Chapter 17 Additional Aspects of

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1 Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 17 Additional Aspects of John D. Bookstaver St. Charles Community College Cottleville, MO Chapter 17 Problems Problems 11, 15, 17, 19, 21, 27, 33, 35, 47, 49, 51, 55, 61 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. 1

2 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. The Common-Ion Effect Calculate the fluoride ion concentration and ph of a solution that is 0.20 M in HF and 0.10 M in HCl. K a for HF is [H 3 O + ] [F ] K a = = [HF] -4 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 x +x +x At Equilibrium 0.20 x x 0.10 x 2

3 The Common-Ion Effect = (0.20) ( ) (0.10) = x = x (0.10) (x) (0.20) The Common-Ion Effect Therefore, [F ] = x = [H 3 O + ] = x = = 0.10 M So, ph = log (0.10) ph = 1.00 Buffers Buffers are solutions of a weak conjugate acid-base pair. They are particularly resistant to ph changes, even when strong acid or base is added. 3

4 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. Buffer Calculations Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H 2 O K a = [H 3 O + ] [A ] [HA] H 3 O + + A 4

5 Buffer Calculations Rearranging slightly, this becomes K a = [H 3 O + ] [A ] [HA] Taking the negative log of both side, we get [A ] log K a = log [H 3 O + ] + log [HA] base pk a ph acid Buffer Calculations So pk a = ph log [base] [acid] Rearranging, this becomes ph = pk a + log [base] [acid] This is the Henderson Hasselbalch equation. Henderson Hasselbalch Equation What is the ph of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is

6 Henderson Hasselbalch Equation ph = pk a + log [base] [acid] ph = log ( ) + log (0.10) (0.12) ph = ( 0.08) ph = 3.77 ph Range The ph range is the range of ph values over which a buffer system works effectively. It is best to choose an acid with a pk a close to the desired ph. 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. 6

7 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 ph of the solution. Calculating ph Changes in Buffers A buffer is made by adding mol HC 2 H 3 O 2 and mol NaC 2 H 3 O 2 to enough water to make 1.00 L of solution. The ph of the buffer is Calculate the ph of this solution after mol of NaOH is added. Calculating ph Changes in Buffers Before the reaction, since mol HC 2 H 3 O 2 = mol C 2 H 3 O 2 ph = pk a = log ( ) =

8 Calculating ph Changes in Buffers The mol NaOH will react with 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 C 2 H 3 O 2 OH Before reaction mol mol mol After reaction mol mol mol Calculating ph Changes in Buffers Now use the Henderson Hasselbalch equation to calculate the new ph: ph = log (0.320) (0.200) ph = ph = 4.80 Titration In this technique a known concentration of base (or acid) is slowly added to a solution of acid (or base). 8

9 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. 9

10 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. 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. 10

11 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. 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. 11

12 Titrations of Polyprotic Acids When one titrates a polyprotic acid with a base there is an equivalence point for each dissociation. The Solubility Product Constant, K sp Many ionic compounds are only slightly soluble in water: ex. Ag salts, sulfides Equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution AgCl(s) Ag + (aq) + Cl (aq) [Ag ][Cl ] Kc but... Kc[AgCl] [Ag ][Cl ] [AgCl] K sp = [Ag + ][Cl ] EOS SAMPLE EXERCISE Writing Solubility-Product (K sp) Expressions Write the expression for the solubility-product constant for CaF 2, and look up the corresponding K sp value in Appendix D. Solution Analyze and Plan: We are asked to write an equilibrium-constant expression for the process by which CaF 2 dissolves in water. We apply the same rules for writing any equilibrium-constant expression, making sure to exclude the solid reactant from the expression. We assume that the compound dissociates completely into its component ions. Solve: Following the italicized rule stated previously, the expression for K sp is In Appendix D we see that this K sp has a value of

