Chapter 17 Acid-Base & Solubility Equilibria

17.1 The Common Ion Effect 17.2 Buffer Solutions 17.3 Acid-Base Titrations (omitted) 17.4 Solubility Equilibria 17.5 Factors Affecting Solubility 17.6 Separation of Ions Using Differences in Solubility (omitted)

The Effect of Common Ion 17.1 A phenomenon known as the common ion effect states that: When a compound containing an ion in common with an already dissolved substance (a weak electrolyte) is added to an aqueous solution at equilibrium, the equilibrium shifts to the left. The ionization of the weak electrolyte is being suppressed by adding the common ion to the solution.

Le Châtelier's principle 17.1 The Common Ion Effect A 1.0 L of 0.10 M solution of CH3COOH. Adding 0.050 mol of CH3COONa: The result is that fewer H+ ions present with a higher pH. CH3COO‒ is a common ion Le Châtelier's principle

Equilibrium Calculation Involving Common Ion Effect 17.1 Equilibrium Calculation Involving Common Ion Effect Before adding the CH3COO‒ ions: 1.0 L of 0.10 M (M) CH3COOH H+ CH3COO‒ Initial conc. 0.10 Change in conc. x + x Equilibrium conc. 0.10 - x x = 1.34  10-3 M Ka < 10-3 [H+] = 1.34  10-3 M pH = -log (1.34  10-3 ) = 2.87

Equilibrium Calculation Involving Common Ion Effect 17.1 Equilibrium Calculation Involving Common Ion Effect After adding the CH3COO‒ (0.05 M) ions: Method II CH3COONa (M) CH3COOH H+ CH3COO‒ Initial conc. 0.10 0.050 Change in conc. - x + x Equilibrium conc. 0.10 - x x 0.050 + x x = 3.6  10-5 M [H+] = x = 3.6  10-5 M pH = -log (3.6  10-5 ) = 4.44

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) 17.1 The Common Ion Effect Exercise: Which of the following when dissolved in aqueous NH3 solution is (are) going to decrease the dissociation of NH3? NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) (a) Ca(OH)2 (b) HNO2 (c) CH3COONa (d) NH4NO3 (e) NH3  

17.2 Buffer Solutions A buffer solution is the one that resists the change in its pH when small amounts of either H+ or OH‒ ions are added. Buffers are useful applications of the common ion effect. Buffer solutions are important for: Biological systems. (some enzymes can only function at a specific pH, pH of blood is always about 7.4, gastric “stomach” juices maintain a pH of about 1.5) Chemical applications. (fermentation processes, dyes used in coloring fabrics, calibration for pH meters).

Buffer Solutions A buffer solution can be: 17.2 Buffer Solutions A buffer solution can be: a solution containing a weak acid and its conjugate base, or CH3COOH (aq) CH3COO‒ (aq) + H+ (aq) It is known as an acidic buffer solution and it maintains a pH value that is less than 7. a solution containing a weak base and its conjugate acid. NH3 (aq) + H2O(l) NH4+ (aq) + OH‒ (aq) It is known as a basic buffer solution and it maintains a pH value that is greater than 7. weak acid conjugate base weak base conjugate acid

Henderson-Hasselbalch Equation How to calculate pH of buffer? Henderson-Hasselbalch Equation For any buffer solution when a valid approximation is applied, its equilibrium expression is: Ka = [H+][A‒] [HA] From H-H equation, when the concentrations of the weak acid and its conjugate base in a buffer are equal, its pH = its pKa . The slight change in the pH of the buffer is due to change in the concentrations of the weak acid and its conjugate base when small amounts of either H+ or OH‒ ions are added to the buffer. HA is the weak acid A‒ is the conjugate base [H+] = Ka [HA] [A‒] ‒ log [H+] = ‒ log Ka + log [HA] [A‒] pH = pKa + log [weak acid] [conjugate base]

Requirement for a solution to be buffer? A solution is only a buffer if it has the capacity to resist pH change when either an acid or a base is added. If the concentrations of a weak acid and conjugate base differ by more than a factor of 10, the solution does not have this capacity.

