Chapter 17 Additional Aspects of Aqueous Equilibria

Chapter 17 Additional Aspects of Aqueous Equilibria
Lecture Presentation Chapter 17 Additional Aspects of Aqueous Equilibria James F. Kirby Quinnipiac University Hamden, CT © 2015 Pearson Education, Inc.

3. Acid-Base Titrations 2

Titration In this technique, an acid (or base) solution of known concentration is slowly added to a base (or acid) solution of unknown concentration. A pH meter or indicators are used to determine when the solution has reached the equivalence point: The amount of acid equals that of base. 3

Equivalence point – the point at which the reaction is complete
In a titration, a solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL The indicator changes color (pink)

Titration Curve Titration Curve – is the plotting of the pH of the solution being analyzed as a function of the amount of titrant added. Equivalence (Stoichiometric) Point – is the point in the titration when enough titrant has been added to react exactly with the substance in solution being titrated.

Units used in Titrations
Since titrations usually involve small quantities, the following units are used: 1 mmol = 1 x 10-3 mol 1 mL = 1 x 10-3 L M = = mol L mmol mL

Strong Acid-Strong Base Titrations
The net ionic reaction for a strong acid-strong base titration is: OH- (aq) + H+ (aq) H2O (l) 7

Neutralization of a Strong Acid with a Strong Base
To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright © Cengage Learning. All rights reserved 8

The pH Curve for the Titration of 50. 0 mL of 0. 200 M HNO3 with 0
The pH Curve for the Titration of 50.0 mL of M HNO3 with M NaOH Copyright © Cengage Learning. All rights reserved 9 9

The pH Curve for the Titration of 100. 0 mL of 0. 50 M NaOH with 1
The pH Curve for the Titration of mL of 0.50 M NaOH with 1.0 M HCI Copyright © Cengage Learning. All rights reserved 10 10

Strong acid-base titrations show:
a sharp change in pH near the equivalence point an equivalence point pH of 7.0 11

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. Just before (and after) the equivalence point, the pH rises rapidly. At the equivalence point, pH = 7. As more base is added, the pH again levels off. 12

Titration of a Strong Base with a Strong Acid
It looks like you “flipped over” the strong acid being titrated by a strong base. Start with a high pH (basic solution); the pH = 7 at the equivalence point; low pH to end.

Weak Acid-Strong Base Titrations
HA (aq) + OH- (aq) A- (aq) + H2O (l) At equivalence point (pH > 7): A- (aq) + H2O (l)  OH- (aq) + HA (aq) 14

Weak Base-Strong Acid Titrations
H+ (aq) + B (aq) BH+ (aq) At equivalence point (pH < 7): BH+ (aq) + H2O (l)  B (aq) + H3O+ (aq) 15

Calculating the pH Curve for Weak Acid–Strong Base Titration
Step 1: A stoichiometry problem. Assume the reaction runs to completion and then determine concentration of acid remaining and conjugate base formed. Step 2: An equilibrium problem. Determine position of weak acid equilibrium and calculate pH. 16

Titration of a Weak Acid with a Strong Base
Use Ka to find initial pH. Find the pH in the “buffer region” using stoichiometry followed by the Henderson–Hasselbalch equation. At the equivalence point the pH is >7. Use the conjugate base of the weak acid to determine the pH. As more base is added, the pH levels off. This is exactly the same as for strong acids.

Ways That a Weak Acid Titration Differs from a Strong Acid Titration
A solution of weak acid has a higher initial pH than a strong acid. The pH change near the equivalence point is smaller for a weak acid. (This is at least partly due to the buffer region.) The pH at the equivalence point is greater than 7 for a weak acid.

