Chapter 18 Advanced Aqueous Equilibria. The Common Ion Effect. The Common Ion Effect. The Common Ion Effect. Stomach Acidity & Acid Base Reactions

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1 1 Chapter 18 More About Chemical Equilibria: Acid Base & Precipitation Reactions Principles of Chemical Reactivity: Other Aects of Aqueous Equilibria Jeffrey Mack California State University, Sacramento Stomach Acidity & Acid Base Reactions The Common Ion Effect In the previous chapter, you examined the behavior of weak acids and bases in terms of equilibrium involving conjugate pairs. The ph of a solution was found via K a or K b. What would happen if you started with a solution of acid that was mixed with a solution of its conjugate base? The change of ph when a significant ammout of conjugate base is present is an example of the Common Ion Effect. The Common Ion Effect What is the effect on the ph of a 0.5M NH (aq) solution when NH 4 Cl is added? NH (aq) H O NH (aq) OH (aq) 4 NH 4 + is an ion that is COMMON to the equilibrium. Le Chatelier predicts that the equilibrium will shift to the left to reduce the disturbance. This results in a reduciton of the hydroxide ion concentration, which will lower the ph. Hint: NH 4 + is an acid! The Common Ion Effect First let s find the ph of a 0.5M NH (aq) Solution: [NH ] [NH 4+ ] [OH - ] Initial Change x +x +x Equilibrium 0.05 x x x

2 The Common Ion Effect First let s find the ph of a 0.5M NH (aq) Solution: [NH ] [NH 4 + ] [OH - ] Initial Change x +x +x Equilibrium 0.05 x x x - 5 [NH 4 ][OH ] x K b 1.8 x 10 [NH ] x The Common Ion Effect First let s find the ph of a 0.5M NH (aq) Solution: - 5 [NH 4 ][OH ] x K b 1.8 x 10 [NH ] x Assuming x is << 0.5, x = [OH ] = 5 OH M poh =.67 ph = = 11. for 0.5 M NH The Common Ion Effect What is the ph of a solution made by adding equal volumes of 0.5M NH (aq) and 0.10M NH 4 Cl(aq)? Since the solutions are mixed with one another, the effect of dilution is cancelled out. One can use the initial concentrations of each ecies in the reaction without calculating new molarities. This also works with mole ratios. The Common Ion Effect What is the ph of a solution made by adding equal volumes of 0.5 M NH (aq) and 0.10 M NH 4 Cl(aq)? NH (aq) H O NH (aq) OH (aq) 4 [NH ] [NH 4 + ] [OH - ] Initial Change x + x +x Equilibrium 0.05 x x x Since there is more ammonia than ammonium present, the RXN will proceed to the right. The Common Ion Effect What is the ph of a solution made by adding equal volumes of 0.5 M NH (aq) and 0.10 M NH 4 Cl(aq)? [NH ] [NH 4+ ] [OH - ] Initial Change x + x +x Equilibrium 0.05 x x x - 5 [NH 4 ][OH ] (0.10 x) x K b 1.8 x 10 [NH ] 0.5 x The Common Ion Effect What is the ph of a solution made by adding equal volumes of 0.5 M NH (aq) and 0.10 M NH 4 Cl(aq)? Assuming x is << 0.5 or [NH 4 ][OH ] (0.10 x) x K b 1.8 x 10 [NH ] 0.5 x x OH M poh = 4.5 ph = 9.65 The ph drops from11. due to the common ion!

