Acid-Base Equilibria Chapter 15

HF(aq) + H2O(l) H3O+(aq) + F-(aq) The common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substance. The presence of a common ion suppresses the ionization of a weak acid or a weak base. HF(aq) + H2O(l) H3O+(aq) + F-(aq) Addition of NaF will shift the equilibrium to the left because of the addition of F-, which is already involved in the equilibrium reaction. A solution of HF and NaF is less acidic than a solution of HF alone.

A buffer solution is a solution of: A weak acid and its salt or a weak base and its salt A buffer must contain relatively large concentration of acid to react with added base (OH-) and must also contain similar concentration of base to react with added acid (H+). The acid and base components of the buffer must not consume each other in a neutralization reaction. A buffer solution has the ability to resist changes in pH upon the addition of small amounts of either acid or base. 16.3

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr (c) Na2CO3/NaHCO3 (d) NaClO4/HClO4 (e) NH3/NH4NO3 Answer (a) HF is a weak acid and KF is its salt. Therefore, this is a buffer system. (b) HBr is a strong acid and hence this is not a buffer system. (c) NaHCO3 contains a weak acid (HCO3-) and Na2CO3 is a salt of weak acid. Therefore, this is a buffer system. (d) HClO4 is a strong acid and hence this is not a buffer system. (e) NH3 is a weak base and NH4NO3 is a salt of weak base, and therefore this is a buffer system. 16.3

Consider an equal molar mixture of CH3COOH and NaCH3COO CH3COOH (aq) H+ (aq) + CH3COO- (aq) Adding more acid creates a shift left IF enough acetate ions are present A buffered solution contains 0.50 M HC2H3O2 (Ka=1.8 x 10-5) and 0.50 M NaC2H3O2.Calculate the pH. List major species: HC2H3O2, Na+, C2H2O2-, and H2O

2. The acetic acid dissociation equilibrium will control the pH. HC2H3O2 H+(aq) + C2H3O2-(aq) I .5 0 .5 C -x +x +x E .5-x x .5+x

List major species: HC2H3O2, Na+, C2H2O2-, OH-, and H2O Calculate the change in pH when 0.010 mol solid NaOH is added to the 1.0 L of the buffered solution. List major species: HC2H3O2, Na+, C2H2O2-, OH-, and H2O (the strong base has an affinity for the p+, which will come from the acetic acid) 2. OH- + HC2H3O2 H2O + C2H3O2-

Even though acetic acid is weak, NaOH is such a strong base The reaction will essentially go to completion. Stoichiometry first. OH- + HC2H3O2 H2O + C2H3O2- before .010mol .5mol .5mol change .010mol-.010mol .5mol-.010mol .5mol+.010mol After 0mol .49mol .51mol **note .010mol HC2H3O2 is converted to C2H3O2- by the addition of NaOH

Equilibrium problem After reaction the major species HC2H3O2, Na+, C2H2O2-, and H2O HC2H3O2 H+ + C2H3O2-

Ice Table I .49 0 .51 C -x +x +x E .49-x x .51+x HC2H3O2 H+(aq) + C2H3O2-(aq) I .49 0 .51 C -x +x +x E .49-x x .51+x

ΔpH=.02 pH = .02

Solving Problems with Buffered Solutions

HCl + CH3COO- CH3COOH + Cl- HCl H+ + Cl- HCl + CH3COO- CH3COOH + Cl- 16.3

Buffers

Adding an Acid to a Buffer

How Does a Buffer Work? Contains large amounts of HA(weak acid) and A-. When OH- added, weak base is best source of p+. OH- + HA H2O + A- OH- are replaced by A- Ka = [H+ ][A-] [H+] = Ka[HA] [HA] [A-] **pH is determined by the ratio [HA]/[A-]

Similar if you added H+ ions. Contains large amounts of HA(weak acid) and A-. H+ added, conjugate base A- is attracted to the p+. H+ + A- HA H+ are replaced by HA Ka = [H+ ][A-] [H+] = Ka[HA] [HA] [A-] **pH is determined by the ratio [HA]/[A-]

Calculate the H+ in a buffered solution of. 10M HF. Ka=7. 2 x 10-4 and Calculate the H+ in a buffered solution of .10M HF. Ka=7.2 x 10-4 and .3M NaF is added. [HF] [H+] = Ka[HA] [A-] [F-] Another useful equation…. Take the –log of both sides pH = pKa – log[HA] or, pH = pKa + log [A-] [A-] [HA]

Henderson–Hasselbalch Equation For a particular buffering system (conjugate acid–base pair), all solutions that have the same ratio [A–] / [HA] will have the same pH.