Exercise 17.7 Calculating pH for a Strong Acid-Strong Base Titration
Calculate the pH when the following quantities of M NaOH solution have been added to 50.0 mL of M HCl solution: (a) 49.0 mL, (b) 51.0 mL. The number of moles of H+ in the original HCl solution is given by the product of the volume of the solution (50.0 mL = L) and its molarity (0.100 M): Analyze: We are asked to calculate the pH at two points in the titration of a strong acid with a strong base. The first point is just before the equivalence point, so we expect the pH to be determined by the small amount of strong acid that has not yet been neutralized. The second point is just after the equivalence point, so we expect this pH to be determined by the small amount of excess strong base. Plan: (a) As the NaOH solution is added to the HCl solution, H+(aq) reacts with OH–(aq)to form H2O. Both Na+ and Cl– are spectator ions, having negligible effect on the pH. To determine the pH of the solution, we must first determine how many moles of H+ were originally present and how many moles of OH– were added. We can then calculate how many moles of each ion remain after the neutralization reaction. To calculate [H+], and hence pH, we must also remember that the volume of the solution increases as we add titrant, thus diluting the concentration of all solutes present. . Likewise, the number of moles of OH– in 49.0 mL of M NaOH is Continued1 19

Solution (Continued) Because we have not yet reached the equivalence point, there are more moles of H+ present than OH–. Each mole of OH– will react with one mole of H+. During the course of the titration, the volume of the reaction mixture increases as the NaOH solution is added to the HCl solution. Thus, at this point in the titration, the total volume of the solutions is (We assume that the total volume is the sum of the volumes of the acid and base solutions.) Thus, the concentration of H+(aq) is The corresponding pH equals Continued1 20

Solution (Continued) Proceed as in part (a) In this case the total volume of the solution is Hence, the concentration of OH–(aq) in the solution is Thus, the pOH of the solution equals and the pH equals Plan: (b)We proceed in the same way as we did in part (a), except we are now past the equivalence point and have more OH– in the solution than H+. As before, the initial number of moles of each reactant is determined from their volumes and concentrations. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess of hydroxide ion. Continued1 21

Exercise 17.8 Calculating pH for a Weak Acid-Strong Base Titration
Calculate the pH of the solution formed when 45.0 mL of M NaOH is added to 50.0 mL of M CH3COOH (Ka = 1.8 ×10-5). Stoichiometry Calculation: The product of the volume and concentration of each solution gives the number of moles of each reactant present before the neutralization: The 4.50 × 10-3 mol of NaOH consumes 4.50 × 10-3 mol of CH3COOH: Analyze: We are asked to calculate the pH before the equivalence point of the titration of a weak acid with a strong base. Plan: We first must determine the number of moles of CH3COOH and CH3COO– that are present after the neutralization reaction. We then calculate pH using Ka together with [CH3COOH] and [CH3COO–]. The total volume of the solution is The resulting molarities of CH3COOH and CH3COO– after the reaction are therefore Continued1 22

Exercise 17.8 Calculating pH for a Weak Acid-Strong Base Titration
Solution (Continued) Equilibrium Calculation: The equilibrium between CH3COOH and CH3COO– must obey the equilibrium-constant expression for CH3COOH Solving for [H+] gives Comment: We could have solved for pH equally well using the Henderson–Hasselbalch equation. Continued1 23

Exercise 17.9 Calculating the pH at Equivalence Point
Calculate the pH at the equivalence point in the titration of 50.0 mL of M CH3COOH with M NaOH. Before addition: The number of moles of acetic acid in the initial solution is: Moles = M × L = (0.100 mol/L)( L) = 5.00 × 10-3 mol CH3COOH Equivalence: Hence 5.00 × 10-3 mol of CH3COO– is formed. It will take 50.0 mL of NaOH to reach the equivalence point. Analyze: We are asked to determine the pH at the equivalence point of the titration of a weak acid with a strong base. Because the neutralization of a weak acid produces its anion, which is a weak base, we expect the pH at the equivalence point to be greater than 7. Plan: The initial number of moles of acetic acid will equal the number of moles of acetate ion at the equivalence point. We use the volume of the solution at the equivalence point to calculate the concentration of acetate ion. Because the acetate ion is a weak base, we can calculate the pH using Kb and [CH3COO–]. The volume of this salt solution at the equivalence point is the sum of the volumes of the acid and base, 50.0 mL mL = mL = L. Thus, the concentration of CH3COO– is Continued1 24

