Applications of Aqueous Equilibria

Buffered Solutions A solution that resists a change in pH when either hydroxide ions or protons are added. Buffered solutions contain either: A weak acid and its salt A weak base and its salt

Acid/Salt Buffering Pairs The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH) Weak Acid Formula of the acid Example of a salt of the weak acid  Hydrofluoric  HF   KF – Potassium fluoride   Formic   HCOOH   KHCOO – Potassium formate   Benzoic   C6H5COOH   NaC6H5COO – Sodium benzoate  Acetic   CH3COOH   NaH3COO – Sodium acetate   Carbonic   H2CO3   NaHCO3 - Sodium bicarbonate  Propanoic   HC3H5O2    NaC3H5O2  - Sodium propanoate  Hydrocyanic   HCN   KCN - potassium cyanide 

Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3) Base Formula of the base Example of a salt of the weak acid Ammonia   NH3  NH4Cl - ammonium chloride  Methylamine  CH3NH2  CH3NH2Cl – methylammonium chloride  Ethylamine  C2H5NH2  C2H5NH3NO3 -  ethylammonium nitrate  Aniline  C6H5NH2  C6H5NH3Cl – aniline hydrochloride  Pyridine  C5H5N    C5H5NHCl – pyridine hydrochloride

Weak Acid/Strong Base Titration Endpoint is above pH 7 A solution that is 0.10 M CH3COOH is titrated with 0.10 M NaOH

Strong Acid/Strong Base Titration Endpoint is at pH 7 A solution that is 0.10 M HCl is titrated with 0.10 M NaOH

Strong Acid/Strong Base Titration A solution that is 0.10 M NaOH is titrated with 0.10 M HCl Endpoint is at pH 7 It is important to recognize that titration curves are not always increasing from left to right.

Strong Acid/Weak Base Titration A solution that is 0.10 M HCl is titrated with 0.10 M NH3 Endpoint is below pH 7

Titration of an Unbuffered Solution A solution that is 0.10 M HC2H3O2 is titrated with 0.10 M NaOH

Titration of a Buffered Solution A solution that is 0.10 M HC2H3O2 and 0.10 M NaC2H3O2 is titrated with 0.10 M NaOH

Comparing Results Unbuffered Buffered In what ways are the graphs different? In what ways are the graphs similar?

Comparing Results Buffered Unbuffered

Buffer capacity The best buffers have a ratio [A-]/[HA] = 1 This is most resistant to change True when [A-] = [HA] Make pH = pKa (since log1=0)

General equation Ka = [H+] [A-] [HA] so [H+] = Ka [HA] [A-] The [H+] depends on the ratio [HA]/[A-] taking the negative log of both sides pH = -log(Ka [HA]/[A-]) pH = -log(Ka)-log([HA]/[A-]) pH = pKa + log([A-]/[HA])

This is called the Henderson-Hasselbalch Equation

Using the Henderson-Hasselbalch Equation pH = pKa + log([A-]/[HA]) pH = pKa + log(base/acid) Calculate the pH of the following mixtures 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4) 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)

Prove they’re buffers What would the pH be if 0.020 mol of HCl is added to 1.0 L of both of the following solutions. 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4) 0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5) What would the pH be if 0.050 mol of solid NaOH is added to each of the proceeding. To do this I must introduce the BAAM table!

Buffer capacity The pH of a buffered solution is determined by the ratio [A-]/[HA]. As long as this doesn’t change much the pH won’t change much. The more concentrated these two are the more H+ and OH- the solution will be able to absorb. Larger concentrations bigger buffer capacity.

Buffer Capacity Calculate the change in pH that occurs when 0.010 mol of HCl(g) is added to 1.0L of each of the following: 5.00 M HAc and 5.00 M NaAc 0.050 M HAc and 0.050 M NaAc Ka= 1.8x10-5

Summary Strong acid and base just stoichiometry. Determine Ka, use for 0 mL base Weak acid before equivalence point Stoichiometry first Then Henderson-Hasselbach Weak acid at equivalence point Kb Weak base after equivalence - leftover strong base.

Summary Determine Ka, use for 0 mL acid. Weak base before equivalence point. Stoichiometry first Then Henderson-Hasselbach Weak base at equivalence point Ka. Weak base after equivalence - leftover strong acid.

Selection of Indicators

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

Watch out Solubility is not the same as solubility product. Solubility product is an equilibrium constant. it doesn’t change except with temperature. