Chapter 17 Additional Aspects of Aqueous Equilibria

17.1 The Common Ion Effect

The Common-Ion Effect Consider a solution of acetic acid: If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left. CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO−(aq)

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.”

Sample Exercise 17.1 What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

Practice 17.1 Calculate the pH of a solution containing 0.085 M nitrous acid (HNO2; Ka = 4.5 × 10-4) and 0.10 M potassium nitrite (KNO2).

Sample Exercise 17.2 Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.

Practice 17.2 HCOOH(aq) H+(aq) + HCOO−(aq) [H+] [HCOO−] [HCOOH] Ka = Calculate the formate ion concentration and pH of a solution that is 0.050M in formic acid (HCOOH; Ka = 1.8 x 10-4) and 0.10 M HNO3. HCOOH(aq) H+(aq) + HCOO−(aq) [H+] [HCOO−] [HCOOH] Ka = = 1.8  10-4

Example HCOOH(aq) H+(aq) + HCOO−(aq) Because HNO3, a strong acid, is also present, the initial [H+] is not 0, but rather 0.10 M. [HCOOH], M [H+], M [HCOO−], M Initially 0.050 0.10 Change −x +x At Equilibrium 0.050 − x  0.050 0.10 + x  0.10 x

Common Ion Effect Also applies to weak bases & strong electrolytes NH3 + H2O ↔ NH4+ + OH- The addition of ammonium chloride, NH4Cl, would cause the reaction to shift left, decreasing [OH-]

p.723 GIST A mixture of 0.10 mol of NH4Cl and 0.12 mol of NH3 is added to enough water to make 1.0 L of solution. A) What are the initial concentrations of the major species in the solution? B) Which of the ions in this solution is a spectator ion in any acid-base chemistry occurring in the solution? C) What equilibrium reaction determines [OH-] and therefore the pH of the solution?

17.2 Buffered Solutions

Buffers Buffers are solutions of a weak conjugate acid-base pair. They are particularly resistant to pH changes, even when small amounts of strong acid or base are added. Contains an acid to neutralize OH- and a base to neutralize H+

Buffers If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

Buffers Similarly, if acid is added, the F− reacts with it to form HF and water.

17.2 GIST Which of the following conjugate acid-base pairs will not function as a buffer and explain: C2H5COOH and C2H5COO- HCO3- and CO32- HNO3 and NO3- What happens when NaOH is added to a buffer composed of CH3COOH and CH3COO-? What happens when HCl is added to this buffer?

Buffer Calculations HA + H2O H3O+ + A− Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H2O H3O+ + A− [H3O+] [A−] [HA] Ka =

Buffer Calculations Taking the negative log of both side, we get Rearranging slightly, this becomes [A−] [HA] Ka = [H3O+] Taking the negative log of both side, we get base [A−] [HA] −log Ka = −log [H3O+] + −log pKa pH acid

Buffer Calculations pKa = pH − log [base] [acid] So pKa = pH − log [base] [acid] Rearranging, this becomes pH = pKa + log [base] [acid] This is the Henderson–Hasselbalch equation. Since the weak acid dissociates very little, use the initial concentration.

Sample Exercise 17.3 What is the pH of a buffer that is 0.12 M in lactic acid (HC3H5O3) and 0.10 M in sodium lactate (NaC3H5O3). For lactic acid, Ka = 1.4 x 10-4.

Practice Exercise 17.3 Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate.

Henderson–Hasselbalch Equation pH = pKa + log [base] [acid] pH = −log (6.3  10−5) + log (0.20) (0.12) pH = 4.20 + 0.22 pH = 4.42

Sample Exercise 17.4 Preparing a Buffer How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.)

Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (C6H5COOH) to produce a pH of 4.00. Practice Exercise

Buffer Capacity & pH Range Buffer capacity is the amount of acid or base the buffer can neutralize before the pH noticeably changes Depends on how much of the acid/base was used to make the buffer greater concentration = greater buffering capacity

Buffer Capacity & pH Range The pH range is the range of pH values over which a buffer system works effectively. Depends on Ka relative concentration of acid & base that make up the buffer

Buffer Capacity & pH Range Buffers work best when concentrations of acid/base are the same – if the concentrations differ by more than 10 fold, the buffer will not work well in both directions Select a buffer by choosing one whose pKa is the same or near the desired pH. Optimal pH of a buffer is when pH = pKa Buffers usually have a usable range between +1/-1 pKa

p.727 GIST What is the optimal pH buffered by a solution containing acetic acid & sodium acetate? (Ka for acetic acid is 1.8 x 10-5)

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.