17.2 Therefore, we consider a solution a buffer, and can use to calculate its pH, only if the following condition is met: Consequently, the log term can only have values from –1 to 1, and the pH of a buffer cannot be more than one pH unit different from the pKa of the weak acid it contains. This is known as the range of the buffer, where pH = pKa 1. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Exercise: Determine the pH at 25°C of a solution prepared by adding 0.050 mole of sodium acetate to 1.0 L of 0.10 M acetic acid. Ka = 1.8 x 10-5 for acetic acid. (Assume that the addition of sodium acetate does not change the volume of the solution.) The solution will work as buffer or not?

pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74 pH = pKa + log [weak acid] [conjugate base] pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74 Log [HA] [A‒] = [0.10] [.050] Log 0.5 -0.3 pH = 4.74 – 0.3 = 4.44

What happens when acid or base is added into a buffer solution? (1.00 M of acetic acid and 1.00 M of sodium acetate) 1.00 mol 1.10 mol

Consider what happens to the pH when we add 0.10 mole of HCl to Pure water The buffer solution prepared by adding 1.00 mole of sodium acetate to 1.0 L of 1.00 M acetic acid. (Ka = 1.8 x 10-5 for acetic acid or pKa = 4.74). (Assume that the addition of HCl causes no change in the volume of the solution.)

pH = 1 A change of only 0.08 pH units. How about the pH of pure water? OR Since we are adding 0.1 M strong acid (HCl); [acid] = [1 M + 0.1 M] = [1.1 M], [conjugate base] = [1 M - .01 M] = [0.9 M] pH = pKa + log [weak acid] [conjugate base] pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74 log [WA] [CB] = [1.10] [0.90] log 0.82 - 0.08 pH = 4.74 – 0.08 = 4.66 A change of only 0.08 pH units. pH = 1 How about the pH of pure water?

Exercise: Starting with 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the pH after the addition of 0.100 mole of NaOH. (Assume that the addition does not change the volume of the solution.) (Ka = 1.8 x 10-5 for acetic acid or pKa = 4.74)

pH = 13 How about the pH of pure water? OR 1.00 mol 1.10 mol OR Since we are adding 0.1 M strong base (NaOH); [acid] = [1 M - 0.1 M] = [0.9 M], [conjugate base] = [1 M + .01 M] = [1.1 M] pKa = ‒ log Ka = ‒ log (1.8 x 10-5) = 4.74 log [WA] [CB] = [0.90] [1.10] log 1.22 + 0.08 pH = 4.74 + 0.08 = 4.82 A change of only 0.08 pH units. pH = 13 How about the pH of pure water?

Exercise: Calculate the pH of 2.0 L of a buffer that is 1.0 mole in both acetic acid and sodium acetate after adding 0.15 M of Ca(OH)2. (pKa = 4.74 for acetic acid)

Concentration = pH = pKa + log (0.8) = 4.74 + log = 4.74 + 0.6 = 5.34 Mole 2 L 1.0 = = 0.5 M pH = pKa + log [CH3COOH] [CH3COO‒] = 4.74 + log (0.2) (0.8) = 4.74 + 0.6 = 5.34

What we learned… Common Ion Effect Buffer solution (properties, composition) How to calculate pH of Buffer (Henderson-Hasselbalch Equation) What happens when we add acid or base in buffer solutions

What we hope to learn today… How to prepare a buffer with specific pH Solubility and Solubility Product Expression (Ksp) Calculations Involving Ksp and Solubility Predicting Precipitation Reactions

How to Prepare a Buffer Solution with a Specific pH 17.2 How to Prepare a Buffer Solution with a Specific pH Choose a weak acid whose pKa is close to the desired pH. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 24

17.2 Substitute the pH and pKa values into the following equation to obtain the necessary ratio of [conjugate base]/[weak acid]. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 25

How you would prepare a buffer with a pH of 9.50.? 17.3 Exercise: How you would prepare a buffer with a pH of 9.50.? 17.3 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 26

17.3 One way to achieve this would be to dissolve 0.41 mol of C6H5ONa and 1.00 mol of C6H5OH in 1 L of water. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 27