Kb = Kw/Ka = (1.0 × 10-14)/(1.8× 10-5) = 5.6 × 10-10.
Exercise 17.9 Calculating the pH at Equivalence Point Solution (Continued) After Equivalence point: The CH3COO– ion , which is a weak base, reacts: The Kb for CH3COO– can be calculated from the Ka value of its conjugate acid, Kb = Kw/Ka = (1.0 × 10-14)/(1.8× 10-5) = 5.6 × Using the Kb expression, we have Making the approximation that – x to , and then solving for x, we have x = [OH–] = 5.3 × 10-6 M, pOH = pH = 8.72 Check: The pH is above 7, as expected for the salt of a weak acid and strong base. Continued1 25

Titrating with an Acid-Base Indicator
An acid-base indicator marks the end point of a titration by changing color. HIn (aq) H+ (aq) + In- (aq)  10 [HIn] [In-] Color of acid (HIn) predominates  10 [HIn] [In-] Color of conjugate base (In-) predominates 26

The equivalence point is not necessarily the same as the end point (but they are ideally as close as possible). Equivalence Point – is defined by the stoichiometry. End point – is defined by the indicator color change. 27

The Acid and Base Forms of the Indicator Phenolphthalein
HIn (aq) H+ (aq) + In- (aq) Colorless Pink 28 28

The Methyl Orange Indicator is Yellow in Basic Solution and Red in Acidic Solution
HIn (aq) H+ (aq) + In- (aq) Red Yellow 29 29

HIn (aq) H+ (aq) + In- (aq)
16.5

Use of Indicators Indicators are weak acids that have a different color than their conjugate base form. Each indicator has its own pH range over which it changes color. An indicator can be used to find the equivalence point in a titration as long as it changes color in the small volume change region where the pH rapidly changes.

Indicator Choice Can Be Critical!

Concept Check: Which indicator(s) would you use for a titration of HNO2 with KOH ?
Weak acid titrated with strong base. At equivalence point, there will be conjugate base of weak acid. At equivalence point, pH > 7 Use cresol red or phenolphthalein 16.5

Titrations of Polyprotic Acids
When a polyprotic acid is titrated with a base, there is an equivalence point for each dissociation. Using the Henderson–Hasselbalch equation, we can see that half way to each equivalence point gives us the pKa for that step.

4. Solubility Equilibria
35

For a slightly soluble salt, an equilibrium is set up between the excess solid (MX) and the ions in solution: MX (s) M+ (aq) + X– (aq) Ksp = [M+] [X–] Ksp is the solubility product constant or simply the solubility constant.

MX (s) M+ (aq) + X– (aq) Ksp= [M+][X–]
Solubility constant (Ksp) – is an equilibrium constant. It has only one value for a given solid at a given temperature. MX (s) M+ (aq) + X– (aq) Ksp= [M+][X–] Solubility (S) – is an equilibrium position. Do not confuse solubility constant and solubility!

Solubility constant (Ksp) – is an equilibrium constant.
MX (s) M+ (aq) + X– (aq) Ksp= [M+] [X–] AgCl (s) Ag+ (aq) + Cl- (aq) Ksp= [Ag+][Cl–] MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp= [Mg2+][F–]2 Ag2CO3 (s) Ag+ (aq) + CO32- (aq) Ksp= [Ag+]2[CO32-] Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq) Ksp= [Ca2+]3[PO43-]2

Solubility (S) – is an equilibrium position.
MX (s) M+ (aq) + X– (aq) Ksp= [M+][X–] AgCl (s) Ag+ (aq) + Cl- (aq) Ksp= S x S = S2 MgF2 (s) Mg2+ (aq) + 2F- (aq) Ksp= S x (2S)2 = 4S3 Ag2CO3 (s) Ag+ (aq) + CO32- (aq) Ksp= (2S)2 x S = 4S3 Ca3(PO4)2 (s) Ca2+ (aq) + 2PO43- (aq) Ksp= (3S)3(2S)2 =108S5