3 Controlling ph: Buffer Solutions HCl is added to pure water. HCl is added to a solution of a weak acid H PO 4 - and its conjugate base HPO 4 -. Controlling ph: Buffer Solutions A Buffer Solution is an example of the common ion effect. From an acid/base standpoint, buffers are solutions that resist changes to ph. A buffer solution requires two components that do not react with one another: 1. An acid capable of consuming OH. The acid s conjugate base capable of consuming H O + Controlling ph: Buffer Solutions Consider the acetic acid / acetate buffer system. The ability for the acid to consume OH is seen from the reverse of the base hydrolysis: CH CO (aq) H O(l) CH CO H(aq) OH (aq) K b 10 K rev is >> 1, indicating that the reaction is product favored. 1 9 Krev K b An hydroxide added will immediately react with the acid so long as it is present. Controlling ph: Buffer Solutions Consider the acetic acid / acetate buffer system. Similarly, the conjugate base (acetate) is readily capable of consuming H O + K rev is >> 1, indicating that the reaction is product favored. 1 4 Krev K An hydronium ion added will immediately react wit the acid so long as it is present. a Buffer Solutions What is the ph of a buffer that has [CH CO H] = M and [CH CO ] = M? Buffer Solutions What is the ph of a buffer that has [CH CO H] = M and [CH CO ] = M? Since the concentration of acid is greater than the base, equilibrium will move the reaction to the right. CH CO H(aq) H O CH CO (aq) H O (aq)

4 4 Buffer Solutions What is the ph of a buffer that has [CH CO H] = M and [CH CO ] = M? [CH CO H] [CH CO ] [H O + ] Initial Change x + x +x Equilibrium x x x Assuming that x << and 0.600, we find: K a 5 [H O ] Buffer Solutions What is the ph of a buffer that has [CH CO H] = M and [CH CO ] = M? K a 5 [H O + ] = ph = 4.68 [H O ] Buffer Solutions The expression for calculating the [H + ] of the buffer reduces to: Orig. conc. of CH CO H [H O ] K a Orig. conc. of CHCO Buffer Solutions Similarly for a basic solution the [OH ] of the buffer reduces to: Orig. conc. of CH CO [OH ] Kb Orig. conc. of CHCOH [Acid] [H O ] K [Conj. base] a [OH ] [Base] [Conj. acid] K b The H O + concentration depends only K a and the ratio of acid to base. The OH concentration depends only K b and the ratio of base to acid. Buffer Solutions: The Henderson- Hasselbalch Equation [Acid] log [HO ] Ka [Conj. base] [Acid] log[ho ] logka log [conj. Base] conj. Base ph pka log Acid The result is known as the Henderson-Hasselbalch equation. The ph of a buffer can be adjusted by manipulating the ratio of acid to base. What is the ph of a solution containing 0.0 M HCOOH and 0.5 M HCOOK? Mixture of weak acid and conjugate base! Initial (M) Change (M) Equilibrium (M) Common ion effect 0.0 x x 0.5 HCOOH pk a =.77 HCOOH (aq) x +x x H + (aq) + HCOO - (aq) x x ph = pk a + log [HCOO- ] [HCOOH] x ph =.77 + log [0.5] [0.0] = 4.01

5 5 Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? A ph of 4.0 correonds to an [H O + ] = M Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? A ph of 4.0 correonds to an [H O + ] = M First choose an acid with a pk a close to the desired ph Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? A ph of 4.0 correonds to an [H O + ] = M First choose an acid with a pk a close to the desired ph Next adjust the ratio of acid to conjugate base to achieve the desired ph. conj. Base ph pka log Acid Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? A ph of 4.0 correonds to an [H O + ] = M First choose an acid with a pk a close to the desired ph Possible Acids HSO / SO 4 4 CH CO H/ CH CO HCN/CN Acetic acid is the best choice. K a Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? A ph of 4.0 correonds to an [H O + ] = M [Acid] [H O ] K [Conj. base] a [Acid] [Conj. base] 5 5 [Acid].78 [Conj. base] 1 Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? A ph of 4.0 correonds to an [H O + ] = M [Acid].78 [Conj. base] 1 Therefore, if you start with mol of acetate ion then add 0.78 mol of acetic acid, will result in a solution with a ph of 4.0.