Solubility Product Expression and Ksp Calculations Involving Ksp and Solubility

Consider – soluble, less soluble, insoluble 17.4 Solubility Product Expression and Ksp The solubility of ionic compounds is important in industry, medicine, and everyday life. Consider – soluble, less soluble, insoluble Solubility, solute, saturated solution, precipitate?? Equilibrium constant is represented by Ksp - called the solubility product constant. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 29

17.4 There are two ways to express the solubility of a substance: The molar solubility: number of moles of solute in 1 L of a saturated solution (mol/L). The solubility: number of grams of solute in 1 L of a saturated solution (g/L). Both of these refer to concentrations at a particular temperature (usually 25°C). How to convert solubility into molar solubility or vice-versa? Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 30

Solid (precipitate) Solvent Solute Saturated Precipitate formation Less saturated

All above compounds are very slightly soluble in water All above compounds are very slightly soluble in water. None of them are soluble in water. The smaller the Ksp value, the less soluble the compound. This is valid for compounds of similar formulas, such as comparing AgCl with CuBr, and CaF2 with Fe(OH)2 .

17.4 Solubility Product Expression and Ksp Ksp is equal to the concentrations of products over the concentrations of reactants, each raised to its coefficient from the balanced chemical equation. Equilibrium constant is represented by Ksp - called the solubility product constant. The smaller the Ksp, the less soluble the compound. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 33

17.4 Calculations Involving Ksp and Solubility To calculate molar solubility from Ksp : The procedure is essentially identical to the procedure for solving weak acid or weak base equilibrium problems Construct an equilibrium or ICE table. Fill in what we know. Figure out what we don’t know. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 34

Calculate the solubility of AgBr? Exercise: Calculate the solubility of AgBr? The Ksp of silver bromide (AgBr) is 7.7 × 10‒13. Let s be the molar solubility (in mol/L) of AgBr. At equilibrium: [Ag+] = [Br‒] = s Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 35

17.4 Calculations Involving Ksp and Solubility The equilibrium expression is Therefore, Is this solubility or molar solubility? Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 36

Calculations Involving Ksp and Solubility The molar solubility of AgBr is 8.8 × 10‒7 M. Express this solubility in g/L by multiplying the molar solubility by the molar mass of AgBr: Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 37

17.4 Exercise: Solubility Equilibria 17.7 (Ksp for Cu(OH)2 = 2.2 × 10‒20) and the molar mass of Cu(OH)2 is 97.57 g/mol. The equation for the dissociation of Cu(OH)2 is Calculate the solubility of copper(II) hydroxide [Cu(OH)2] in g/L. Calculations Involving Ksp and Solubility Ksp = [Cu2+][OH‒]2 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 38

Remember to raise an entire term to the appropriate power! 17.7 Therefore, Ksp = [Cu2+][OH‒]2 Remember to raise an entire term to the appropriate power! Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 39

The molar solubility of Cu(OH)2 is 1.8 × 10‒7 M. 17.7 The molar solubility of Cu(OH)2 is 1.8 × 10‒7 M. Multiplying by its molar mass gives: Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 40

Exercise: 17.8 The solubility of calcium sulfate (CaSO4) is measured experimentally and found to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate. The molar mass of CaSO4 is 136.2 g/mol. The molar solubility of CaSO4 is Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 41

17.8 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 42

Predicting Precipitation Reactions To predict whether a precipitation will form or not, we calculate the reaction quotient (Q) for the possible precipitation for the initial state of mixing two solutions. If Q < Ksp, no precipitation is going to form. If Q > Ksp, precipitation is going to form. If Q = Ksp, the solution is saturated.

Ag2SO4(s) + H2O(l) 2Ag+(aq) + SO42-(aq) Exercise: Predict whether a precipitate will form when 0.001 M Ag2SO4 is prepared? Ag2SO4, Ksp = 1.5 × 10‒5 Ag2SO4(s) + H2O(l) 2Ag+(aq) + SO42-(aq) 0.001M 0.002M Q = [Ag+]2 [SO4-] = [0.002]2 [0.001] 4 10-9  Q < Ksp, (No precipitate)

Exercise: Predict whether a precipitate will form when 0.002 M AgNO3 is added to 0.100 M K2SO4? (Assume that the addition does not change the volume of the solution.) Ksp values of Ag2SO4, Ksp = 1.5 × 10‒5

When we have 2 solutes… Q = [Ag+] [Cl‒] Q = [Ag+]2 [SO4‒] AgNO3 (completely soluble) Ag2SO4 (slightly soluble) K2SO4 (completely soluble) Q = [Ag+] [Cl‒] Q = [Ag+]2 [SO4‒] At the first stage, you should be able to determine which compound is soluble in water and which is very slightly soluble in water. You calculate Q for the latter compound and then compare it with the listed Ksp values.

