Chapter 17: Additional Aspects of Aqueous Equilibria Jennie L. Borders

Section 17.1 – The Common-Ion Effect The common-ion effect state that whenever a weak electrolyte and a strong electrolyte ionizes less than it would if it were alone in solution.

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 1.0L of solution?

Practice Exercise Calculate the pH of a solution containing 0.085M nitrous acid (Ka = 4.5 x 10-4) and 0.1M potassium nitrite.

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

Practice Exercise 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 0.10M in HNO3.

Section 17.2 – Buffered Solutions A buffered solution or buffer contains a weak conjugate acid-base pair. Since a buffer contains both acidic and basic ions, it will resist a change in pH if an acid or base is added.

Adding an Acid to a Buffer If we add acid to a buffer made of acetic acid and sodium acetate, the following reaction would occur. H+ + C2H3O2-  HC2H3O2 The added acid (H+) reacts with the conjugate base (acetate ion) to neutralize the acid.

Adding an Acid to a Buffer

Adding a Base to a Buffer If we add base to a buffer made of acetic acid and sodium acetate, the following reaction would occur: HC2H3O2 + OH-  H2O + C2H3O2- The added base (OH-) reacts with the weak acid (acetic acid) to neutralize the base.

Adding a Base to a Buffer

Calculating the pH of a Buffer When calculating the pH of a buffer solution, we can go though the long ICE chart method or we can use the Henderson-Hasselbalch equation: pH = pKa + log[base] [acid]

Ideal Buffer An ideal buffer contains equal concentrations of the weak acid and conjugate base. When the concentrations are equal, the Henderson-Hasselbalch equation shows that pH = pKa.

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

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

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

Practice Exercise Calculate the concentration of sodium benzoate that must be present in a 0.2M solution of benzoic acid (C6H5COOH) to produce a pH of 4.00.

Buffer Capacity Buffer capacity is the amount of acid or base that the buffer can neutralize before the pH begins to change drastically. The greater the amounts of the conjugate acid-base pair, the more resistant is the ratio of their concentrations and the pH is to change.

pH Range The pH range of a buffer is the pH range over which the buffer acts effectively. Since buffers are most effective when [acid] = [base] and pH = pKa, then we pick a weak acid with a pKa close to the desired pH.

Strong Acids and Bases When a strong acid or base is added to a buffer, we assume that it is entirely consumed as long as buffer capacity is not exceeded.

Calculations There are 2 steps to calculations involving the addition of a strong acid or base: Stoichiometry of the neutralization Equilibrium calculation using Henderson-Hasselbalch equation

Sample Exercise 17.5 A buffer is made by adding 0.300 mol CH3COOH and 0.300 mole CH3COONa to enough water to make 1.00L 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.

Sample Exercise 17.5 con’t b. For comparison, calculate the pH that would result if 0.020 mol of NaOH were added to 1.00L of pure water (neglect any volume changes).

Practice Exercise Determine the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl. Determine the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00L of pure water.

Section 17.3 – Acid-Base Titrations Acid-base titrations have 2 important points: Equivalence point – the point at which stoichiometrically equivalent amounts of acid and base have been added. End point – the point at which the indicator changes color.

Titration Curve A titration curve is a graph of the pH as a function of the volume of the added titrant.

Strong Acid-Strong Base Titration Curves There are 4 important points in a strong acid-strong base titration curve: Initial pH – The pH of the initial solution is determined by the [H+]. Between the initial pH and the equivalence point – The pH is determined by the amount of acid not yet neutralized.

Strong Acid-Strong Base Titration Curves 3. The equivalence point – For a strong acid-strong base titration, the pH is 7.00 at the equivalence point. 4. After the equivalence point – the pH is determined by the amount of excess base.

Sample Exercise 17.6 Calculate the pH when the following quantities of 0.100M NaOH solution have been added to 50.0mL of 0.100M HCl solution 49.0mL 51.0mL

Practice Exercise Calculate the pH when the following quantities of 0.100M HNO3 have been added to 25.0mL of 0.100M KOH solution: 24.9mL 25.1mL

Weak Acid-Strong Base Titrations There are 4 important points on a weak acid-strong base titration curve: The initial pH – the initial pH is calculated using the Ka of the weak acid. Between the initial pH and the equivalence point – the pH is calculated based on the neutralized acid. (Buffer)

Weak Acid-Strong Base Titration 3. The equivalence point – the pH is calculated based on the neutralization of the acid. It will be higher than 7. 4. After the equivalence point – the pH is calculated based on the excess base.

Sample Exercise 17.7 Calculate the pH of the solution formed when 45.0mL of 0.100M NaOH is added to 50.0mL of 0.100M CH3COOH (Ka = 1.8 x 10-5).