Addition of Strong Acid or Base to a Buffer Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution. Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

Sample Exercise 17.5 Calculating pH Changes in Buffers a) A buffer is made by adding 0.300 mol HC2H3O2 and 0.300 mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. a) Calculate the pH of this solution after 0.020 mol of NaOH is added. b) For comparison, calculate the pH that would result if 0.020 mol of NaOH were added to pure water.

Calculating pH Changes in Buffers Before the reaction, since mol HC2H3O2 = mol C2H3O2− pH = pKa = −log (1.8  10−5) = 4.74

Calculating pH Changes in Buffers The 0.020 mol NaOH will react with 0.020 mol of the acetic acid: HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l) HC2H3O2 OH− C2H3O2− Before reaction 0.300 mol 0.000 mol Addition 0.020 mol After reaction 0.280 mol 0.320 mol

Calculating pH Changes in Buffers Now use the Henderson–Hasselbalch equation to calculate the new pH: pH = 4.74 + log (0.320) (0.280) pH pH = 4.80 pH = 4.74 + 0.06

Practice Exercise 17.5 Determine (a) the pH of the original buffer in the previous example after the addition of 0.020 mol HCl and (b) the pH of the solution that would result from 0.020 mol HCl to 1.00 L of pure water.

Practice A buffer solution contains 0.10 mol of propionic acid, C2H5OOH and 0.13 mol of sodium propionate, C2H5OONa in 1.50 L. (a) what is the pH of this buffer? (b) what is the pH of this buffer after the addition of 0.01 mol of NaOH? (c) what is the pH of the buffer after the addition of 0.01 mol of HI?

17.3 Acid-Base Titrations

Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base). p.730 GIST: For this setup, will pH increase or decrease as titrant is added?

Titration A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

Titration of a Strong Acid with a Strong Base From the start of the titration to near the equivalence point, the pH goes up slowly.

Titration of a Strong Acid with a Strong Base Just before (and after) the equivalence point, the pH increases rapidly.

Titration of a Strong Acid with a Strong Base At the equivalence point, moles acid = moles base, and the 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 As more base is added, the increase in pH again levels off.

Sample Exercise 17.6 Calculating pH for a Strong Acid-Strong Base Titration Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution: (a) 49.0 mL, (b) 51.0 mL.

Calculate the pH when the following quantities of 0 Calculate the pH when the following quantities of 0.100 M HNO3 have been added to 25.0 mL of 0.100 M KOH solution: (a) 24.9 mL, (b) 25.1 mL. Practice Exercise

p.733 GIST What is the pH at the equivalence point when 0.10 M HNO3 is added to a solution containing 0.30 g of KOH?

Titration of a Weak Acid with a Strong Base Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed. At the equivalence point the pH is >7. Phenolphthalein is commonly used as an indicator in these titrations.

Titration of a Weak Acid with a Strong Base At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

Titration of a Weak Acid with a Strong Base With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

Sample Exercise 17.7 Calculating pH for a Weak Acid-Strong Base Titration Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH (Ka = 1.8 ×10-5).

(a) Calculate the pH in the solution formed by adding 10. 0 mL of 0 (a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C6H5COOH, Ka = 6.3 × 10-5) . (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to 20.0 mL of 0.100 M NH3. Practice Exercise

Sample Exercise 17.8 Calculating the pH at the Equivalence Point Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.

Calculate the pH at the equivalence point when (a) 40. 0 mL of 0 Calculate the pH at the equivalence point when (a) 40.0 mL of 0.025 M benzoic acid (C6H5COOH, Ka = 6.3 × 10-5 ) is titrated with 0.050 M NaOH; (b) 40.0 mL of 0.100 M NH3 is titrated with 0.100 M HCl. Practice Exercise

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

Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation.

p.737 GIST Why is the choice of indicator more crucial for a weak acid-strong base titration than for a strong acid-strong base titration? Sketch the titration curve for the titration of Na2CO3 with HCl.

17.4 Solubility Equilibria

Solubility Products BaSO4(s) Ba2+(aq) + SO42−(aq) Consider the equilibrium that exists in a saturated solution of BaSO4 in water: BaSO4(s) Ba2+(aq) + SO42−(aq)

Solubility Products The equilibrium constant expression for this equilibrium is Ksp = [Ba2+] [SO42−] where the equilibrium constant, Ksp, is called the solubility product.

Sample Exercise 17.9 Writing Solubility-Product (Ksp) Expressions Write the expression for the solubility-product constant for CaF2, and look up the corresponding Ksp value in Appendix D.