AgNO3(s) + H2O(l) Ag+(aq) + NO3-(aq) 0.002 M 0.002 M 0.002 M AgNO3(s) + H2O(l) Ag+(aq) + NO3-(aq) 0.1 M 0.2 M 0.1 M K2SO4(s) + H2O(l) 2K+(aq) + SO42-(aq) Ag+ reacts with SO42- to form slightly soluble Ag2SO4 Ag2SO4(s) + H2O(l) 2Ag+(aq) + SO42-(aq) Q = [Ag+]2 [SO4‒] = [0.002]2[0.1] = 4  10-7 Ksp Ag2SO4 = 1.5 × 10‒5 Q < Ksp, (No precipitate)

Exercise: 17.9 Predict whether a precipitate will form when 250 mL of 0.0040 M BaCl2 is added to 650 mL of 0.0080 M K2SO4: The Ksp values of BaSO4, Ksp = 1.1 × 10‒10 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 48

Q = [Ba2+][SO42‒] = (0.0011)(0.0058) = 6.4 × 10‒6 Solution 17.9 Concentrations of the constituent ions of BaSO4 are: Q = [Ba2+][SO42‒] = (0.0011)(0.0058) = 6.4 × 10‒6 Q > Ksp (1.1 × 10‒10) Ksp BaSO4 = 1.1 × 10‒10 (BaSO4 will precipitate) Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 49

Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0  10-15 mol/L at 25C. Answer : 1.1  10-73

What we learned in last lecture… How to prepare a buffer with specific pH Solubility and Solubility Product Expression (Ksp) Calculations Involving Ksp and Solubility Predicting Precipitation Reactions

What we hope to learn today… Factors affecting solubility

Factors Affecting Solubility 17.5 Factors Affecting Solubility There are some factors that affect solubility. We will be discussing: The common ion effect. The pH. Complex ion formation

Any factor that will shift the equilibrium left (reverse) will decrease the solubility. Any factor that will shift the equilibrium right (forward) will increase the solubility.

Factors Affecting Solubility 17.5 The common ion effect. Consider dissolving AgCl salt in pure water to get a saturated aqueous solution of AgCl. The solubility of AgCl in water at 25C is 1.3  10-5 M. AgCl salt The solubility of AgCl in water can be calculated from its solubility product constant (Ksp = 1.6  10-10) 1.0 L Pure water Cl‒ Ag+ Cl‒ Cl‒ Ag+ Ag+ Cl‒ Cl‒ Ag+ Ag+ Ag+ Cl‒ So how about if we dissolve AgCl in presence of other salts? Cl‒ Ag+ AgCl (s)

Factors Affecting Solubility 17.5 Factors Affecting Solubility The common ion effect. The common ion effect is an example of Le Châtelier’s principle. The presence of a second salt (normally very soluble in water) that produces an ion common to a solubility equilibrium will reduce solubility. AgCl (s) Ag+ (aq) + Cl‒ (aq) AgNO3 solution NaCl solution What about NaNO3?

Factors Affecting Solubility 17.5 Factors Affecting Solubility (less soluble) (soluble) Example: Calculate the molar solubility of BaSO4 in 0.0010 M Na2SO4. BaSO4 (s) Ba2+ (aq) + SO42‒ (aq) Ksp of BaSO4 = 1.1  10-10 (M) BaSO4 (s) Ba2+ (aq) SO42‒ (aq) Initial conc. 1  10-3 Change in conc. + s Equilibrium conc. s 1  10-3 + s For comparison, the solubility in pure water is: s = (1.1  10-10)1/2 = 1.0  10-5 M Ksp = 1.1  10-10 = [Ba2+][SO42‒] = (s)(1  10-3 + s) 1.1  10-10 ≈ (s)(1  10-3) s = 1  10-7 M

In general, the common ion effect decreases the solubility of a substance.

