Homework Problems Chapter 16 Homework Problems 2 4

Homework Problems Chapter 16 Homework Problems: 2, 4, 16, 18, 24, 27, 30 a, 36, 39 a, b, 46, 54, 67, 78, 84, 86, 90, 96, 97 (but for Cu(OH)2), 106

CHAPTER 16 Aqueous Ionic Equilibrium

Neutralization Reaction A neutralization reaction is the reaction of an acid with a base, usually giving salt plus water as products. For three types of neutralization reactions the reaction goes essentially to completion: strong acid + strong base HCl(aq) + Na. OH(aq) Na. Cl(aq) + H 2 O( ) strong acid + weak base HCl(aq) + NH 3(aq) NH 4 Cl(aq) weak acid + strong base HF(aq) + Na. OH(aq) Na. F(aq) + H 2 O( )

Buffers A buffer solution is a solution whose p. H does not change significantly upon the addition of a small amount of a strong acid or strong base. Buffer solutions contain a substantial amount of a weak acid and a weak base, usually a weak acid-weak base conjugate pair. Buffer solutions are important for systems where a particular value for p. H is required, such as in biological systems.

How Buffers Work Consider a buffer solution containing a weak acid “HA” and the conjugate base of the weak acid “A-”. What will happen if we add a small amount of a strong acid (HCl) or strong base (Na. OH)? strong acid HCl(aq) + A-(aq) HA(aq) + Cl-(aq) strong base Na. OH(aq) + HA(aq) Na+(aq) + A-(aq) + H 2 O( ) Buffer solution converts the strong acid or base into a weak acid or base. By doing so, they minimize the effect of these additions on p. H.

Example of Buffering Consider 1. 000 L of each of the following two solutions. Solution A. 0. 100 M CH 3 COOH and 0. 200 M CH 3 COONa. Solution B. Pure water Find the initial p. H and the p. H after addition of 0. 0010 moles HCl to each of the above solutions.

Solution A. 0. 100 M CH 3 COOH and 0. 200 M CH 3 COONa(aq) Na+(aq) + CH 3 COO-(aq) CH 3 COOH(aq) + H 2 O( ) H 3 O+(aq) + CH 3 COO-(aq) Ka = [H 3 O+] [CH 3 COO-] = 1. 8 x 10 -5 [CH 3 COOH] Substance Initial Change Equilibrium H 3 O+ 0. 000 x x CH 3 COO- 0. 200 x 0. 200 + x CH 3 COOH 0. 100 -x 0. 100 - x

Ka = [H 3 O+] [CH 3 COO-] = (x) (0. 200 + x) = 1. 8 x 10 -5 [CH 3 COOH] (0. 100 - x) Assume x << 0. 100 (x) (0. 200) = 1. 8 x 10 -5 (0. 100) x = (1. 8 x 10 -5) (0. 100) = 9. 0 x 10 -6 (0. 200) p. H = - log 10(9. 0 x 10 -6) = 5. 05 Is 9. 0 x 10 -6 << 0. 100? Yes (at least 10 times smaller).

Now, consider the effect of adding 0. 0010 moles of HCl (equivalent to 0. 0010 M HCl, since 1. 00 L solution). initial after rxn. HCl(aq) + CH 3 COO-(aq) CH 3 COOH(aq) + Cl-(aq) 0. 0010 M 0. 0000 M 0. 2000 M 0. 1990 M 0. 1000 M 0. 1010 M 0. 0010 M CH 3 COOH(aq) + H 2 O( ) H 3 O+(aq) + CH 3 COO-(aq) Ka = [H 3 O+] [CH 3 COO-] = 1. 8 x 10 -5 [CH 3 COOH] Substance H 3 O+ CH 3 COOH Initial After HCl 0. 000 0. 200 0. 199 0. 100 0. 101 Change x x -x Equilibrium x 0. 199 + x 0. 101 – x

Ka = (x) (0. 199 + x) = 1. 8 x 10 -5 (0. 101 – x) Assume x << 0. 100, then (x) (0. 199) = 1. 8 x 10 -5 (0. 101) x = (0. 101) (1. 8 x 10 -5) = 9. 1 x 10 -6 (0. 199) p. H = - log 10(9. 1 x 10 -6) = 5. 04 The p. H decreased from 5. 05 to 5. 04, a change of 0. 01 p. H unit.