6 6 Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? Therefore, if you start with mol of acetate ion then add 0.78 mol of acetic acid, will result in a solution with a ph of 4.0. Since both ecies are in the same solution (the same volume), the mole ratios are equal to the concentration ratios! Preparing a Buffer Suppose you wish to prepare a buffer a solution at ph of 4.0. How would you proceed? So, by adding 8.0 g of sodium acetate to 780 ml of a M acetic acid solution, one would make a buffer of ph g mol NaCHCO 8.0g NaCHCO 1mol 1L 10 ml 0.78 mols CHCOH 780 ml 0.100mol 1L Preparing a Buffer Buffer prepared from 8.4 g NaHCO weak acid Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to: a) 1.00 L of pure water b) 1.00 L of buffer that has [CH CO H] = M and [CH CO ] = M (ph = 4.68) 16.0 g Na CO conjugate base What is the ph? Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to: a) 1.00 L of pure water ml L mols H O + M(H O + ) ph 1 L 1.00 mols HCl 1 mol H O 1 10 ml 1 L 1 mol HCl 1.00 L 1.00 ml M H O ph log( ).00 Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to: b) 1.00 L of buffer that has [CH CO H] = M and [CH CO ] = M (ph = 4.68) 1. The acid added will immediately be consumed by the acetate ion.. This in turn increases the acetic acid concentration.. The acetic acid then will react with water to reestablish equilibrium.

7 7 Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to: b) 1.00 L of buffer that has [CH CO H] = M and [CH CO ] = M (ph = 4.68) 1. The acid added will immediately be consumed by the acetate ion. H O (aq) CH CO (aq) CH CO H(aq) Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to b) 1.00 L of buffer that has [CH CO H] = M and [CH CO ] = M (ph = 4.68) H O (aq) CH CO (aq) CH CO H(aq) [H O + ] [CH CO ] [CH CO H] Before Change After Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to: b) 1.00 L of buffer that has [CH CO H] = M and [CH CO ] = M (ph = 4.68). This in turn increases the acetic acid concentration.. The acetic acid then will react with water to reestablish equilibrium. CH CO H(aq) H O(l) H O (aq) CH CO (aq) Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to b) 1.00 L of buffer that has [CH CO H] = M and [CH CO ] = M (ph = 4.68) CH CO H(aq) H O(l) H O (aq) CH CO (aq) [CH CO H] [H O + ] [CH CO ] Initial Change x + x +x Equilibrium x x x [HOAc] [OAc ] [HO ] K - a (1.8 x 10 ) Adding an Acid to a Buffer What is the ph when 1.00 ml of 1.00 M HCl is added to b) 1.00 L of buffer that has [CH CO H] = M and [CH CO ] = M (ph = 4.68) [HOAc] [OAc ] [HO ] K - a ( ) [H O ].110 ph Commercial Buffers The solid acid and conjugate base in the packet are mixed with water to give the ecified ph. Note that the quantity of water does not affect the ph of the buffer. The ph does not change! The solution absorbs the added acid. It is a buffer!

8 8 Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na CO /NaHCO (a) KF is a weak acid and F - is its conjugate base buffer solution (b) HBr is a strong acid not a buffer solution (c) CO - is a weak base and HCO - is its conjugate acid buffer solution 4 44 Calculate the ph of the 0.0 M NH /0.6 M NH 4 Cl buffer system. What is the ph after the addition of 0.0 ml of M NaOH to 80.0 ml of the buffer solution? NH 4 + (aq) H + (aq) + NH (aq) ph = pk a + log [NH ] [NH 4+ ] pk a = 9.5 ph = log [0.0] [0.6] = 9.17 start (moles) end (moles) NH + 4 (aq) + OH - (aq) H O (l) + NH (aq) final volume = 80.0 ml ml = 100 ml [NH ph = log [0.5] 4+ ] = [NH = ] = 0.10 [0.8] Acid Base Titrations Acid Base Titrations Adding NaOH from the buret to acetic acid in the flask, a weak acid. In the beginning the ph increases very slowly. Additional NaOH is added. ph rises as equivalence point is approached.