Factors Affecting Solubility 17.5 Factors Affecting Solubility The pH Adding weak acids (HF, CH3COOH etc.) Adding weak base (NH3) Adding strong acids (HCl, HBr, HNO3, H2SO4) Adding strong bases (NaOH, Ca(OH)2)

Exercise: Which of the following compounds will be more soluble in acidic solution than in water: CaF2 CuS AgCl HNO3 H2SO4 HSO4-   Think about it: How solubility will change under basic solution?

Exercise: Which of the following compounds will be more soluble in acidic solution than in water: Cu(OH)2 CuCN PbBr2 PbSO4 KClO4 KClO2     Think about it: How solubility will change under basic solution?

Example: Calculate the solution pH, above which the solubility of Ca(OH)2 will decrease (Ksp of Ca(OH)2 = 8.0  10-6 ). pH = 12.41

Factors Affecting Solubility 17.5 Factors Affecting Solubility Ca(OH)2 (s) Ca2+ (aq) + 2OH‒ (aq) Ksp = 8.0  10-6 Ksp = 8.0  10-6 = [Ca2+][OH‒]2 = (s)(2s)2 s = 1.3  10-2 M [OH‒] = 2(1.3  10-2 M) = 2.6  10-2 M pOH= - log (2.6  10-2) = 1.59 Ca(OH)2 solubility decreases 14.00 pH = 14.00 – 1.59 = 12.41 Ca(OH)2 solubility increases 1.00

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation. A Complex ion is an ion that involves a central metal cation (mostly are transition metal ions) bonded to one or more ions or molecules ( ligands). Tetra-ammine-copper(II) cation , Cu(NH3)42+ , is one example of complex ions. - Complex ions exhibit colors when transition metal ions are contained at the central position. - Different types of ligands give different color Co(H2O)62+ CoCl42

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation. Consider adding aqueous ammonia to a saturated AgCl solution. AgCl dissolves, and Ag+ ions form Ag(NH3)2+ complex NH3 (aq) AgCl in water Kf =

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation. We measure the tendency of a metal ion to form a complex ion using the formation constant, Kf , (or stability constant).

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation. Let’s write equilibrium equations for the previous experiment. AgCl (s) Ag+ (aq) + Cl‒ (aq) Ag+ (aq) + 2NH3 (aq) Ag(NH3)2+ (aq) AgCl (s) + 2NH3 (aq) Ag(NH3)2+ (aq) + Cl‒ (aq) K” = Ksp  Kf = 1.6  10-10  1.5  107 = 2.4  10-3 K” >> Ksp In general, the formation of complex ions increases the solubility of a substance. We can also apply Le Châtelier’s principle to understand! Ksp = 1.6  10-10 Kf = 1.5  107 K ” = ?

Exercise: Determine whether the solubility of Cu(OH)2 will increase or decrease when mixed with NaCN? Ksp of Cu(OH)2 = 1.6  10-19 Kf Cu(CN)42- = 1 1025 Solubility will increase because CN- will react with Cu2+ making Cu(CN)42- and shifting the equilibrium to the right.

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation Determine the molar conc. of free Cu2+ ion in solution when 0.10 mole of Cu(NO3)2 is dissolved in a L of 3.0 M NH3 to form Cu(NH3)42+. Kf for Cu(NH3)42+ = 5.0  1014 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 69

Consider the following reaction Solving an equilibrium problem involving complex ion formation is complicated by the magnitude of Kf and stoichiometry. Consider the following reaction Kf is so large that we cannot ignore “change in concentration (x)”. A new approach is needed

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation Assume that all the copper(II) ion is consumed to form the complex ion (since Kf is so large.) Consider the equilibrium in terms of the reverse reaction; the dissociation of Cu(NH3)42+, for which the equilibrium constant is the reciprocal of Kf. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 71