Solution B (pure water) pure water. p. H = 7. 00 After addition of 0. 0010 moles of HCl we simply have a 0. 001 M solution of HCl, a strong acid. So p. H = - log 10(0. 0010) = 3. 00 We may summarize the results as follows: solution A solution B initial p. H 5. 05 7. 00 p. H after addn. of 5. 04 3. 00 0. 0010 mol HCl The p. H of solution A (the buffer solution) changed by only a small amount after addition of the HCl. The p. H of solution B (unbuffered water) changed by 4 p. H units (a change in H 3 O+ concentration of a factor of 10, 000).

Henderson Equation The Henderson (or Henderson-Hasselbalch) equation applies for solutions of a weak acid/conjugate base pair. We may develop the equation as follows. Consider a weak acid HA HA(aq) + H 2 O( ) H 3 O+(aq) + A-(aq) Ka = [H 3 O+] [A-] so [H 3 O+] = Ka [HA] [A-] Therefore - log 10[H 3 O+] = - log 10 Ka - log 10([HA]/[A-]) p. H = p. Ka - log 10([HA]/[A-]) , or p. H = p. Ka + log 10([A-]/[HA]) , the Henderson equation.

Properties of the Henderson Equation 1) The Henderson equation applies for solutions of a weak acid/conjugate base pair. 2) There must be appreciable amounts (at least 0. 001 M) of both the weak acid and the conjugate base present in the system. 3) If the weak acid and conjugate base concentrations are the same ([HA] = [A-]), then p. H = p. Ka + log 10([A-]/[HA]) = p. Ka + log 10(1) = p. Ka This suggests that the best buffers are those where p. H ~ p. Ka. In fact, buffers work best if the p. Ka of the weak acid is within about 1 p. H unit of the desired buffer p. H. 4) While buffers are often prepared by combining a weak acid and conjugate base pair, they can also be prepared by adding a small amount of strong base to a weak acid solution, or a small amount of strong acid to a weak base solution.

Example: We have 0. 500 M stock solutions of the following: CH 3 COOH HF HOCl Na. CH 3 COO Na. F Na. OCl Ka = 1. 8 x 10 -5 Ka = 3. 5 x 10 -4 Ka = 3. 5 x 10 -8 p. Ka = 4. 74 p. Ka = 3. 46 p. Ka = 7. 46 How would you prepare 1. 000 L of a p. H = 3. 75 buffer solution?

Example: We have 0. 500 M stock solutions of the following: CH 3 COOH HF HOCl Na. CH 3 COO Na. F Na. OCl Ka = 1. 8 x 10 -5 Ka = 3. 5 x 10 -4 Ka = 3. 5 x 10 -8 p. Ka = 4. 74 p. Ka = 3. 46 p. Ka = 7. 46 How would you prepare 1. 000 L of a p. H = 3. 75 buffer solution? Which acid/conjugate base pair should we use? We should use the one where the p. Ka is as close as possible to the p. H of the buffer we wish to make. That would be the HF/Na. F system. Now, let x = m. L of the HF stock solution (1000 - x) = m. L of the Na. F stock solution

Then, in the 1. 000 L buffer solution (using Mi. Vi = Mf. Vf) [HF] = (0. 500 M) x [F-] = (0. 500 M) (1000 - x) 1000 p. H = p. Ka + log 10([A-]/[HA]) 3. 75 = 3. 46 + log 10 (0. 500)[(1000 - x)/1000] (0. 500) (x/1000) 3. 75 = 3. 46 + log 10[(1000 - x)/x] 3. 75 - 3. 46 = 0. 29 = log 10[(1000 - x)/x] 100. 29 = 1. 95 = (1000 - x)/x 1. 95 x = 1000 - x 2. 95 x = 1000 ; x = 1000/2. 95 = 339. So 339 m. L of the HF solution and 661 m. L of the Na. F solution.

Titration By titration, we mean the reaction of a fixed amount of one substance by the slow addition of a second substance. In an acid-base titration we slowly add an acid to a fixed amount of base, or we slowly add a base to a fixed amount of acid. We use titrations to determine the concentration of acid or base in a stock solution. We may also be interested in the titration curve, a plot of p. H vs m. L of added titrant (solution being slowly added). Equivalence point. The equivalence point in a titration is the point where sufficient titrant has been added to have a complete reaction with no excess acid or base.