Practice Problem: Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 × 10–15 mol/L at 25oC Bi2S3 (s) 2Bi3+ (aq) + 3S2– (aq) Ksp= [Bi3+]2[S2–]3 Ksp= (2S)2 x (3S)3 Ksp= 108S5 Initial (M) Change (M) Equilibrium (M) 0.00 +2s +3s 2s 3s Ksp= 108 (1.0 × 10–15 mol/L)5 = 1.1 × 10–73 40

Practice Problem: Calculate the solubility of silver chloride in water
Practice Problem: Calculate the solubility of silver chloride in water. Ksp = 1.6 × 10–10 AgCl (s) Ag+ (aq) + Cl- (aq) Ksp= 1.6 × 10–10 Ksp= [Ag+][Cl–] Ksp= S x S = S2 Initial (M) Change (M) Equilibrium (M) 0.00 +s s S = (Ksp)1/2 S = (1.6 × 10–10)1/2 S = 1.3×10-5 M a) 1.3×10-5 M b) 1.6×10-5 M [Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M 41

Practice Problem: Calculate the solubility of silver phosphate in water. Ksp = 1.8 × 10–18
Ag3PO4 (s) Ag+ (aq) + PO43- (aq) Ksp= 1.8 × 10–18 Ksp= [Ag+]3[PO43-] Ksp= (3S)3 x S Ksp= 27 S4 Initial (M) Change (M) Equilibrium (M) 0.00 +3s +s 3s s a) 1.3×10-5 M b) 1.6×10-5 M S = (Ksp/ 27)1/4 S = (1.8 × 10–18 / 27)1/4 S = 1.6×10-5 M [Ag+] = 4.8 x 10-5 M [PO43-] = 1.6 x 10-5 M 42

Solubility Equilibria
Because ionic compounds are strong electrolytes, they dissociate completely to the extent that they dissolve. When an equilibrium equation is written, the solid is the reactant and the ions in solution are the products. The equilibrium constant expression is called the solubility-product constant. It is represented as Ksp.

The equilibrium constant expression is
Concept Check: Write the equilibrium constant expression BaSO4(s) Ba2+ (aq) + SO42- (aq) The equilibrium constant expression is Ksp = [Ba2+] [SO42] Ba3(PO4)2(s) Ba2+ (aq) + 2PO43- (aq) The equilibrium constant expression is Ksp = [Ba2+]3 [PO43]2 44

Solubility vs. Solubility Product
Ksp is not the same as solubility. Solubility is the quantity of a substance that dissolves to form a saturated solution Common units for solubility: Grams per liter (g/L) Moles per liter (mol/L)

Practice Problem: Calculating Solubility from Ksp
Practice Problem: Calculating Solubility from Ksp. The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. What is its molar solubility? CaF2 (s) Ca2+ (aq) + 2F- (aq) Ksp= [Ca2+] [F-]2 = 3.9 × 10–11 CaF2(s) [Ca2+](M) [F–](M) Initial concentration (M) --- Change in concentration +x +2x Equilibrium concentration x 2x a) 1.3×10-5 M b) 1.6×10-5 M Ksp= (x) (2x)2 = 4x3 = 3.9 × 10–11 x = 2.1 × 10–4 M 46

Practice Problem: Calculating Solubility from Ksp
Practice Problem: Calculating Solubility from Ksp. The Ksp for CaF2 is 3.9 × 10–11 at 25 °C. What is its molar solubility? Solution (Continued) If you want the answer in g/L, multiply by molar mass; this would give g/L. a) 1.3×10-5 M b) 1.6×10-5 M 47

5. Factors that Affect Solubility
48

Three factors that affect the solubility of ionic compounds are:
Common ion pH Complex formation

This is the common ion effect!
The presence of a common ion (either M+ or X-) decreases the solubility of the salt. This is the common ion effect! MX (s) M+ (aq) X– (aq) common ion NaX (s) Na+ (aq) X- (aq) 50

Common-Ion Effect: If one of the ions in a solution equilibrium is already dissolved in the solution, the solubility of the salt will decrease. If either calcium ions or fluoride ions are present, then calcium fluoride will be less soluble. 51

