CHAPTER 17 AcidBase Equilibriu and Solubility Equilibria Neutralization

CHAPTER 17 Acid-Base Equilibriu and Solubility Equilibria

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( ) For weak acid + weak base an equilibrium is established between reactants and products. This case is mathematically difficult.

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. Examples: CH 3 COOH + CH 3 COO- acetate buffer H 2 CO 3 + HCO 3 - carbonate buffer NH 3 + NH 4+ ammonia buffer

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, at T = 25. C. The value for Ka for CH 3 COOH at this temperature is Ka = 1. 8 x 10 -5.

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).

Buffer Preparation There are two common methods of preparing buffers: 1) Directly add a conjugate acid/base pair to a solution Example: Acetate buffer – add CH 3 COOH(aq) and Na. CH 3 COO(s) Na+(aq) + CH 3 COO-(aq) 2) Add a weak acid or weak base and partially neutralize with a strong base or acid Example: Acetate buffer Begin with a solution of acetic acid, and add sufficient strong base to partially neutralize the acid. CH 3 COOH(aq) + Na. OH(aq) Na+(aq) + CH 3 COO-(aq) + H 2 O(l)

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]) = p. Ka + log 10([base]/[acid]) 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) Buffers can be prepared by combining a weak acid and conjugate base pair, and 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: Consider 1. 000 L of a 0. 100 M solution of acetic acid (CH 3 COOH). If we add 0. 050 moles of Na. OH, then H 2 O( ) CH 3 COOH(aq) + Na. OH(s) CH 3 COO-(aq) + Na+(aq) + initial (. 100 mol) (. 050 mol) (. 000 mol) final (. 050 mol) (. 000 mol) (. 050 mol) The neutralization reaction is used to produce the conjugate base needed for the buffer. Note that we require that there be a larger number of moles of weak acid than strong base.

Buffer Calculation (Example) We have 0. 0500 M solutions of the following weak acids, and the sodium salts of their conjugate bases. CH 3 COOH HF HOCl 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 Na. CH 3 COO Na. F Na. OCl MW = 82. 0 g/mol MW = 42. 0 g/mol MW = 74. 5 g/mol We wish to prepare 1. 000 L of a p. H = 4. 50 buffer, by adding the appropriate solid sodium salt to the weak acid solution. 1) Which weak acid/conjugate base pair should we use? 2) How much of the sodium salt should we add to prepare the buffer?

Buffer Calculation (Example) We have 0. 0500 M solutions of the following weak acids, and the sodium salts of their conjugate bases. CH 3 COOH HF HOCl 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 Na. CH 3 COO Na. F Na. OCl MW = 82. 0 g/mol MW = 42. 0 g/mol MW = 74. 5 g/mol We wish to prepare 1. 000 L of a p. H = 4. 50 buffer, by adding the appropriate solid sodium salt to the weak acid solution. 1) Which weak acid/conjugate base pair should we use? ANSWER: CH 3 COOH + Na. CH 3 COO

Buffer Calculation (Example) 2) How much of the sodium salt should we add to prepare the buffer (1. 000 L, 0. 0500 M CH 3 COOH, p. H = 4. 50)? CH 3 COOH Ka = 1. 8 x 10 -5 ; p. Ka = 4. 74 Na. CH 3 COOH MW = 82. 0 g/mol p. H = p. Ka + log 10([A-]/[HA]) 4. 50 = 4. 74 + log 10([CH 3 COO-]/[CH 3 COOH]) = 4. 50 - 4. 74 = - 0. 24 ([CH 3 COO-]/[CH 3 COOH]) = 10 -0. 24 = 0. 575 [CH 3 COO-] = [CH 3 COOH] (0. 575) = (0. 0500 M) (0. 575) = 0. 0288 M

Buffer Calculation (Example) 2) How much of the sodium salt should we add to prepare the buffer (1. 000 L, 0. 0500 M CH 3 COOH, p. H = 4. 50)? Na. CH 3 COOH MW = 82. 0 g/mol [CH 3 COO-] = [CH 3 COOH] (0. 575) = (0. 0500 M) (0. 575) = 0. 0288 M So grams of Na. CH 3 COO is: g salt = 1. 000 L soln 0. 0288 mol 82. 0 g = 2. 36 g salt L mol In doing this calculation we assume that adding a small amount of solid to the solution will have a negligible effect on the volume. In real life, we would add the approximate amount of sodium salt and then use small amounts of strong acid or base to adjust the p. H.