Practice 17.9 Give the solubility-product-constant expressions and the values of the solubility-product constants (from Appendix D) for the following compounds: A) barium carbonate B) silver sulfate

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 mL (g/mL) of solution, or in mol/L (M).

p.739 GIST Without doing a calculation, predict which of the following compounds will have the greatest molar solubility in water: AgCl (Ksp = 1.8 x 10-10) AgBr (Ksp = 5.0 x 10-13) AgI (Ksp = 8.3 x 10-17)

Sample Exercise 17.10 Calculating Ksp from Solubility Solid silver chromate is added to pure water at 25 ºC. Some of the solid remains undissolved at the bottom of the flask. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10-4 M. Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving the Ag+ or CrO42– ions in the solution, calculate Ksp for this compound.

A saturated solution of Mg(OH)2 in contact with undissolved solid is prepared at 25 ºC. The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH– ions in the solution, calculate Ksp for this compound. Practice Exercise

Sample Exercise 17.11 Calculating Solubility from Ksp The Ksp for CaF2 is 3.9 ×10-11 at 25 ºC. Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.

The Ksp for LaF3 is 2 × 10-19. What is the solubility of LaF3 in water in moles per liter? Practice Exercise

17.5 Factors that Affect Solubility

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. BaSO4(s) Ba2+(aq) + SO42−(aq)

Sample Exercise 17.12 Calculating the Effect of a Common Ion on Solubility Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) 0.010 M in Ca(NO3)2, (b) 0.010 M in NaF.

The value for Ksp for manganese(II) hydroxide, Mn(OH)2, is 1. 6 ×10-13 The value for Ksp for manganese(II) hydroxide, Mn(OH)2, is 1.6 ×10-13. Calculate the molar solubility of Mn(OH)2 in a solution that contains 0.020 M NaOH. Practice Exercise

Factors Affecting Solubility pH If a substance has a basic anion, it will be more soluble in an acidic solution. Common basic anions: OH-, F-, CO32-, PO43-, CN-, S2- Substances with acidic cations are more soluble in basic solutions.

(a) Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)? Sample Exercise 17.13 Predicting the Effect of Acid on Solubility Which of the following substances will be more soluble in acidic solution than in basic solution: (a) Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)?

Practice 17.13 Write the net ionic equation for the reaction of the following copper(II) compounds with acid: (a) CuS, (b) Cu(N3)2.

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

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

Sample Exercise 17.14 Evaluating an Equilibrium Involving a Complex Ion Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 0.010 M solution of AgNO3 to give an equilibrium concentration of [NH3] = 0.20 M. Neglect the small volume change that occurs when NH3 is added.

in a liter of solution buffered at pH 10.0. Sample Exercise 17.14 Evaluating an Equilibrium Involving a Complex Ion Calculate [Cr3+] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0. Practice Exercise

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 Al3+, Zn2+, and Sn2+. p.750 GIST: What kind of behavior characterizes an amphoteric oxide or an amphoteric hydroxide?

17.6 Precipitation and Separation of Ions

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 Exercise 17.15 Predicting Whether a Precipitate Will Form Will a precipitate form when 0.10 L of 8.0 × 10-3 M Pb(NO3)2 is added to 0.40 L of 5.0 ×10-3 M Na2SO4?

Sample Exercise 17.15 Predicting Whether a Precipitate Will Form Will a precipitate form when 0.050 L of 2.0 ×10-2 M NaF is mixed with 0.010 L of 1.0 × 10-2 M Ca(NO3)2? Practice Exercise

Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation A solution contains 1.0 × 10-2 M Ag+ and 2.0 ×10-2 M Pb2+. When Cl– is added to the solution, both AgCl (Ksp = 1.8× 10-10) and PbCl2 (Ksp = 1.7×10-5) precipitate from the solution. What concentration of Cl– is necessary to begin the precipitation of each salt? Which salt precipitates first?

Sample Exercise 17.16 Calculating Ion Concentrations for Precipitation A solution consists of 0.050 M Mg2+ and 0.020 M Cu2+. Which ion will precipitate first as OH– is added to the solution? What concentration of OH– is necessary to begin the precipitation of each cation? [Ksp = 1.8 × 10-11 for Mg(OH)2, and Ksp = 4.8 ×10-20 for Cu(OH)2.] Practice Exercise

17.7 Qualitative Analysis for Metallic Elements

Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture.

p.754 GIST If a precipitate forms upon addition of HCl to an aqueous solution, what conclusions can you draw about the contents of the solution?

Integrative Exercise A sample of 1.25 L of HCl gas at 21ºC and 9.50 atm is bubbled through 0.500 L of 0.150 M NH3 solution. Calculate the pH of the resulting solution assuming that all the HCl dissolves and that the volume of the solution remains 0.500 L.