Mixture of weak acid and conjugate base! What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? (formic acid) Mixture of weak acid and conjugate base! HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x Ka for HCOOH = 1.8 x 10 -4 x = 1.038 X 10 -4 [H+] [HCOO-] Ka = [HCOOH] pH = 3.98 16.2

Henderson-Hasselbach equation OR…… Use the Henderson-Hasselbach equation Consider mixture of salt NaA and weak acid HA. NaA (s) Na+ (aq) + A- (aq) Ka = [H+][A-] [HA] pKa = -log Ka HA (aq) H+ (aq) + A- (aq) [H+] = Ka [HA] [A-] Henderson-Hasselbach equation -log [H+] = -log Ka - log [HA] [A-] pH = pKa + log [conjugate base] [acid] -log [H+] = -log Ka + log [A-] [HA] pH = pKa + log [A-] [HA] 16.2

What is the pH of a solution containing 0.30 M HCOOH and 0.52 M HCOOK? Ka for HCOOH = 1.8 x 10 -4 Mixture of weak acid and conjugate base! HCOOH (aq) H+ (aq) + HCOO- (aq) Initial (M) Change (M) Equilibrium (M) 0.30 0.00 0.52 -x +x +x 0.30 - x x 0.52 + x pH = pKa + log [HCOO-] [HCOOH] Common ion effect 0.30 – x  0.30 pH = 3.77 + log [0.52] [0.30] = 4.01 0.52 + x  0.52 HCOOH pKa = 3.77 16.2

NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+] [OH-] [NH3] Kb = = 1.8 X 10-5 0.36 Initial 0.30 + x + x Change - x End 0.30 - x 0.36 + x x (.36 + x)(x) (.30 – x) 1.8 X 10-5 = 0.36x 0.30 1.8 X 10-5  x = 1.5 X 10-5 pOH = 4.82 pH= 9.18 16.3

final volume = 80.0 mL + 20.0 mL = 100 mL Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? final volume = 80.0 mL + 20.0 mL = 100 mL NH4+ 0.36 M x 0.080 L = 0.029 mol / .1 L = 0.29 M OH- 0.050 x 0.020 L = 0.001 mol / .1 L = 0.01M NH3 0.30 M x 0.080 = 0.024 mol / .1 L = 0.24M start (M) 0.29 0.01 0.24 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (M) 0.28 0.0 0.25 [H+] [NH3] [NH4+] Ka= = 5.6 X 10-10 [H+] = 6.27 X 10 -10 [H+] 0.25 0.28 pH = 9.20 = 5.6 X 10-10 16.3

NH4+ (aq) H+ (aq) + NH3 (aq) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution? NH4+ (aq) H+ (aq) + NH3 (aq) pH = pKa + log [NH3] [NH4+] pH = 9.25 + log [0.30] [0.36] pKa = 9.25 = 9.17 final volume = 80.0 mL + 20.0 mL = 100 mL start (M) 0.29 0.01 0.24 NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq) end (M) 0.28 0.0 0.25 pH = 9.25 + log [0.25] [0.28] = 9.20 16.3

Chemistry In Action: Maintaining the pH of Blood 16.3

Equivalence point – the point at which the reaction is complete Titrations 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 the endpoint (hopefully close to the equivalence point) Slowly add base to unknown acid UNTIL The indicator changes color (pink) 4.7

Strong Acid-Strong Base Titrations 100% ionization! No equilibrium NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq) OH- (aq) + H+ (aq) H2O (l) 16.4

Weak Acid-Strong Base Titrations CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l) CH3COOH (aq) + OH- (aq) CH3COO- (aq) + H2O (l) At equivalence point (pH > 7): CH3COO- (aq) + H2O (l) OH- (aq) + CH3COOH (aq) 16.4