9 9 Acid Base Titrations Titration of a Strong Acid with a Additional NaOH is added. ph increases and then levels off as NaOH is added beyond the equivalence point. Titration of a Strong Acid with a The reaction of a strong acid and strong base produces a salt and water. The Net Ionic Equation is: H O (aq) OH (aq) H O(l) At the equivalence point, the moles of base added equal the moles of acid titrated. H O OH H O OH K w H O H O K w H O K w ph ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? At the equivalence point, all of the acid is converted to its conjugate base. The conjugate base will then react with water to reestablish equilibrium. The ph can be determined from K b. Equivalence Point (moles base = moles acid) HBz (aq) + OH (aq) Bz (aq) + H O(l) C 6 H 5 CO H = HBz Benzoate ion = Bz -

10 ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? Solution: At the equivalence point, all of the HBz is converted to Bz - by the strong base. The conjugate base of a weak acid (Bz - ) will hydrolyze to reform the weak acid (K b ). The ph will be > 7 This will yield the [H O + ] and ph ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? Volume of OH - added to the eq. point: HBz (aq) + OH (aq) Bz (aq) + H O(l) 1L 0.05mol HBz 1mol OH 1L 10 ml 10 ml 1L 1molHbz 0.100molOH 1L 100.0mL 5mL The new total volume of the solution is 15 ml 100. ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? Moles of OH - & Bz - at the eq. point: HBz (aq) + OH (aq) Bz (aq) + H O(l) ml 1 L 0.05 mol HBz 1 mol OH 1 mol Bz mols Bz 10 ml 1 L 1 mol Hbz 1 mol OH The concentration of Bz - at the eq. point is: mols Bz 10 ml 0.00 M Bz 15 ml 1 L 100. ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? [OH - ] at the eq. point: Bz (aq) H O(l) HBz (aq) OH (aq) K b [Bz - ] [HBz] [OH - ] Initial Change - x + x + x Equilibrium x x x 100. ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? [OH - ] at the eq. point: [Bz - ] [HBz] [OH - ] Initial Change - x + x + x Equilibrium x x x K 1.6 x 10 b 10 x 0.00 x 100. ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? [OH - ] at the eq. point: K b Assuming that x << 0.00, x 0.00 x 6 X = [OH ] = poh = 5.75 ph = 8.5

11 11 Conclusion: At the equivalence point of the titration, unlike the titration of a strong acid and strong base, the ph is > 7. This is due to the production of the conjugate base of a week acid. Equivalence point ph = 8.5 Conclusion: What would the ph equal at the half-way point of the titration? Hint: Only ½ of the moles of weak acid have been converted to its conjugate base! Half-way point ph =?? 100. ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? At the half-way point, moles of Hbz and Bz - are equal. This is a BUFFER SOLUTION! 100. ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? At the half-way point, moles of Hbz and Bz - are equal. This is a BUFFER SOLUTION! conj. Base ph pka log Acid 100. ml of a 0.05 M solution of benzoic acid is titrated with M NaOH to the equivalence point. What is the ph of the final solution? At the half-way point, moles of Hbz and Bz - are equal. This is a BUFFER SOLUTION! conj. Base ph pka log pka log(1) pka 0 pk a Acid Acetic Acid Titrated with NaOH 5 ph log

12 1 Titration of a Weak Polyprotic Acid with a In the case of a titration of a weak polyprotic acid (H n A) there are n equivalence points. In the case of the diprotic oxalic acid, (H C O 4 ) there are two equivalence points. Titration of a Weak Polyprotic Acid with a The titration of a polyprotic weak acid follows the same process a monoprotic weak acid. As the acid is titrated, buffering occurs until the last eq. point is reached. K a(1) HA(aq) OH (aq) HA (aq) HO(l) The ph is relative to the amounts of conjugate acids and bases. K a( ) HA (aq) OH (aq) A (aq) HO(l) At the second eq. point all of the acid has been converted to A -, ph is determined by K b. Kw Kb K a() Titration of a Weak Base with a Strong Acid ph Indicators for Acid Base Titrations In the case of a titration of a weak base, the process follows that of a weak acid in reverse. There exists a region of buffering followed by a rapid drop in ph at the eq. point. ph Indicators for Acid Base Titrations An acid/base indicator is a substance that changes color at a ecific ph. HInd (acid) has another color than Ind (base) These are usually organic compounds that have conjugated pi-bonds, often they are dyes or compounds that occur in nature such as red cabbage pigment or tannins in tea. Care must be taken when choosing an appropriate indicator so that the color change (end point of the titration) is close to the steep portion of the titration curve where the equivalence point is found. ph Indicators for Acid Base Titrations