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([Ind-]/[HInd]) we expect the color change will occur at p. H p. KInd.

In choosing an indicator we would like the color change to be as close as possible to the equivalence point for the titration.

Phenolphthalein is a suitable indicator for both the strong acid+strong base titration (blue line) and weak acid+strong base titration (red line). Methyl red is a suitable indicator only for the strong acid+strong base titration. Other indicators would also be suitable choices.

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).

p. H Calculation During Titration 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 What is the p. H after addition of 10. 00 m. L of HCl solution? Strategy: 1) Find the moles of acid and base present, and the total volume. 2) Find the species present after the neutralization reaction. 3) Find the p. H (as in a regular acid/base calculation)

mol Na. OH = 0. 02500 L (0. 0100 M) = 2. 5 x 10 -4 mol Na. OH mol HCl = 0. 0100 L (0. 0100 M) = 1. 0 x 10 -4 mol HCl Volume - 0. 0250 L + 0. 0100 L = 0. 0350 L Na. OH(aq) + HCl(aq) Na. Cl(aq) + H 2 O( ) neutralization After neutralization, excess Na. OH is Excess Na. OH = 2. 5 x 10 -4 mol - 1. 0 x 10 -4 mol = 1. 5 x 10 -4 mol [Na. OH]excess = (1. 5 x 10 -4 mol)/(0. 0350 L) = 0. 00429 M Na. OH(aq) Na+(aq) + OH-(aq) p. OH = - log 10(. 00429) = 2. 37 p. H = 14. 00 - 2. 47 = 11. 63 if excess base

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

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. CH 3 COOH(aq) + Na. OH(aq) H 2 O( ) + Na+(aq) + CH 3 COO-(aq) If there is excess base then the p. H will be controlled by the concentration of excess strong base. Na. OH(aq) Na+(aq) + OH-(aq) 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

Example of Weak Acid + Strong Base Calculation 25. 00 m. L of a 0. 100 M solution of formic acid (HCOOH, Ka = 1. 7 x 10 -4) is titrated with a 0. 100 M solution of KOH, a strong base. What is the p. H after 10. 00 m. L of KOH solution has been added?

25. 00 m. L of a 0. 100 M solution of formic acid (HCOOH, Ka = 1. 7 x 10 -4) is titrated with a 0. 100 M solution of KOH, a strong base. What is the p. H after 10. 00 m. L of KOH solution has been added? Moles HCOOH = 0. 02500 L (0. 100 mol/L) = 2. 50 x 10 -3 mol Moles Na. OH = 0. 01000 L (0. 100 mol/L) 1. 00 x 10 -3 mol Volume = 25. 00 m. L + 10. 00 m. L = 35. 00 m. L HCOOH(aq) + Na. OH(aq) Na+(aq) + HCOO-(aq) + H 2 O( ) Initial moles Moles after neutralization Concentration after neutralization HCOOH 2. 50 x 10 -3 1. 50 x 10 -3 0. 04286 HCOO- 0 1. 00 x 10 -3 0. 02857 Na. OH 1. 00 x 10 -3 0 0

We can now use the reaction for a weak acid in equilibrium with its conjugate base. HCOOH(aq) + H 2 O( ) H 3 O+(aq) + HCOO-(aq) Ka = [H 3 O+][HCOO-] = 1. 7 x 10 -4 [HCOOH] After neutralization Change Equilibrium HCOOH 0. 04286 -x 0. 04286 - x COOH- 0. 02857 x 0. 02857 + x x x H 3 O+ 0 (0. 02857 + x)(x) = 1. 7 x 10 -4 If we assume x << 0. 028, then (0. 04286 - x) x = (0. 04286)(1. 7 x 10 -4) = 2. 6 x 10 -4 (0. 02857) p. H = 3. 59

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). So we can also say at this point p. OH p. Kb.

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 Ksp = [Ag+] [Cl-] Ca. F 2(s) Ca 2+(aq) + 2 F-(aq) Ksp = [Ca 2+] [F-]2 Since our reactant is a solid, equilibrium requires only that there be some solid present in the system.