Practice Problem: Calculating Solubility with a common ion
Practice Problem: Calculating Solubility with a common ion.. What is the molar solubility of CaF2 in M Ca(NO3)2? CaF2 (s) Ca2+ (aq) + 2F- (aq) Ksp= [Ca2+] [F-]2 = 3.9 × 10–11 CaF2(s) [Ca2+](M) [F–](M) Initial concentration (M) --- 0.010 Change in concentration +x +2x Equilibrium concentration x 2x a) 1.3×10-5 M b) 1.6×10-5 M Ksp= ( x) (2x)2 = 3.9 × 10–11  (0.010)(2x)2 x = 3.1 × 10–5 M Assume that x << 0.010, so that x = 0.010! 52

Practice Problem: What is the molar solubility of AgBr in (a) pure water and (b) M NaBr? Ksp= 7.7 × 10–13 (a) pure water AgBr (s) Ag+ (aq) + Br- (aq) Ksp= 7.7 × 10–13 Ksp= [Ag+][Br–] Ksp= S x S = S2 Initial (M) Change (M) Equilibrium (M) 0.00 +s s S = (Ksp)1/2 S = 8.8 ×10-7 M Continued1 53

Practice Problem: What is the molar solubility of AgBr in (a) pure water and (b) M NaBr? Ksp= 7.7 × 10–13 (b) M NaBr Ksp= 7.7 × 10–13 Ksp= [Ag+][Br–] Ksp= S x ( S) Ksp  S x AgBr (s) Ag+ (aq) + Br- (aq) NaBr (s) Na+ (aq) + Br- (aq) AgBr (s) Ag+ (aq) + Br- (aq) Initial (M) Change (M) Equilibrium (M) 0.00 +s 0.0010 s s S = (Ksp/0.0010) S = 7.7 ×10-10 M 54

Exercise 17.13 Calculating the Effect of a Common Ion on Solubility
Calculate the molar solubility of CaF2 at 25 °C in a solution that is (a) M in Ca(NO3)2, (b) M in NaF. The value of Ksp is unchanged by the presence of additional solutes. However, the solubility of the salt will decrease in the presence of common ions because of the common-ion effect. Analyze: We are asked to determine the solubility of CaF2 in the presence of two strong electrolytes, each of which contains an ion common to CaF2. In (a) the common ion is Ca2+, and NO3– is a spectator ion. In (b) the common ion is F–, and Na+ is a spectator ion. Plan: Because the slightly soluble compound is CaF2, we need to use the Ksp for this compound, which is available in Appendix D: (a) The initial concentration of Ca2+ is M because of the dissolved Ca(NO3)2: Substituting into the solubility-product expression gives Continued1 55

Solution (Continued) The solubility of CaF2 is very small (2.1 × 10-4 M). Thus, we assume that the M concentration of Ca2+ from Ca(NO3)2 is very much greater than the small additional concentration resulting from the solubility of CaF2; that is, x is small compared to 0.010 M, and x = . Continued1 (b) In this case the common ion is F–, and at equilibrium we have Assuming that 2x is small compared to M (that is, x ), we have Thus, 3.9 × 10-7 mol of solid CaF2 should dissolve per liter of M NaF solution. 56

Solution (Continued) The molar solubility of CaF2 in pure water is 2.1 × 10-4 M in the presence of M Ca2+ is 3.1 × 10-5 M, and in the presence of M F– ion it is 3.9 × 10-7 M. Thus, the addition of either Ca2+ or F– to a solution of CaF2 decreases the solubility. However, the effect of F- on the solubility is more pronounced than that of Ca2+ because [F–] appears to the second power in the Ksp expression for CaF2, whereas Ca2+ appears to the first power. 57

pH and Solubility The pH of a solution can greatly affect a salt’s solubility. Insoluble bases dissolve in acidic solutions Insoluble acids dissolve in basic solutions 58

Consider the salt, MX: MX (s) M+ (aq) + X– (aq)
HX is a weak acid HZ (aq) H+ (aq) Z- (aq) The salt MX will show increased solubility in an acidic solution if the anion X- is an effective base – that is, HX is a weak acid.

pH and solubility If a substance has a basic anion, it will be more soluble in an acidic solution. Remember that buffers control pH. When a buffer is used, there is no change in concentration of hydroxide ion!