Strong Acid-Weak Base Titrations HCl (aq) + NH3 (aq) NH4Cl (aq) H+ (aq) + NH3 (aq) NH4Cl (aq) At equivalence point (pH < 7): NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq) 16.4

Acid-Base Indicators 16.5

pH 16.5

The titration curve of a strong acid with a strong base. 16.5

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

Finding the Equivalence Point (calculation method) Strong Acid vs. Strong Base 100 % ionized! pH = 7 No equilibrium! Weak Acid vs. Strong Base Acid is neutralized; Need Kb for conjugate base equilibrium Strong Acid vs. Weak Base Base is neutralized; Need Ka for conjugate acid equilibrium Weak Acid vs. Weak Base Depends on the strength of both; could be conjugate acid, conjugate base, or pH 7

Weak Acid - Strong Base Titration Step 1 - A stoichiometry problem - reaction is assumed to run to completion - then determine remaining species. Step 2 - An equilibrium problem - determine position of weak acid equilibrium and calculate pH.

H+(aq) + OH-(aq) H2O(l) Millimole (mmol) since titrations are done in mL. A 1.0 M solution contains contains 1.0mmol of solute/mL solution. Number of mmol= volume (mL) x molarity

Strong Acid-Strong Base Titration of 50.0mL of 0.200 M HNO3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. a. no NaOH added b. 10.0 mL NaOH c. 20.0 mL NaOH d. 50.0 mL NaOH e. 100.0 mL NaOH

0 NaOH Strong acid goes to completion. [H+] = .200M and pH = .699

Titration of 50. 0mL of 0. 200 M HNO3 with 0. 100 M NaOH Titration of 50.0mL of 0.200 M HNO3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. b) 10.0 mL NaOH Major species: H+, NO3-, Na+, OH-, and H2O H+ + OH-  H2O Before 50.0 mL x 0.200M 10.0 mL x 0.100 M 10.0 mmol 1.00mmol After 9.0mmol 0

Titration of 50. 0mL of 0. 200 M HNO3 with 0. 100 M NaOH Titration of 50.0mL of 0.200 M HNO3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. c. 20.0 mL of NaOH H+ + OH- H2O Before 50.0 mL x 0.200M 20.0 mL x 0.100 M = 10.0 mmol =2.00mmol After =8.0mmol 0

Titration of 50. 0mL of 0. 200 M HNO3 with 0. 100 M NaOH Titration of 50.0mL of 0.200 M HNO3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. d. 50.0 mL NaOH H+ + OH-  H2O Before 50.0 mL x 0.200M 50.0 mL x 0.100 M = 10.0 mmol =5.00mmol After =5.0mmol 0 [H+]= 5.0 mmol/100 mL =0.05M pH = 1.301

Titration of 50. 0mL of 0. 200 M HNO3 with 0. 100 M NaOH Titration of 50.0mL of 0.200 M HNO3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. e. 100 mL NaOH At this point the amount of NaOH that has been added is 100.0mL x 0.100M = 10.mmol Original amount of nitric acid was 50.0mL x 0.200 M = 10.0 mmol Look at what’s left in solution: Na+, NO3-, and H2O (none affect the pH) Equivalence point pH = 7.00

Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Titration of 50. 0mL of 0. 200 M HNO3 with 0. 100 M NaOH Titration of 50.0mL of 0.200 M HNO3 with 0.100 M NaOH. Calculate the pH when 0.100M NaOH has been added. d. 150.0 mL NaOH H+ + OH-  H2O Before 50.0 mL x 0.200M 150.0 mL x 0.100 M = 10.0 mmol =15.00mmol After =0.0mmol =5.0mmol [OH-]= 5.0 mmol/200 mL =0.025M [H+] [OH-]= 1.0 x 10-14 [H+]= 1.0 x 10-14 / .025 = 4.0 x 10-13M pH = 12.40