Ksp(Al(OH)3) = [Al 3+][OH-]3 = 1. 8 x 10 -33 Ksp(Ca. F) = [Ca 2+][F-]2 = 4. 0 x 10 -11 Ksp(Ag. Br) = [Ag+][Br-] = 7. 7 x 10 -13 Ksp(Ag. Cl) = [Ag+][Cl-] = 1. 6 x 10 -6 Ksp(Zn. S) = [Zn 2+][S 2 -] = 3. 0 x 10 -23 The solid compound does not appear in the expression for Ksp, but must be present in some amount for equilibrium for a saturated solution.

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 in water?

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 in water? 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

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 there is no solid present, the solution is stable, but not at equilibrium. 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).

Consider the solubility of lead II bromide, Pb. Br 2(s) Pb 2+(aq) + 2 Br-(aq) Qsp = [Pb] [Br]2 Qsp < Ksp Qsp = Ksp Qsp > Ksp unsaturated supersaturated “stable” equilibrium “unstable”

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 equilibrium concentration of Ni 2+ ion in a buffer solution with p. H = 10. 00? Ksp(Ni(OH)2) = 5. 5 x 10 -16.

Example: What is the maximum equilibrium concentration of Ni 2+ ion in a buffer 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 = 5. 5 x 10 -16 = 5. 5 x 10 -8 M [OH-]2 (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).

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. For example, consider a solution that contains 0. 010 M of Ag+, Ca 2+, and Fe 3+ ion. Since Ag. Cl is an insoluble (or slightly soluble) ionic compound, and Ca. Cl 2 and Fe. Cl 3 are soluble ionic compounds, adding Na. Cl to the solution will cause Ag. Cl to precipitate, while Ca 2+ and Fe ions will remain in solution.

Selective Precipitation (Example) Consider a solution containing 0. 0100 M Mg 2+ ion and 0. 0100 M Cu 2+ ion. Can we separate these ions? How?

Consider a solution containing 0. 0100 M Mg 2+ ion and 0. 0100 M Cu 2+ ion. Can we separate these ions? How? 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

Consider a solution containing 0. 0100 M Mg 2+ ion and 0. 0100 M Cu 2+ ion. Can we separate these ions? How? 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 (p. H = 5. 60). No Mg(OH)2 will precipitate until [OH-] reaches 2. 4 x 10 -5 M (p. H = 9. 38). 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). We may do equilibrium calculations in the same way we do with other systems at equilibrium, but because Kf is large, it is best to start by assuming the maximum amount of product has formed.

Kf(Cu(CN)42 -) = [Cu(CN 4)2 -] [Cu 2+][CN-]4 Kf(CO(NH 3)63+) = [CO(NH 3)63+] [Co 3+][NH 3]6

Example: Copper ion (Cu 2+) will form a complex ion with ammonia (NH 3) by the reaction Cu 2+(aq) + 4 NH 3(aq) Cu(NH 3)42+(aq) Kf = 5. 0 x 1013 A solution initially contains 0. 0100 M Cu 2+ ion and 0. 200 M NH 3. What will be the concentration of Cu 2+ ion at equilibrium?

Example: Copper ion (Cu 2+) will form a complex ion with ammonia (NH 3) by the reaction Cu 2+(aq) + 4 NH 3(aq) Cu(NH 3)42+(aq) Kf = 5. 0 x 1013 A solution initially contains 0. 0100 M Cu 2+ ion and 0. 200 M NH 3. What will be the concentration of Cu 2+ ion at equilibrium? Kf = [Cu(NH 3)42+] [Cu 2+] [NH 3]4 Initial Cu(NH 3)42+ 0 Cu 2+ 0. 0100 NH 3 0. 200 If all complex 0. 0100 Change Equilibrium -x 0. 0100 - x 0 x x 0. 160 4 x 0. 160 + 4 x

So Kf = (0. 0100 – x) = 5. 0 x 1013 (x) (0. 160 + 4 x)4 If we assume x << 0. 01, then we get (0. 0100) = 5. 0 x 1013 x= (x) (0. 160)4 (0. 0100) (0. 160)4 (5. 0 x 1013) x = 3. 1 x 10 -13 x << 0. 010 is a good assumption. So at equilibrium, Cu 2+ = 3. 1 x 10 -13 M

End of Chapter 17 “If you are not part of the solution, you are part of the precipitate. ” - Unknown