13 1 Natural Indicators: Red Rose Extract in Methanol Neutral ph ph<<7 ph >7 Buffer <7 ph >>7 Solubility of Salts Prior to this chapter, exchange reactions which formed ionic salts were governed by the solubility rules. A compound was either soluble, insoluble or slightly soluble. So how do we differentiate between these? The answer lies in equilibrium. It turns out that equilibrium governs the solubility of inorganic salts. Lead(II) iodide Solubility of Salts The extent of solubility can be measured by the equilibrium process of the salt s ion concentrations in solution, K. K is called the solubility constant for an ionic compound. It is the product of the ion s solubilities. For the salt: A x B y (s) xa y+ (aq) + yb x- (aq) Solubility of Salts K = [A y+ ] x [B x- ] y Solubility of Salts Analysis of Silver Group Consider the solubility of a salt MX: If MX is added to water then: H MX(s) O(l) M (aq) X (aq) K [M ][X ] Generally eaking, If K >> 1 then MX is considered to be soluble If K << 1 then MX is considered to be insoluble If K 1 then MX is slightly soluble Ag + Pb + Hg + AgCl PbCl Hg Cl All salts formed in this experiment are said to be INSOLUBLE. They form when mixing moderately concentrated solutions of the metal ion with chloride ions.

14 14 Analysis of Silver Group Ag + Pb + Hg + Analysis of Silver Group Ag + Pb + Hg + AgCl PbCl Hg Cl Although all salts formed in this experiment are said to be insoluble, they do dissolve to some SLIGHT extent. AgCl(s) e Ag + (aq) Cl - (aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED. AgCl PbCl Hg Cl AgCl(s) Ag + (aq) + Cl - (aq) When solution is SATURATED, expt. shows that [Ag + ] = 1.67 x 10-5 M. This is equivalent to the SOLUBILITY of AgCl. What is [Cl - ]? [Cl - ] = [Ag + ] = 1.67 x 10-5 M Analysis of Silver Group Ag + Pb + Hg + Analysis of Silver Group Ag + Pb + Hg + AgCl PbCl Hg Cl AgCl(s) Ag + (aq) + Cl - (aq) Saturated solution has [Ag + ] = [Cl - ] = M Use this to calculate K c K c = [Ag + ] [Cl - ] = ( )( ) = AgCl PbCl Hg Cl AgCl(s) Ag + (aq) + Cl - (aq) K c = [Ag + ] [Cl - ] = Because this is the product of solubilities, we call it: K = solubility product constant Lead(II) Chloride PbCl (s) Pb + (aq) + Cl - (aq) K = = [Pb + ][Cl ] Relating Solubility & K The solubility of lead (II) iodide is found to be M. What is the K for PbI?

15 15 Relating Solubility & K The solubility of lead (II) iodide is found to be M. What is the K for PbI? Recall that lead (II) iodide dissociates via: K PbI (s) Pb (aq) I (aq) K [Pb ][I ] Relating Solubility & K The solubility of lead (II) iodide is found to be M. What is the K for PbI? From the reaction stoichiometry, K PbI (s) Pb (aq) I (aq) [Pb + ] = M & [I ] = M = M Relating Solubility & K The solubility of lead (II) iodide is found to be M. What is the K for PbI? Entering these values into the K expression, Relating Solubility & K K for MgF = Calculate the solubility in: (a) moles/l & (b) in g/l K [Mg ][F ] K [Pb ][I ] K K For PbI, K = 4 (solubility) Relating Solubility & K K for MgF = Calculate the solubility in: (a) moles/l & (b) in g/l K [Mg ][F ] (a) The solubility of the salt is governed by equilibrium so let s set up an ICE table: [MgF (s)] [Mg + ] [F ] Initial Change - + x + x Equilibrium - x x Relating Solubility & K K for MgF = Calculate the solubility in: (a) moles/l & (b) in g/l K [Mg ][F ] (a) Entering the values into the K expression: K Mg I The solubility of MgF = mols/l K x x 4x K 5.10 x