Exactly 100 mL of 0. 10 M HNO2 are titrated with 100 mL of a 0 Exactly 100 mL of 0.10 M HNO2 are titrated with 100 mL of a 0.10 M NaOH solution. What is the pH at the equivalence point ? start (moles) 0.01 0.01 HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l) end (moles) 0.0 0.0 0.01 [NO2-] = 0.01 0.200 = 0.05 M Final volume = 200 mL NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq) Initial (M) Change (M) Equilibrium (M) 0.05 0.00 0.00 -x +x +x 0.05 - x x x Kb = [OH-][HNO2] [NO2-] = x2 0.05-x pOH = 5.98 = 2.2 x 10-11 pH = 14 – pOH = 8.02 0.05 – x  0.05 x  1.05 x 10-6 = [OH-]

Half way point, how many mmols of HCN? Example: HCN Ka=6.2 x 10-10 If 50.0mL of .10M HCN is titrated with .100 M NaOH calculate the pH at the equivalence point and the half way point. Half way point, how many mmols of HCN? How many mmols of NaOH are needed?How many mL are needed? When does the pH = pKa ?

Example: HCN Ka=6. 2 x 10-10 If 50. 0mL of. 10M HCN is titrated w/ Example: HCN Ka=6.2 x 10-10 If 50.0mL of .10M HCN is titrated w/ .100 M NaOH calculate the pH at the equivalence point and the half way point. At the halfway point: How many mmol of OH? How many mL of NaOH are needed? 50.0 MmL x 0.100M = 5.00 mmol HCN 2.5 mmol OH- is the halfway point Volume of NaOH = 25.0 mL pH = pKa = -log(6.2 x 10-10) = 9.21

CN-, Na+, H2O Kb =Kw/Ka 1.6 x 10-5 = [HCN][OH-] CN- + H2O  HCN + OH- Example: HCN Ka=6.2 x 10-10 If 50.0mL of .10M HCN is titrated w/ .100 M NaOH calculate the pH at the equivalence point and the half way point. At the equivalence point: How many mmol of OH? How many mL of NaOH are needed? 5.00 mmol OH- which is 50.0 mL Major species: CN-, Na+, H2O Kb =Kw/Ka 1.6 x 10-5 = [HCN][OH-] [CN-] CN- + H2O  HCN + OH- [OH-] = x = 8.9 x 10-4 [H+] = 1.1 x 10-11 pH = 10.96

Weak acid and a strong base Reacts essentially to completion The equivalence point is always greater than 7. At the half way point [H+] = Ka pH = pKa It is the amount of acid, not the strength that determines the equivalence point. pH values at equivalence point is effected by acid strength Acid strength effects the shape of the curve.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. The weaker the Acid the greater The pH at the Equivalence point. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Copyright©2000 by Houghton Mifflin Company. All rights reserved. Acid-Base Indicator . . . marks the end point of a titration by changing color. The equivalence point is not necessarily the same as the end point. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

Solubility Equilibria Will it all dissolve, and if not, how much?

All dissolving is an equilibrium. If there is not much solid it will all dissolve. As more solid is added the solution will become saturated. Solid dissolved The solid will precipitate as fast as it dissolves . Equilibrium

General equation M+ stands for the cation (usually metal). Nm- stands for the anion (a nonmetal). MaNmb(s) aM+(aq) + bNm- (aq) K = [M+]a[Nm-]b/[MaNmb] But the concentration of a solid doesn’t change. Ksp = [M+]a[Nm-]b Called the solubility product for each compound.

Look out Solubility is not the same as solubility product. Solubility product is an equilibrium constant. (amount nor size of particle matters) (it doesn’t change except with temperature.) Solubility is an equilibrium position for how much can dissolve. A common ion can change this.

Calculating Solubility The solubility is determined by equilibrium. Its an equilibrium problem. Watch the coefficients

Copper(I) bromide has a measured solubility of 2.0 x 10-4mol/L at 25C°. Calculate the Ksp. Write the equation. What is the Ksp equation? What are the initial and final concentrations? CuBr(s)  Cu+(aq) + Br-(aq) Ksp= [Cu+][Br-] Initial : [Cu+]0=0 [Br-]0=0

CuBr(s)  Cu+(aq) + Br-(aq) [Cu+] = [Cu+]0 + change to reach equilibrium [Br-] = [Br-]0 + change to reach equilibrium Solubility of CuBr(s) is 2.0 x 10-4mol/L. 0 + 2.0 x 10-4 mol/L of each ion. Ksp= [Cu+][Br-] Ksp = (2.0 x 10-4 m/L)(2.0 x 10-4 mol/L) Units are usually omitted =(4.0 x 10-8)