Practice Problem: How does the solubility of silver chloride in water compare to that of silver chloride in an acidic solution (made by adding nitric acid to the solution)? Explain AgCl (s) Ag+ (aq) + Cl- (aq) HNO3 (aq) H+ (aq) + NO3- (aq) HCl (aq) H+ (aq) + Cl- (aq) The solubilities are the same. Since HCl is a strong acid, it is completely dissociated in water. The solubilities are the same. Since HCl is a strong acid, it is completely dissociated in water. 61

Practice Problem: How does the solubility of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)? Explain Ag3PO4 (s) Ag+ (aq) + PO43- (aq) HNO3 (aq) H+ (aq) + NO3- (aq) H3PO4 (aq) H+ (aq) + PO43- (aq) The silver phosphate is more soluble in an acidic solution. This is because the phosphate ion is a relatively good base and will react with the proton from the acid (essentially to completion). The phosphate ion does not react nearly as well with water. This is an example of the effect of LeChâtelier's principle on the position of the solubility equilibrium. The silver phosphate is more soluble in an acidic solution. This is because the phosphate ion is a relatively good base and will react with the proton from the acid. 62

The Ksp values are the same (assuming the temperature is constant).
Practice Problem: How does the Ksp of silver phosphate in water compare to that of silver phosphate in an acidic solution (made by adding nitric acid to the solution)? Explain Ag3PO4 (s) Ag+ (aq) + PO43- (aq) HNO3 (aq) H+ (aq) + NO3- (aq) H3PO4 (aq) H+ (aq) + PO43- (aq) The Ksp values are the same (assuming the temperature is constant). The Ksp values are the same (assuming the temperature is constant). 63

(a) Ni(OH)2(s), (b) CaCO3(s), (c) BaF2(s), (d) AgCl(s)?
Exercise 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)? Analyze: The problem lists four sparingly soluble salts, and we are asked to determine which will be more soluble at low pH than at high pH. Plan: Ionic compounds that dissociate to produce a basic anion will be more soluble in acid solution. The reaction between CO32– and H+ occurs in a stepwise fashion, first forming HCO3–. H2CO3 forms in appreciable amounts only when the concentration of H+ is sufficiently high. Continued1 64

Exercise 17.13 Predicting the Effect of Acid on Solubility
Solution (Continued) (d) The solubility of AgCl is unaffected by changes in pH because Cl– is the anion of a strong acid and therefore has negligible basicity. 65

Complex Ion Formation Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent. The formation of these complex ions increases the solubility of these salts. 66

How Complex Ion Formation Affects Solubility
Silver chloride is insoluble. It has a Ksp of 1.6 × 10–10. In the presence of NH3, the solubility greatly increases because Ag+ will form complex ions with NH3.

Amphoterism and Solubility
Amphoteric oxides and hydroxides are soluble in strong acids or base, because they can act either as acids or bases. Examples are oxides and hydroxides of Al3+, Cr3+, Zn2+, and Sn2+. 68

Exercise 17.15 Evaluating an Equilibrium Involving a Complex Ion
Calculate the concentration of Ag+ present in solution at equilibrium when concentrated ammonia is added to a 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. If [Ag+] is M initially, then [Ag(NH3)2+ will be M following addition of the NH3 because Kf for the formation of Ag(NH3)2+ is quite large. Note that the NH3 concentration given in the problem is an equilibrium concentration rather than an initial concentration. Analyze: When NH3(aq) is added to Ag+(aq) , a reaction occurs forming Ag(NH3)2+ as shown in Equation We are asked to determine what concentration of Ag+(aq) will remain uncombined when the NH3 concentration is brought to 0.20 M in a solution originally M in AgNO3. Plan: We first assume that the AgNO3 is completely dissociated, giving 0.10 M Ag+. Because Kf for the formation of Ag(NH3)2+ is quite large, we assume that essentially all the Ag+ is then converted to Ag(NH3)2+ and approach the problem as though we are concerned with the dissociation of Ag(NH3)2+ rather than its formation. To facilitate this approach, we will need to reverse the equation to represent the formation of Ag+ and NH3 from Ag(NH3)2+ and also make the corresponding change to the equilibrium constant. Continued1 69