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation Because this K is so small, we can expect x to be insignificant. Note that [NH3], which had been 3.0 M, has been diminished by 4 × 0.10 M due to the amount required to complex 0.10 mole of copper(II) ion. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 72

Factors Affecting Solubility 17.5 Factors Affecting Solubility Complex Ion Formation Neglecting x, x = 4.4 × 10‒18 M Because Kf is so large, the amount of copper that remains uncomplexed is extremely small. Think about it! What is the conc. of NH3 and Cu(NH3)42+ at equilibrium?? Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 73

Kf for the complex ion Cd(CN)42‒ is 7.1 × 1016. The reverse process 17.12 In the presence of aqueous cyanide, cadmium(II) forms the complex ion Cd(CN)42‒. Determine the molar concentration of free (uncomplexed) cadmium(II) ion in solution when 0.20 mole of Cd(NO3)2 is dissolved in a liter of 2.0 M sodium cyanide (NaCN). Setup Kf for the complex ion Cd(CN)42‒ is 7.1 × 1016. The reverse process has an equilibrium constant of 1/Kf = 1.4 × 10 ‒17 Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 74

The equilibrium expression for the dissociation is 17.12 The equilibrium expression for the dissociation is Stoichiometry indicates that four CN‒ ions are required to react with one Cd2+ ion. Therefore, [CN‒] will be [2.0 M ‒ 4(0.20 M)] = 1.2 M. Solution Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 75

17.12 Neglecting x, x = 1.4 × 10‒18 M Don’t forget to adjust the concentration of the complexing agent before entering it in the equilibrium table. Copyright © 2014, The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 76

Review… Common Ion Effect Buffer solution (properties, composition, required ratio for solution to be buffer) How to calculate pH of Buffer (Henderson-Hasselbalch Equation) What happens when we add acid or base in buffer solutions

How to prepare a buffer with specific pH Solubility and Solubility Product Expression (Ksp) Calculations Involving Ksp and Solubility Predicting Precipitation Reactions Factors affecting solubility (common ions, solution pH, complex ion formation)

Examples

Which of the following pairs will show common ion effect? BaF2, Ba(NO3)2 NaCl, HCl NH4OH,NH4Cl NaOH, NaCl   Which of the following will give buffer solutions i) NH4Cl + NH3 solution ii) HCl + NaOH solution iii) HA + NaA solution iv) NaHCO3 and Na2CO3 solution v) NaHSO4/H2SO4 A) ii and iii only B) i and iii only C) iii only D) i, iii iv and v only E) i and ii only 

Which one of the following aqueous solutions, when mixed with an equal volume of 0.10 mol L-1 aqueous NH3, will produce a buffer solution?  a. 0.10 mol L-1 HCl  b. 0.10 mol L-1 NH4Cl c. 0.050 mol L-1 NaOH  d. 1.10 mol L-1 NH4Cl    Think about it – how will you calculate the pH? How will calculate pH after adding acid or base? Calculate the pH of the buffer 0.20 M Na2HPO4 / 0.30 M KH2PO4 [Ka (Phosphoric acid) = 6.2 × 10–8] A) 3.07 B) 5.05 C) 9.01 D) 7.03 E) 2.06  Think about it –How will calculate pH after adding acid or base?

Calculate the molar solubility and the solubility of SnS in g/L at 25°C. Ksp for SnS is 1.0  10-26. Molar mass of SnS = 150.77 g

Solubility Calculations 17.4 Solubility Calculations Exercise: Calculate the molar solubility and the solubility of SnS in g/L at 25°C. Ksp for SnS is 1.0  10-26. SnS (s) Sn2+ (aq) + S2‒ (aq) (M) SnS Sn2+ S2‒ Initial conc. Change in conc. + x Equilibrium conc. x Molar solubility Ksp = [Sn2+][S2‒] = (x)(x) = x2 = 1.0  10-26 s = 1.0  10-13 M Molar solubility of SnS is then 1.0  10-13 mol/L 1 L 1.0  10-13 mol 1 mol 150.77 g Solubility =  = 1.5  10-11 g/L

Calculate the molar solubility of CaF2 salt (Ksp = 4.0  10-11) in a 0.025 M NaF solution. Answer : 6.4  10-8 mol/L