Indicator An indicator is a substance (usually a weak acid) that changes colors at a particular p. H. HInd(aq) + H 2 O( ) H 3 O+(aq) + Ind-(aq) color 1 color 2 We often use indicators in titrations. We choose an indicator so that the end point (point where color change occurs) is as close as possible to the equivalence point for the titration. Based on the Henderson equation p. H = p. Ka + log 10([A-]/[HA]) we expect the color change will occur at p. H p. KInd.

Table 16. 1, page 747 of Tro. In choosing an indicator we would like the color change to be as close as possible to the equivalence point for the titration.

Titration of a Strong Base With a Strong Acid Consider the titration of 25. 00 m. L of a 0. 0100 M solution of Na. OH (a strong base) with a 0. 0100 M solution of HCl (a strong acid). The reactions that will occur are as follows: Na. OH(aq) + HCl(aq) Na. Cl(aq) + H 2 O( ) neutralization Na. OH(aq) Na+(aq) + OH-(aq) if excess base HCl(aq) H+(aq) + Cl-(aq) if excess acid If there is excess Na. OH the p. H will be greater than 7. 0. If there is excess HCl the p. H will be less than 7. 0. At the equivalence point the moles of Na. OH and HCl will be equal, and the solution will be neutral (p. H = 7. 0).

The equivalence point occurs at p. H = 7. 0. A good indicator would be one where the color change occurs near p. H = 7. 0 (such as bromthymol blue). Notice that the titration curve is steep (nearly vertical) near the equivalence point. Note that we can calculate the p. H at every point in the titration (which was done in generating the above titration curve).

Titration of a Strong Acid With a Strong Base Consider the same system, but the case where 25. 00 m. L of a 0. 0100 M solution of HCl is titrated with a 0. 0100 M solution of Na. OH. The reactions taking place are the same, but now the solution is initially acidic.

Titration of a Weak Acid With a Strong Base Consider the titration of 25. 00 m. L of a 0. 0100 M solution of CH 3 COOH (a weak acid; Ka = 1. 8 x 10 -5) with a 0. 0100 M solution of Na. OH (a strong base). The reactions that will occur are as follows: CH 3 COOH(aq) + Na. OH(aq) H 2 O( ) + Na+(aq) + CH 3 COO-(aq) neutralization CH 3 COOH(aq) + H 2 O( ) H 3 O+(aq) + CH 3 COO-(aq) if excess weak acid Na. OH(aq) Na+(aq) + OH-(aq) if excess base

CH 3 COOH(aq) + Na. OH(aq) H 2 O( ) + Na+(aq) + CH 3 COO-(aq) neutralization CH 3 COOH(aq) + H 2 O( ) H 3 O+(aq) + CH 3 COO-(aq) if excess weak acid Na. OH(aq) Na+(aq) + OH-(aq) if excess base If there is excess CH 3 COOH then there will usually be a buffer solution with both CH 3 COOH and CH 3 COO- present. If there is excess base then the p. H will be controlled by the concentration of excess strong base. Note that at the half-equivalence point p. H p. Ka, since p. H = p. Ka + log 10([A-]/[HA])

At the half-equivalence point (point where we have added half the titrant needed to reach the equivalence point) p. H p. Ka for the weak acid. Notice that there will be a region where the solution forms a buffer solution. Note that the p. H at the equivalence point is greater than 7. 0. Indicator - phenolphthalein

Titration of a Weak Base With a Strong Acid Consider the titration of 25. 00 m. L of a 0. 0100 M solution of NH 3 (a weak base; Kb = 1. 8 x 10 -5) with a 0. 0100 M solution of HCl (a strong acid). The reactions that will occur are as follows: NH 3(aq) + HCl(aq) NH 4+(aq) + Cl-(aq) neutralization NH 3(aq) + H 2 O( ) NH 4+(aq) + OH-(aq) if excess weak base HCl(aq) H+(aq) + Cl-(aq) if excess acid If there is excess NH 3 then there will usually be a buffer solution with both NH 3 and NH 4+ present. If there is excess acid then the p. H will be controlled by the concentration of excess strong acid. As was the case for the titration of a weak acid with a strong base, at the half equivalence point p. H p. Ka (this will be for NH 4+, the conjugate acid of NH 3).