16 16 Relating Solubility & K K for MgF = Calculate the solubility in: (a) moles/l & (b) in g/l K [Mg ][F ] (b) the solubility of the salt in g/l is found using the formula weight: Relating Solubility & K What is the maximum [Cl ] in solution with M Hg + without forming Hg Cl (s)? Hg Cl (s) Hg (aq) Cl (aq) K mols MgF 6.g MgF MgF L mol 0.015g L Relating Solubility & K What is the maximum [Cl ] in solution with M Hg + without forming Hg Cl (s)? Hg Cl (s) Hg (aq) Cl (aq) K Precipitation will initiate when the product of the concentrations exceeds the K. Relating Solubility & K What is the maximum [Cl ] in solution with M Hg + without forming Hg Cl (s)? The maximum chloride concentration can be found from the K expression. Hg Cl (s) Hg (aq) Cl (aq) 18 K = Hg Cl K Cl 1.10 Hg Solubility & the Common Ion Effect Adding an ion common to an equilibrium causes the equilibrium to shift towards reactants according to Le Chatelier s principle. Common Ion Effect PbCl (s) Pb (aq) Cl (aq) K

17 17 Solubility & the Common Ion Effect What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in M Ba(NO )? K for BaSO 4 = Solubility & the Common Ion Effect What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in M Ba(NO )? K for BaSO 4 = (a) BaSO (s) Ba (aq) SO (aq) 4 4 K Ba SO x 4 5 x K M Solubility & the Common Ion Effect What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in M Ba(NO )? K for BaSO 4 = Solubility & the Common Ion Effect What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in M Ba(NO )? K for BaSO 4 = (b) [BaSO 4 (s)] [Ba + ] [SO 4 ] Initial Change - + x + x Equilibrium x x (b) [BaSO 4 (s)] [Ba + ] [SO 4 ] Initial Change - + x + x Equilibrium x x K Ba SO K x x Solubility & the Common Ion Effect What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in M Ba(NO )? K for BaSO 4 = Solubility & the Common Ion Effect What is the the solubility of BaSO 4 (s) in (a) pure water and (b) in M Ba(NO )? K for BaSO 4 = (b) K Ba SO K x x (b) K Ba SO K x x Since x << 0.010, K x K x x M 10 Since x << 0.010, K x K x x M 10 Solubility is significantly decreased by the presence of a common ion.

18 18 Effect of Basic Salts on Solubility Some anions of precipitates are the conjugate bases of weak acids. In solution these anions can hydrolyze to form the weak acid. X (aq) H O(l) HX(aq) OH (aq) Addition of additional base can shift equilibrium to the right thus reducing the concentration of X. This in turn would increase the solubility of the ppt. Effect of Basic Salts on Solubility X (aq) H O(l) HX(aq) OH (aq) Add a base: Decrease the concentration of HX Effect of Basic Salts on Solubility Effect of Basic Salts on Solubility X (aq) H O(l) HX(aq) OH (aq) Eq. Shifts to the right Eq. Shifts to the right X (aq) H O(l) HX(aq) OH (aq) The concentration of X drops Effect of Basic Salts on Solubility K & the Reaction Quotient Relating Q to K Eq. Shifts to the right Q = K The solution is saturated MX(s) M (aq) X (aq) The solubility of MX increases! Q < K Q = K The solution is not saturated The solution is over saturated, precipitation will occur

