by adding 0. 300 mol HC 2 H 3 O 2 and 0. 300 mol Na. C 2 H 3 O 2 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. Aqueous Equilibria

Calculating p. H Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2− p. H = p. Ka = −log (1. 8 10− 5) = 4. 74 Aqueous Equilibria

Calculating p. H Changes in Buffers The 0. 020 mol Na. OH will react with 0. 020 mol of the acetic acid: HC 2 H 3 O 2(aq) + OH−(aq) C 2 H 3 O 2−(aq) + H 2 O(l) HC 2 H 3 O 2− OH− Before reaction 0. 300 mol 0. 020 mol After reaction 0. 280 mol 0. 320 mol 0. 000 mol Aqueous Equilibria

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. 200) p. H = 4. 74 + 0. 06 p. H = 4. 80 Aqueous Equilibria

Titration • A known concentration of base (or acid) is slowly added from a buret to an unknown solution of acid (or base) just until the amount of moles added is exactly equal to the amount of moles originally present in the unknown. Aqueous Equilibria

Titration The known solution is the titrant (buret). The unknown solution is the analyte (flask). The purpose of doing a titration is to figure out the concentration of analyte. The analyte concentration is determined using stoichiometry based on the amount of titrant Aqueous Equilibria used.

Titration the equivalence point is the name of the point when the stoichiometric amount (moles) of titrant added equals the moles of analyte. An indicator that changes color near the p. H at the equivalence point OR a p. H meter (more precise) is used to see determine when the titration is complete. Aqueous Equilibria

Titration curve for a Strong Acid w/ a Strong Base • A plot of titrant added vs p. H. • Shape depends on type of analyte (weak or strong) • Direction depends of analyte (acid or base) • Different points/regions on the graph correspnd to Aqueous different ratios Equilibria of acid/base

Titration of a Strong Acid with a Strong Base Chemical reaction is HCl (in flask) + Na. OH (in buret) Na. Cl + H 2 O (in flask) Net ionic equation is H+ + OH- H 2 O Aqueous Equilibria From the start of the titration to near the equivalence point, the p. H goes up slowly.

Titration of a Strong Acid with a Strong Base Aqueous Equilibria Just before and after the equivalence point, the p. H increases rapidly.

Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the Aqueous Equilibria solution contains only water and the salt from the cation of the base and the anion of the acid.

Titration of a Strong Acid with a Strong Base Aqueous Equilibria As more base is added, the increase in p. H again levels off.

• Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the p. H when it is formed. • The p. H at the equivalence point will be >7. • Phenolphthalein is commonly used as an indicator in these titrations. Chemical reaction is • HA (in flask) + Na. OH (in buret) Na. A + H 2 O (in flask) Net ionic equation is • HA + OH- H 2 O + A- Aqueous Equilibria

Titration of a Weak Acid with a Strong Base • How is the shape similar and different from the strong-strong curve? • What would a weak base titrated with a strong acid curve look like? • What would be the p. H at the equivalence point be (>, <, or =7)? • Would methyl red or phenolphthalein be a better indicator? Why? Aqueous Equilibria

Titration of a Weak Base with a Strong Acid • The p. H at the equivalence point in these titrations is < 7. • Methyl red is the indicator of choice. Aqueous Equilibria

Titration of a Weak Acid with a Strong Base • The weaker acids, the initial p. H is higher and p. H changes near the equivalence point are more subtle. • The p. H at the halfequivalence point (the volume of titrant added equals half of the volume required to reach the equivalence Aqueous Equilibria point) = p. Ka

Titration Problems 1. Determine the concentration or amount of analyte originally present (a stoichiometry problem) Ø Same for strong-strong and strong-weak titrations 2. Determine the p. H of the solution at a particular point in the titration (a stoichiometry and p. H/equilibrium problem) Ø Slightly different for strong-strong and strong-weak titrations Aqueous Equilibria