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

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

Practice Exercise Calculate the pH at the equivalence point when 40.0mL of 0.025M benzoic acid (C6H5COOH, Ka = 6.3 x 10-5). 40.0mL of 0.100M NH3 is titrated with 0.100M HCl.

Comparison of Titration Curves

Comparison of Titration Curves There are 3 main differences in a strong acid-strong base titration and a weak acid-strong base titration: The weak acid has a higher initial pH than a strong acid of the same concentration. A weak acid has a smaller sharp slope than the strong acid.

Comparison of Titration Curves 3. The pH at the equivalence point is above 7.00 for a weak acid and equals 7.00 for a strong acid.

Polyprotic Acid The titration curve for a polyprotic acid has and equivalence point for each hydrogen removed.

Section 17.4 – Solubility Equilibria The solubility-product constant is written like an equilibrium constant. Remember pure liquids and solids are not listed. BaSO4(s)  Ba2+(aq) + SO42-(aq) Ksp = [Ba2+][SO42-]

Sample Exercise 17.9 Write the expression for the solubility-product constant for CaF2.

Practice Exercise Give the solubility-product-constant expressions: Barium carbonate Silver (I) sulfate

Molar Solubility Molar solubility is the number of moles of the solute that dissolve in forming a liter of saturated solution of the solute. The Ksp is the equilibrium constant between the ionic solid and its aqueous ions.

Sample Exercise 17.10 Solid silver chromate is added to pure water at 25oC. 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 x 10-4M.

Sample Exercise 17.10 con’t Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibriua involving the Ag+ of CrO42- ions in the solution, calculate Ksp for this compound.

Practice Exercise A saturate solution of Mg(OH)2 in contact with undissolved solid is prepared at 25oC. 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.

Sample Exercise 17.11 The Ksp for CaF2 is 3.9 x 10-11 at 25oC. 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.

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

Section 17.5 – Factors that Affect Solubility The solubility of a substance is affected by temperature as well as by the presence of other solutes. 3 other factors that affect solubility: The presence of common ions The pH of the solution The presence of complexing agents

Common-Ion Effect In general, the solubility of a slightly soluble salt is decreased by the presence of a second solute that furnishes a common ion.

Sample Exercise 17.12 Calculate the molar solubility of CaF2 at 25oC in a solution that is 0.010M Ca(NO3)2 0.010M NaF

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

Solubility and pH The pH of a solution will affect the solubility of any substance whose anion is basic. In general, if a compound contains a basic anion (the anion of a weak acid) its solubility will increase as the solution becomes more acidic.

Solubility and pH

Sample Exercise 17.13 Which of the following substances will be more soluble in acidic solution than in basic solution: Ni(OH)2 CaCO3 BaF2 AgCl

Complex Ions WE DO NOT HAVE TO DO COMPLEX IONS!!! YEA!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Amphoterism Amphoteric oxides and hydroxides can act as acids or bases. Amphoteric is a substance that can be make to dissolve in either acidic or basic solutions. Amphiprotic is a particle that can act as an acid or base by losing or gaining a proton.

Amphoterism Amphoteric species dissolve in acidic solutions because they contain basic anions. They can also dissolve in basic solutions because they can form complex ions.

Section 17.6 – Precipitation and Separation of Ions You can use Q and Ksp to decide whether or not a precipitate will form: If Q > Ksp, precipitation will occur until Q = Ksp. If Q = Ksp, equilibrium exists (saturated solution) If Q < Ksp, solid dissolves until Q = Ksp.

Sample Exercise 17.15 Will a precipitate form when 0.10L of 8.0 x 10-3M Pb(NO3)2 is added to 0.40L of 5.0 x 10-3M Na2SO4?

Practice Exercise Will a precipitate form when 0.050L of 2.0 x 10-2M NaF is mixed with 0.010L or 1.0 x 10-2M Ca(NO3)2?

Selective Precipitation Selective precipitation is separation of ions in an aqueous solution by using a reagent that form a precipitate with one or a few of the ions. The substance that will precipitate first is the one with the lowest concentration of ions required.

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

Practice Exercise A solution consists of 0.050M Mg2+ and 0.020M Cu2+. Which ion will precipitate first as OH- is added to the solution? What concentration of OH- is necessary to begin precipitation of each cation? (Mg(OH)2, Ksp = 1.8 x 10-11; Cu(OH)2, Ksp = 4.8 x 10-20)

Sample Integrative Exercise A sample of 1.25L of HCl gas at 21oC and 0.950 atm is bubbled through 0.500L of 0.150M NH3 solution. Calculate the pH of the resulting solution assuming that all the HCl dissolves and that the volume of the solution remains 0.500L.

THE END of EQUILIBRIUM… or is it?????