19 19 K & the Reaction Quotient Suppose you have a solution that is M Mg +. Enough NaOH(s) is added to give a [OH ] = M. Will a precipitate form? K & the Reaction Quotient Suppose you have a solution that is M Mg +. Enough NaOH(s) is added to give a [OH ] = M. Will a precipitate form? Solution: Calculate Q and compare to K. K & the Reaction Quotient M Mg + & [OH ] = M. Solution: Calculate Q and compare to K. Mg(OH) (s) Mg (aq) OH (aq) Q Mg OH K & the Reaction Quotient M Mg + & [OH ] = M. Solution: Calculate Q and compare to K. Mg(OH) (s) Mg (aq) OH (aq) 6 4 Q Q K & the Reaction Quotient M Mg + & [OH ] = M. Solution: Calculate Q and compare to K. Q K Since Q < K, no precipitation will occur. K & the Reaction Quotient What concentration of hydroxide ion will precipitate a M Mg + solution. Mg(OH) (s) Mg (aq) OH (aq) 6 K x

20 0 K & the Reaction Quotient What concentration of hydroxide ion will precipitate a M Mg + solution. x K OH 6 6 OH Equilibria & Complex Ions Metal ions exist in solution as complex ions. A complex ion involves a metal ion bound to molecules or ions called ligands. Ligands are Lewis bases that form coordinate covalent bonds with the metal. Examples are: 6 Ni(H O) or Cu(NH ) Equilibria & Complex Ions Equilibria & Complex Ions The equilibrium constants for complex ion are very product favored: Cu (aq) + 4 H O(l) [Cu(NH ) ] K.110 f 1 K f is known as the formation equilibrium constant. Equilibria & Complex Ions The extent of dissociation of a complex ions is given by the dissociation constant, K D Dissolving Precipitates by forming Complex Ions The presence of a ligand dramatically affect the solubility of a precipitate: [Cu(NH ) ] Cu (aq) + 4 H O(l) K K.110 D 1 f 14 AgCl(s) NH (aq) Ag(NH ) Cl (aq)

21 1 Solubility of Complex Ions The presence of a ligand dramatically affect the solubility of a precipitate: + AgCl(s) NH (aq) Ag(NH ) Cl (aq) Solubility of Complex Ions The presence of a ligand dramatically affect the solubility of a precipitate: + AgCl(s) NH (aq) Ag(NH ) Cl (aq) K + AgCl(s) Ag (aq) Cl (aq) Ag (aq) NH (aq) Ag(NH ) K f + Solubility of Complex Ions The presence of a ligand dramatically affect the solubility of a precipitate: + AgCl(s) NH (aq) Ag(NH ) Cl (aq) K + AgCl(s) Ag (aq) Cl (aq) net f K f Ag (aq) NH (aq) Ag(NH ) K K K K Ag(NH ) Cl net NH Solubility of Complex Ions What is the solubility of AgCl(s) in grams per liter in a 0.010M NH (aq) solution: + AgCl(s) NH (aq) Ag(NH ) Cl (aq) K Ag(NH ) Cl net NH Solubility of Complex Ions What is the solubility of AgCl(s) in grams per liter in a 0.010M NH (aq) solution: + AgCl(s) NH (aq) Ag(NH ) Cl (aq) K Ag(NH ) Cl net NH [NH ] [[Ag(NH ) ] + ] [Cl ] Initial Change - x + x + x Equilibrium x x x Solubility of Complex Ions What is the solubility of AgCl(s) in grams per liter in a 0.010M NH (aq) solution: [NH ] [[Ag(NH ) ] + ] [Cl ] Initial Change - x + x + x Equilibrium x x x x Knet x

22 Solubility of Complex Ions What is the solubility of AgCl(s) in grams per liter in a 0.010M NH (aq) solution: x Knet x Solving using the quadratic equation: X = 0.00M = [Cl ] 5 x Solubility of Complex Ions What is the solubility of AgCl(s) in grams per liter in a 0.010M NH (aq) solution: + AgCl(s) NH (aq) Ag(NH ) Cl (aq) 0.00 mol Cl 1 mol AgCl 14.g AgCl 1 L g 1 L 1 mol Cl 1 mol AgCl In pure water, the solubility of AgCl is only g/l Separating Metal Ions Cu +, Ag +, Pb + K Values AgCl 1.8 x PbCl 1.7 x 10-5 PbCrO x 10-14