Characterizing the analyte example Titration reveals that 11. 6 m. L of 3. 00 M sulfuric acid are required to neutralize the sodium hydroxide in 25. 00 m. L of unknown Na. OH solution. (a) What is the molarity of the Na. OH solution? (b) how many grams of Na. OH were in this sample? • Write titration equation • Determine moles of titrant used • Use mole ratio to determine moles of analyte • (a) Calculate molarity of analyte • (b) calculate mass of analyte Aqueous Equilibria

Determining the p. H example (strong -strong) What is the p. H of an aqueous solution to which 26. 0 m. L of 0. 10 M Na. OH have been added to 50. 0 m. L 0. 050 M HCl? • Write titration equation • Determine moles of analyte (H+ or OH-) originally present • Determine moles of titrant added (H+ or OH-) • Use stoichiometry to figure out moles of excess titrant OR analyte (H+ or OH-) remaining after the reaction • Calculate molarity of excess H+ or OH- using total volume • Determine p. H Aqueous Equilibria

Titration of a Weak Acid with a Strong Base At each point below or at the equivalence point, the p. H of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time (basically a buffer). If excess titrant is added after the equivalence point. Aqueous Equilibria the amount of excess is used to calculate p. H

Determining the p. H example (strong -weak) What is the p. H of an aqueous solution to which 30. 0 m. L of 0. 10 M Na. OH have been added to 40. 0 m. L 0. 120 M HC 2 H 3 O 2? • Write titration equation and Ka expression • Determine moles of analyte (weak acid or base) originally present • Determine moles of titrant added (H+ or OH-) • Use stoichiometry to figure out moles of conjugate acid and base present after reaction • Calculate H+ from ICE and Ka (and then find Aqueous p. H) or use H-H equation to get p. H Equilibria

Titrations of Polyprotic Acids In these cases there is an equivalence point for each dissociation. Aqueous Equilibria

Solubility Products • Guess what? There is no such thing as a completely insoluble ionic compound. • The compounds we know as insoluble are actually very slightly soluble. • When these solids are put in water a little tiny bit dissolves, but the vast majority remains solid (much like a tiny bit of a weak acid ionizes, while most of the molecules remain in their original unionized form) • Consider the equilibrium that exists in a saturated solution of Pb(OH)2 in water: Pb(OH)2(s) Pb 2+(aq) + 2 OH−(aq) Aqueous • Write the Keq expression for this equilibrium Equilibria

Solubility Products The equilibrium constant expression for this equilibrium is Keq = Ksp = [Pb 2+] [OH−]2 where the equilibrium constant, Ksp, is called the solubility product. The larger the value of Ksp is the more soluble the compound is. Aqueous Ksp is very temperature dependent! Equilibria

Solubility Products • Ksp is not the same thing as solubility, but they are directly related and the value of one can be found from the other. • Solubility is the maximum mass of solute dissolved in 1 L(g/L) or 100 m. L (g/m. L) of solution, or in mol/L (M). • The solubility of silver chromate Ag 2 Cr. O 4 is. 0215 g/L • What is the molar solubility of silver chromate? • What is the molar conc. of silver ions? Aqueous • What is the molarity of chromate ions? Equilibria • What is the Ksp of silver chromate?

Solubility Products • The solubility of silver chromate Ag 2 Cr 2 O 7 is. 0215 g/L To find the molar solubility convert grams to moles! find molar mass 331. 73 g/mol SO • The molar solubility of silver chromate is 6. 5 X 10 -5 mol/L (M) Aqueous Equilibria

Solubility Products What is the molar conc. of silver ions? What is the molarity of chromate ions? • You know the solubility of silver chromate is 6. 5 X 105 mol/L, so know use the ratio of ions in the compound to find the molar conc. of each ion. • There are 2 Ag+ per silver chromate so [Ag+] is 2 * (6. 5 X 10 -5 M)= 1. 3 x 10 -4 M. • There is 1 Cr. O 4 - per silver chromate so Aqueous Equilibria [Cr. O 4 -] = 6. 5 X 10 -5 M