At the half-equivalence point (point where we have added half the titrant needed to reach the equivalence point) p. H p. Ka for the conjugate acid of the weak base. Notice that there will be a region where the solution forms a buffer solution. Note that the p. H at the equivalence point is less than 7. 0. Indicator - chlorophenol red

Titration of a Polyprotic Acid With a Strong Base Consider the titration of 1. 000 L of a 1. 000 M solution of a diprotic acid (H 2 A) with Na. OH, a strong base.

Solubility Product We previously divided ionic compounds into two general categories: insoluble - does not dissolve in water soluble - dissolves in water However, “insoluble” ionic compounds usually dissolve in water to a very small extent. The solubility product, Ksp, is defined as the equilibrium constant for the solubility reaction. Examples: Ag. Cl(s) Ag+(aq) + Cl-(aq) Ca. F 2(s) Ca 2+(aq) + 2 F-(aq) Ksp = [Ag+] [Cl-] Ksp = [Ca 2+] [F-]2 Since our reactant is a solid, equilibrium requires only that there be some solid present in the system.

Solubility and Molar Solubility There are two terms that are related to the solubility product. Solubility (solubility by mass) - The number of grams of a compound that will dissolve per 1. 00 L of water. Molar solubility - The number of moles of a compound that will dissolve per 1. 00 L of water. Of course, our previous methods for indicating solubility (such as grams per 100 grams solvent) are still valid. Example: The solubility product for lead II chloride (Pb. Cl 2, MW = 278. 1 g/mol) is Ksp = 1. 2 x 10 -5 at T = 25. C. What are the solubility by mass and the molar solubility of lead II chloride?

Example: The solubility product for lead II chloride (Pb. Cl 2, MW = 278. 1 g/mol) is Ksp = 1. 2 x 10 -5 at T = 25. C. What are the solubility by mass and the molar solubility of lead II chloride? Pb. Cl 2(s) Pb 2+(aq) + 2 Cl-(aq) Ksp = [Pb 2+] [Cl-]2 = 1. 2 x 10 -5 Substance Initial Change Equilibrium Pb 2+ 0. 00 +x x Cl- 0. 00 + 2 x 2 x (x) (2 x)2 = 4 x 3 = 1. 2 x 10 -5 x 3 = 3. 0 x 10 -6 ; x = (3. 0 x 10 -6)1/3 = 1. 44 x 10 -2 molar solubility = 1. 44 x 10 -2 moles/L solubility = 1. 44 x 10 -2 mol 278. 1 g = 4. 01 g 1 L 1 mol L

Factors Affecting Solubility 1) Temperature. By Le Chatlier’s principle, we may make the following general statements If the solubility reaction is exothermic ( H < 0) then solubility decreases as temperature increases. AB(s) A+(aq) + B-(aq) + “heat” If the solubility reaction is endothermic ( H > 0) then solubility increases as temperature increases. AB(s) + “heat” A+(aq) + B-(aq) 2) p. H. This will be true for hydroxide compounds (those containing OH- ion) or compounds containing weak base anions (F-, CN-, CO 32 -, S 2 -, PO 33 -, and other conjugate bases of weak acids). Example: What is the maximum concentration of Ni 2+ ion in a solution with p. H = 10. 00? Ksp(Ni(OH)2) = 5. 5 x 10 -16.

Example: What is the maximum concentration of Ni 2+ ion in a solution with p. H = 10. 00? Ksp(Ni(OH)2) = 5. 5 x 10 -16. Ni(OH)2 Ni 2+(aq) + 2 OH-(aq) Ksp = [Ni 2+] [OH-]2 = 5. 5 x 10 -16 Since p. H = 10. 00, p. OH = 14. 00 - 10. 00 = 4. 00. So [OH-] = 10 -4. 00 = 1. 0 x 10 -4 M Ksp = [Ni 2+] [OH-]2 [Ni 2+] = Ksp [OH-]2 = 5. 5 x 10 -16 = 5. 5 x 10 -8 M (1. 0 x 10 -4)2 Note that it is possible for [Ni 2+] to be 5. 5 x 10 -8 M.