Solubility Product Calculate the Ksp value for bismuth sulfide (Bi2S3), which has a solubility of 1.0 x 10-15 mol/L at 25C. Hint**coefficients** Ksp = [Bi3+]2[S2-]3 1.0 x 10-15  [0 + 2(1.0 x 10-15)] + [0 + 3(1.0 x 10-15)] Ksp = [2.0 X 10-15]2[3.0 X 10 -15]3 1.1 x 10-73

Relative solubilities Ksp will only allow us to compare the solubility of solids that fall apart into the same number of ions. The bigger the Ksp of those the more soluble. If they fall apart into different number of ions you have to do the math.

Common Ion Effect If we try to dissolve the solid in a solution with either the cation or anion already present less will dissolve.

Common Ion Calculate the solubility of solid CaF2 (Ksp=4.0 x 10 -11) in a 0.025 M NaF solution. CaF2  Ca2+(aq) + 2F-(aq) Ksp=4.0 x 10 -11 = [Ca2+] [F-]2 [Ca2+]0 = 0 [Ca2+] = x [F-]20 = 0.025M [F-] 2 = .025 + 2x Ksp=4.0 x 10 -11 = [Ca2+] [F-] 2 (x)(.025 + 2x)2 =(x)(.025)2 x = 6.4 x 10-8 Approximation is ok.

pH and solubility OH- can be a common ion. More soluble in acid. For other anions if they come from a weak acid they are more soluble in an acidic solution than in water because they produce strong conjugate bases.. CaC2O4 Ca+2 + C2O4-2 H+ + C2O4-2 HC2O4- Reduces [C2O4-2] in acidic solution.

General Rule for pH and solubility If the anion X- is an effective base (HX is a weak acid) the salt MX will show increased solubility in an acidic solution. OH-, S2-, CO32-, C2O42-, and CrO42-

Precipitation Ion Product, Q =[M+]a[Nm-]b Q = [Ca2+]0[F-]02 If Q>Ksp a precipitate forms. If Q<Ksp No precipitate. If Q = Ksp equilibrium.

A solution of 750. 0 mL of 4. 00 x 10-3M Ce(NO3)3 is added to 300 A solution of 750.0 mL of 4.00 x 10-3M Ce(NO3)3 is added to 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10) precipitate and if so, what is the concentration of the ions? [Ce3+]0 = (750.0 mL)(4.00 x 10-3M) 1050mL [IO3-]03 = (300.0 mL)(2.00 x 10-2M) Q = [Ce3+]0[IO3-]03 = 2.86 x 10-3M 5.71 x 10-3M (2.86 x 10-3M) (5.71 x 10-3M)3 (5.32 x 10-10M) Q>Ksp (precip)

Selective Precipitations Used to separate mixtures of metal ions in solutions…. Adding anions that will only precipitate certain metals at a time. Used to purify mixtures. Solution containing Ba2+ and Ag+. If NaCl is added, AgCl precips, but Ba2+ does not.

A solution contains 1. 0 x 10-4 M Cu+ and 2. 0 x 10-3 M Pb2+ A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) o CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentrations of I- necessary to begin precipitation for each salt. PbI2 Ksp = 1.4 x 10-8 =[Pb2+][I-]2 = (2.0 x 10-3 ) (I-)2 (I-) = 2.6 x 10-3M CuI Ksp = 5.3 x 10-12 = [Cu+][I-] =(1.0 x 10-4 ) (I-) = 5.3 x 10-8M

Selective Precipitation Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate. Then add OH-solution [S-2] will increase so more soluble sulfides will precipitate. Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3

Selective precipitation Follow the steps With dilute HCl insoluble chlorides (Ag, Pb, Hg, Ba) Add H2S most insoluble sulfides (Hg, Cd, Bi, Cu, Sn) in acid. Then sulfides are in basic solution. (Co, Zn, Mn, Ni, Fe, Cr, Al) Then insoluble carbonate (Ca, Ba, Mg) Alkali metals and NH4+ remain in solution.