Exercise 17.15 Evaluating an Equilibrium Involving a Complex Ion
Solution (Continued) Because the concentration of Ag+ is very small, we can ignore x in comparison with Thus, – x = M . Substituting these values into the equilibrium constant expression for the dissociation of Ag(NH3)2+, we obtain Solving for x, we obtain x = 1.5 × 10-8 M = [Ag+] . Thus, formation of the Ag(NH3)2+ complex drastically reduces the concentration of free Ag+ ion in solution. 70

6. Precipitation and Separation of Ions
71

Ion product (Q) – is defined like the expression for Ksp for a given solid except that initial concentrations are used instead of equilibrium concentrations. MX (s) M+ (aq) + X– (aq) Q = [M+]0[X–]0 If a solution containing M+ or X- ions is added, a precipitate may or may not form, depending on the concentrations of these ions in the resulting mixed solution!

Precipitation (Mixing Two Solutions of Ions)
Q < Ksp Unsaturated solution No precipitate Q = Ksp Saturated solution Q > Ksp Supersaturated solution Precipitate will form Q > Ksp; precipitation occurs and will continue until the concentrations are reduced to the point that they satisfy Ksp.

Concept Check: How to determine whether a precipitate will form?
To decide, we calculate the reaction quotient, Q, and compare it to the solubility product constant, Ksp. If Q = Ksp, the system is at equilibrium and the solution is saturated. If Q < Ksp, more solid can dissolve, so no precipitate forms. If Q > Ksp, a precipitate will form. 74

precipitate from this solution?
Practice Problem: A solution is prepared by adding mL of 4.00 x 10-3 M Ce(NO3)3 to mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 × 10–10) precipitate from this solution? Continued

precipitate from this solution?
Practice Problem: A solution is prepared by adding mL of 4.00 x 10-3 M Ce(NO3)3 to mL of 2.00 x 10-2 M KIO3. Will Ce(IO3)3 (Ksp = 1.9 × 10–10) precipitate from this solution? Step 1: Determine the initial concentrations [Ce3+]0 = = 2.86 x 10-3 M (750.0 mL)(4.00 x 10-3 mmol/mL) ( ) mL [IO3-]0 = = 5.71 x 10-3 M (300.0 mL)(2.00 x 10-2 mmol/mL) ( ) mL Step 2: Find Q Q = [Ce3+]0 [IO3-]03 = 2.86 x 10-3 x (5.71 x 10-3)3 = 5.32 x 10-10 Step 3: Is Q > Ksp? Since Q > Ksp, Ce(IO3)3 will precipitate from the solution 76

Exercise 17.16 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? When the two solutions are mixed, the total volume becomes 0.10 L L = 0.50 L. The number of moles of Pb2+ is The concentration of Pb2+ is The number of moles of SO42– is The concentration of SO42– is Analyze: The problem asks us to determine whether a precipitate will form when two salt solutions are combined. Plan: We should determine the concentrations of all ions immediately upon mixing of the solutions and compare the value of the reaction quotient, Q, to the solubility-product constant, Ksp, for any potentially insoluble product. The possible metathesis products are PbSO4 and NaNO3. Sodium salts are quite soluble; PbSO4 has a Ksp of 6.3 × 10-7 (Appendix D), however, and will precipitate if the Pb2+ and SO42– concentrations are high enough for Q to exceed Ksp for the salt. Because Q > Ksp , PbSO4 will precipitate. Continued1 77

Exercise 17.17 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? For AgCl we have Because [Ag+] = 1.0 × 10-2 M, the greatest concentration of Cl– that can be present without causing precipitation of AgCl can be calculated from the Ksp expression Any Cl– in excess of this very small concentration will cause AgCl to precipitate from solution. Analyze: We are asked to determine the concentration of Cl– necessary to begin the precipitation from a solution containing Ag+ and Pb2+, and to predict which metal chloride will begin to precipitate first. Plan: We are given Ksp values for the two possible precipitates. Using these and the metal ion concentrations, we can calculate what concentration of Cl– ion would be necessary to begin precipitation of each. The salt requiring the lower Cl– ion concentration will precipitate first. Continued1 78