3) Common ion effect. The common ion effect refers to the fact that in equilibrium reactions all that matters are the concentrations of the ions involved in the equilibrium. The source of the ions does not matter. If a common is already present in solution this can affect the solubility of a compound. Example: What is the molar solubility of Ag. Cl in a) pure water and b) in a 0. 100 M solution of Na. Cl. Ksp(Ag. Cl) = 1. 8 x 10 -10.

Example: What is the molar solubility of Ag. Cl in a) pure water and b) in a 0. 100 M solution of Na. Cl. Ksp(Ag. Cl) = 1. 8 x 10 -10. Ag. Cl(s) Ag+(aq) + Cl-(aq) Ksp = [Ag+] [Cl-] = 1. 8 x 10 -10 a) pure water compound initial change equilibrium Ag+ 0. 00 +x x Cl- 0. 00 +x x (x) = x 2 = 1. 8 x 10 -10 x = (1. 8 x 10 -10)1/2 = 1. 34 x 10 -5 molar solubility = 1. 34 x 10 -5 mol/L.

b) 0. 100 M solution of Na. Cl(s) Na+(aq) + Cl-(aq) compound initial change equilibrium Ag+ 0. 00 +x x Cl- 0. 100 +x 0. 100 + x (x) (0. 100 + x) = 1. 8 x 10 -10 Assume x << 0. 100 (x) (0. 100) = 1. 8 x 10 -10 x= (1. 8 x 10 -10) = 1. 8 x 10 -9 (0. 100) molar solubility = 1. 8 x 10 -9 mol/L. This is much smaller than the molar solubility we found for silver chloride in pure water (1. 34 x 10 -5 mol/L).

Solubility and the Reaction Quotient Consider a solution where only ions are present (no solid initially present). As a specific example consider a 1: 1 ionic compound MX MX(s) M+(aq) + X-(aq) Qsp = [M+][X-] There are three possibilities for Qsp, the reaction quotient If Qsp < Ksp the solution is unsaturated (more solid could dissolve). If Qsp = Ksp the solution is saturated (no more solid could dissolve). If Qsp > Ksp the solution is supersaturated (a precipitate will form as the system goes to equilibrium).

Selective Precipitation We may use differences in solubility to remove specific cations from a mixture by selective precipitation, the formation of a precipitate with one ion in the mixture. Example: Consider a solution containing 0. 0100 M Mg 2+ ion and 0. 0100 M Cu 2+ ion. Can we separate these ions?

Example: Consider a solution containing 0. 0100 M Mg 2+ ion and 0. 0100 M Cu 2+ ion. Can we separate these ions? Consider adding OH- ion. Mg(OH)2(s) Mg 2+(aq) + 2 OH-(aq) Ksp = [Mg 2+] [OH-]2 = 5. 6 x 10 -12 Cu(OH)2(s) Cu 2+(aq) + 2 OH-(aq) Ksp = [Cu 2+] [OH-]2 = 1. 6 x 10 -19 Mg [OH-] = {Ksp/[Mg 2+]}1/2 = {5. 6 x 10 -12/0. 01}1/2 = 2. 4 x 10 -5 M Cu [OH-] = {Ksp/[Cu 2+]}1/2 = {1. 6 x 10 -19/0. 01}1/2 = 4. 0 x 10 -9 M So if OH- is added to the solution Cu(OH)2 will begin to precipitate when [OH-] reaches 4. 0 x 10 -9. No Mg(OH)2 will precipitate until [OH-] reaches 2. 4 x 10 -5 M. So we can separate most of the Cu 2+ ion out of solution.

Complex Ion Formation Some metal cations will react with small molecules or ions (NH 3, CN-, OH-) to form complex ions. A complex ion is defined as a metal cation bonded to one or more small molecules or ions. For example Ag+(aq) + 2 NH 3(aq) Ag(NH 3)2+(aq) The equilibrium constant for the formation of a complex ion is called the formation constant (or stability constant), Kf Kf = [Ag(NH 3)2+] [Ag+] [NH 3]2 Formation constants are usually large numbers (meaning complex ion formation is favored when the reactants are present).

End of Chapter 16 Tony Cortino: Where did you learn to dance like that? Pepper: Julliard. I wanted to be a research chemist, but my legs were too long. - dialogue from the movie Mafia!