We have a solution that contains the following cations: Ag+, Cu2+, and Mg2+ We wish to isolate each ion by causing one at a time to precipitate out of solution. Once a solid precipitate forms, we can filter the solid precipitate out, leaving the other ions still in solution. We are given the following solutions to use: Na2S, NaCl, and NaOH.

To help us analyze our situation, we'll prepare a chart, with the cations we need to separate along one axis and anions along the other axis.

Complex ion Equilibria A charged ion surrounded by ligands. Ligands are Lewis bases using their lone pair to stabilize the charged metal ions. Common ligands are NH3, H2O, Cl-,CN- Coordination number is the number of attached ligands. Cu(NH3)42+ has a coordination # of 4

The addition of each ligand has its own equilibrium Usually the ligand is in large excess. And the individual K’s will be large so we can treat them as if they go to completion. The complex ion will be the biggest ion in solution.

Calculate the concentrations of Ag+, Ag(S2O3)-, and Ag(S2O3)-3 in a solution made by mixing 150.0 mL of 0.010 M AgNO3 with 200.0 mL of 5.00 M Na2S2O3 Ag+ + S2O3-2 Ag(S2O3)- K1=7.4 x 108 Ag(S2O3)- + S2O3-2 Ag(S2O3)2-3 K2=3.9 x 104

Complex Ion Equilibria and Solubility A complex ion is an ion containing a central metal cation bonded to one or more molecules or ions. Co2+ (aq) + 4Cl- (aq) CoCl4 (aq) 2- Co(H2O)6 2+ CoCl4 2- 16.10

Equilibria Involving Complex Ions Complex Ion: A charged species consisting of a metal ion surrounded by ligands (Lewis bases). Coordination Number: Number of ligands attached to a metal ion. (Most common are 6 and 4.) Formation (Stability) Constants: The equilibrium constants characterizing the stepwise addition of ligands to metal ions. Copyright©2000 by Houghton Mifflin Company. All rights reserved.

16.10

Complex Ion Formation These are usually formed from a transition metal surrounded by ligands (polar molecules or negative ions). As a "rule of thumb" you place twice the number of ligands around an ion as the charge on the ion... example: the dark blue Cu(NH3)42+ (ammonia is used as a test for Cu2+ ions), and Ag(NH3)2+. Memorize the common ligands.

Common Ligands H2O aqua NH3 ammine OH- hydroxy Cl- chloro Br- bromo Names used in the ion H2O aqua NH3 ammine OH- hydroxy Cl- chloro Br- bromo CN- cyano SCN- thiocyanato (bonded through sulphur) isothiocyanato (bonded through nitrogen)

Names Names: ligand first, then cation Examples: tetraamminecopper(II) ion: Cu(NH3)42+ diamminesilver(I) ion: Ag(NH3)2+. tetrahydroxyzinc(II) ion: Zn(OH)4 2- The charge is the sum of the parts (2+) + 4(-1)= -2.

When Complexes Form Aluminum also forms complex ions as do some post transitions metals. Ex: Al(H2O)63+ Transitional metals, such as Iron, Zinc and Chromium, can form complex ions. The odd complex ion, FeSCN2+, shows up once in a while Acid-base reactions may change NH3 into NH4+ (or vice versa) which will alter its ability to act as a ligand. Visually, a precipitate may go back into solution as a complex ion is formed. For example, Cu2+ + a little NH4OH will form the light blue precipitate, Cu(OH)2. With excess ammonia, the complex, Cu(NH3)42+, forms. Keywords such as "excess" and "concentrated" of any solution may indicate complex ions. AgNO3 + HCl forms the white precipitate, AgCl. With excess, concentrated HCl, the complex ion, AgCl2-, forms and the solution clears.

Coordination Number Total number of bonds from the ligands to the metal atom. Coordination numbers generally range between 2 and 12, with 4 (tetracoordinate) and 6 (hexacoordinate) being the most common.

Some Coordination Complexes molecular formula Lewis base/ligand Lewis acid donor atom coordination number Ag(NH3)2+ NH3 Ag+ N 2 [Zn(CN)4]2- CN- Zn2+ C 4 [Ni(CN)4]2- Ni2+ [PtCl6] 2- Cl- Pt4+ Cl 6 [Ni(NH3)6]2+