Exercise 17.15 Calculating Ion Concentrations for Precipitation
Solution (Continued) Proceeding similarly for PbCl2, we have Thus, a concentration of Cl– in excess of 2.9 × 10-2 M will cause PbCl2 to precipitate AgCl will precipitate first because it requires a much smaller concentration of Cl–. Thus, Ag+ can be separated from by slowly adding Cl– so [Cl–] is between 1.8 × 10-8 M and 2.9 × 10-2 M. 79

7. Precipitation and Separation of Ions
80

Selective Precipitation (Mixtures of Metal Ions)
A mixture of ions can be separated by selective precipitation. Selective precipitation involves the use of a reagent whose anion forms a precipitate with only one or a few of the metal ions in the mixture. Example: Solution contains Ba2+ and Ag+ ions. Adding NaCl will form a precipitate with Ag+ (AgCl), while still leaving Ba2+ in solution. 81

Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture. This has been used for qualitative analysis of the presence of ions in a solution.

Qualitative Analysis The classic scheme of qualitative analysis of a mixture containing all the common cations involves first separating them into five major groups based on solubilities: Group I – Insoluble chlorides Group II – Acid-Insoluble sulfides Group III – Base-insoluble sulfides Group IV – Insoluble carbonates Group V – Alkali metal and ammonium ions 83

The End 85

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. Solution Analyze: We are given the equilibrium concentration of Ag+ in a saturated solution of Ag2CrO4. From this information, we are asked to determine the value of the solubilityproduct constant, Ksp, for Ag2CrO4. Plan: The equilibrium equation and the expression for Ksp are To calculate Ksp, we need the equilibrium concentrations of Ag+ and CrO42–. We know that at equilibrium [Ag+] = 1.3 × 10-4 M. All the Ag+ and CrO42– ions in the solution come from the Ag2CrO4 that dissolves. Thus, we can use [Ag+] to calculate [CrO42–]. Solve: From the chemical formula of silver chromate, we know that there must be 2 Ag+ ions in solution for each CrO42– ion in solution. Consequently, the concentration Of CrO42– is half the concentration of Ag+: We can now calculate the value of Ksp. 86

Sample Exercise 17.10 Calculating Ksp from Solubility
Solution (Continued) Check: We obtain a small value, as expected for a slightly soluble salt. Furthermore, the calculated value agrees well with the one given in Appendix D, 1.2 × 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 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. Answer: 1.6 ×10-12 Practice Exercise 87

Sample Integrative Exercise Putting Concepts Together
A sample of 1.25 L of HCl gas at 21 ºC and atm is bubbled through L of M NH3 solution. Calculate the pH of the resulting solution assuming that all the HCl dissolves and that the volume of the solution remains L. Solution The number of moles of HCl gas is calculated from the ideal-gas law. The number of moles of NH3 in the solution is given by the product of the volume of the solution and its concentration. The acid HCl and base NH3 react, transferring a proton from HCl to NH3, producing NH4+ and Cl– ions. To determine the pH of the solution, we first calculate the amount of each reactant and each product present at the completion of the reaction. 88

Sample Integrative Exercise Putting Concepts Together
Solution (Continued) Thus, the reaction produces a solution containing a mixture of NH3, NH4+ , and Cl–. The NH3 is a weak base (Kb = 1.8 ×10-5), NH4+ is its conjugate acid, and Cl– is neither acidic nor basic. Consequently, the pH depends on [NH3] and [NH4+] . We can calculate the pH using either Kb for NH3 or Ka for NH4+. Using the Kb expression, we have Hence, pOH = –log(9.4 × 10-6) = 5.03 and pH = – pOH = – 5.03 = 8.97. 89

Chapter 17 Additional Aspects of Aqueous Equilibria

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Chapter 17 Additional Aspects of Aqueous Equilibria

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