13 K sp and Molar Solubility The solubility product constant is related to the solubility of an ionic solute K sp = [Ag + ][Cl ]; solubility given by [Ag + ] From stoichiometry, the ion ratio is 1:1, so [Ag + ] = [Cl ], both of which are unknown (x) Ag + Cl Ag + + Cl K sp = x 2 and [Ag + ] = (K sp ) 1/2 EOS SAMPLE EXERCISE Calculating K sp from Solubility Solid silver chromate is added to pure water at 25 C. 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 Ag 2CrO 4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is M. Assuming that Ag 2CrO 4 dissociates completely in water and that there are no other important equilibria involving the Ag + or CrO 4 2 ions in the solution, calculate K sp for this compound. Solution Analyze: We are given the equilibrium concentration of Ag + in a saturated solution of Ag 2CrO 4. From this, we are asked to determine the value of the solubility-product constant for the dissolution of Ag 2CrO 4. Plan: The equilibrium equation and the expression for K sp are To calculate K sp, we need the equilibrium concentrations of Ag + and CrO 4 2. We know that at equilibrium [Ag + ] = M. All the Ag + and CrO 4 2 ions in the solution come from the Ag 2CrO 4 that dissolves. Thus, we can use [Ag + ] to calculate [CrO 4 2 ]. Solve: From the chemical formula of silver chromate, we know that there must be 2 Ag + ions in solution for each CrO 4 2 ion in solution. Consequently, the concentration of CrO 4 2 is half the concentration of Ag +. We can now calculate the value of K sp. SAMPLE EXERCISE Calculating Solubility from K sp The K sp for CaF 2 is at 25 C. Assuming that CaF 2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF 2 in grams per liter. Solution Analyze: We are given K sp for CaF 2 and are asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, K sp, is an equilibrium constant. Plan: We can approach this problem by using our standard techniques for solving equilibrium problems. We write the chemical equation for the dissolution process and set up a table of the initial and equilibrium concentrations. We then use the equilibrium-constant expression. In this case we know K sp, and so we solve for the concentrations of the ions in solution. Solve: Assume initially that none of the salt has dissolved, and then allow x moles/liter of CaF 2 to dissociate completely when equilibrium is achieved. 13

14 SAMPLE EXERCISE continued The stoichiometry of the equilibrium dictates that 2x moles/liter of F are produced for each x moles/liter of CaF 2 that dissolve. We now use the expression for K sp and substitute the equilibrium concentrations to solve for the value of x: (member that calculator, with to calculate the cube root of a number, you can use the y x function on your ) Thus, the molar solubility of CaF 2 is mol/l. The mass of CaF 2 that dissolves in water to form a liter of solution is Solubility Products Consider the equilibrium that exists in a saturated solution of BaSO 4 in water: BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq) Solubility Products The equilibrium constant expression for this equilibrium is K sp = [Ba 2+ ] [SO 4 2 ] where the equilibrium constant, K sp, is called the solubility product. 14

15 K sp s (25 o C) EOS Solubility Products K sp 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). 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. BaSO 4 (s) Ba 2+ (aq) + SO 4 2 (aq) 15

16 Factors Affecting Solubility ph 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. 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. 16

17 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+. Will a Precipitate Form? In a solution, If Q = K sp, the system is at equilibrium and the solution is saturated. If Q < K sp, more solid can dissolve until Q = K sp. If Q > K sp, the salt will precipitate until Q = K sp. EXAMPLE Applying the Criteria for Precipitation of a Slightly Soluble Solute. Three drops of 0.20 M KI are added to ml of M Pb(NO 3 ) 2. Will a precipitate of lead iodide form? (1 drop 0.05 ml) PbI 2 (s) Pb 2+ (aq) + 2 I - (aq) K sp = Determine the amount of I - in the solution: n I - = 3 drops 0.05 ml 1 L 0.20 mol KI 1 mol I - 1 drop 1000 ml 1 L 1 mol KI = mol I - 17

18 EXAMPLE Determine the concentration of I - in the solution: [I - ] = mol I - = M I L Apply the Precipitation Criteria: Q = [Pb 2+ ][I - ] 2 = (0.010)( ) 2 = < K sp = Selective Precipitation of Ions One can use differences in solubilities of salts to separate ions in a mixture. 18