Ksp = [F‒]2 [Ca2+]= (0.025 + s)2 (s) = 4.0 × 10–11 CaF2 (s) Ca2+ (aq) + 2F‒ (aq) (M) CaF2 Ca2+ 2F‒ Initial conc. 0.025 Change in conc. + s + 2s Equilibrium conc. s 0.025 + 2s Ksp = [F‒]2 [Ca2+]= (0.025 + s)2 (s) = 4.0 × 10–11 Because Ksp is small, we can ignore 2s = (0.025)2 (s) = 4.0 × 10–11 s = 6.4  10-8 M Since the Molar solubility of Ca2+ is = 6.4  10-8 M mol/L, the Molar solubility of CaF2 will same (6.4  10-8 M mol/L)

How many grams of CaCO3 will dissolve in 4 How many grams of CaCO3 will dissolve in 4.1 × 102 mL water containing 0.06 M Ca (NO3)2? [Ksp (Calcium Carbonate) = 8.7 × 10–9 at room temperature] A) 6.0 × 10–6 g B) 9.1 × 10–9 g C) 5.1 × 10–5 g D) 4.9 × 10–6 g E) 3.6 × 10–7 g Molar mass of CaCO3 = 110.1 g

1 mol 100.1 g Solubility = 1.45  10-7 mol/L  = 1.45  10-5 g/L 1 L CaCO3 Ca2+ CO32‒ Initial conc. 0.06 Change in conc. + s Equilibrium conc. 0.06 + s s Ksp = [Ca2+][CO32‒] = (0.06 + s)(s) = 8.7 × 10–9 = (0.06) (s) = 8.7 × 10–9 s = 1.45  10-7 M Molar solubility of CaCO3 is = 1.45  10-7 mol/L 1 mol 100.1 g Solubility = 1.45  10-7 mol/L  = 1.45  10-5 g/L 1 L 1.45  10-5 g Solubility = .410 L  = 6.0  10-6 g

Examples on Solubility Equilibria Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0  10-15 mol/L at 25C. Answer : 1.1  10-73 Precipitation of bismuth sulfide

Predict whether a precipitate will form when 0 Predict whether a precipitate will form when 0.001 M Ag2SO4 is prepared? Ag2SO4, Ksp = 1.5 × 10‒5

Ag2SO4(s) + H2O(l) 2Ag+(aq) + SO42-(aq) 0.001M 0.002M Q = [Ag+]2 [SO4-] = [0.002]2 [0.001] 4 10-9  Ag2SO4, Ksp = 1.5 × 10‒5 Q < Ksp, (No precipitate)

Examples on Solubility Equilibria A solution is prepared by adding 750.0 mL of 4.00  10-3 M Ce(NO3)3 to 300.0 mL of 2.00  10-2 M KIO3. Will Ce(IO3)3 solid (Ksp = 1.9  10-10) form from this solution? Think about it – What will be the concentration of IO3-1, if we use K2(IO3)2 ?

Ce(IO3)3 (s) Ce3+ (aq) + 3IO3‒ (aq) Qsp = [Ce3+] [IO3-]3 M1V1 = M2V2 1050 mL 4.0  10-3 M  750 mL 2.857  10-3 M = [Ce(NO3)3] = 1050 mL 2.0  10-2 M  300 mL 5.714  10-3 M = [KIO3] = Qsp = [Ce3+] [IO3-]3] = [2.857  10-3 M] [5.714  10-3 M]3 = 5.33  10-10 Qsp > Ksp(1.910-10), precipitate will form

Cu(OH)2(s) ⇄ Cu2+(aq) + 2OH - (aq) Which will shift the following equilibrium in the left (reverse) direction. Cu(OH)2(s) ⇄ Cu2+(aq) + 2OH - (aq) (i) addition of Cu(NO3)2. (ii) addition of NaNO3. (iii) due to increase in reaction pH. (iv) decrease in reaction pH. (v) addition of NH3 (Kf of Cu(NH3)42+ = 5.0 × 1013). A) (iii), (iv) B) (ii), (iii), (iv) C) (i), (iii) D) (i), (iv) E) (i), (ii), (v) 