Solubility Products • Ksp is not the same as solubility. • Solubility is the maximum mass of solute dissolved in 1 L (g/L) or 100 m. L (g/m. L) of solution, or in mol/L (M). • What is the Ksp of silver chromate? • Write Ksp expression, Ksp = [Ag+]2[Cr. O 4 -] • Plug in conc. values from previous step • So Ksp = (1. 3 x 10 -4 M)2(6. 5 X 10 -5 M) =1. 1 X 10 -12 Aqueous Equilibria

Solubility Products • Ksp is not the same as solubility. • Solubility is the maximum mass of solute dissolved in 1 L (g/L) or 100 m. L (g/m. L) of solution, or in mol/L (M). SUMMARY • The solubility of silver chromate is. 0215 g/L • The molar solubility of silver chromate is 6. 5 X 10 -5 M • The molarity of silver ions is 1. 3 X 10 -4 M, the molarity of chromate ions is 6. 5 X 10 -5 M Aqueous Equilibria • The Ksp of silver chromate is 1. 1 X 10 -12

Factors Affecting Solubility of mostly insoluble solids is affected by several factors that depend on the particular compound involved. 1. The Common-Ion Effect How does the amount of Barium sulfate that will dissolve in. Ba. SO 1 liter of water compare to the (s) 2+(aq) + SO 2−(aq) 4 amount that will dissolve in Ba 0. 1 M Ba. Cl 42? � 2. p. H If a substance has a basic anion (such as OH-), it will be more soluble in an acidic solution. Substances with acidic cations are more soluble in basic solutions. Aqueous Equilibria

Factors Affecting Solubility • p. H Ø If a substance has a basic anion, it will be more soluble in an acidic solution. Ø Substances with acidic cations are more soluble in basic solutions. Aqueous Equilibria

Factors Affecting Solubility • Complex Ions Ø Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. Aqueous Equilibria

Factors Affecting Solubility • Complex Ions Ø The formation of these complex ions increases the solubility of these salts. Aqueous Equilibria

Factors Affecting Solubility • Amphoterism Ø Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases. Ø Examples are Al 3+, Zn 2+, and Sn 2+. Aqueous Equilibria

Will a Precipitate Form? To determine if a precipitate will form in a solution determine the concentration of both ions. Then plug into and compare the value (Q) to the Ksp of the precipitate Ø If Q = Ksp, the system is at equilibrium and the solution is saturated but no precip. Ø If Q < Ksp, more solid can dissolve until Q = Ksp no precip. Ø If Q > Ksp, the salt will precipitate until Q = Ksp. Aqueous Equilibria

Q vs Ksp example If 0. 1 g of Na. Cl is added to 10 m. L of a solution with [Ag+] =0. 02 M will silver chloride Ag. Cl precipitate? • Find Ksp in Appendix D, write Ksp expression, • Determine moles Cl- and then [Cl-] • Plug values for [Cl-] and [Ag+] into Ksp expression • Compare value to Ksp Aqueous Equilibria

Q vs Ksp example If 0. 1 g of Na. Cl is added to 10 m. L of a solution with [Ag+] =0. 02 M will silver chloride Ag. Cl precipitate? • Find Ksp in Appendix D, write Ksp expression, Ksp = 1. 8 X 10 -10, Ksp = [Ag+] [Cl-], • Determine moles Cl- and then [Cl-] 0. 1 g/58. 44 g =0. 0017 mol Na. Cl, 1: 1 Na. Cl: Cl 0. 0017 mol Cl- /0. 010 L = 0. 17 M • Plug values for [Cl-] and [Ag+] into Ksp expression Q =0. 02*0. 17 = 0. 0034 Aqueous • Compare value to Ksp Equilibria Q> Ksp so the Ag. Cl will precipitate

Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture. Look at chart on page 757. How can Mg 2+ and Pb 2+ be separated? How about Ag+ Zn 2+ and Na+? Aqueous Equilibria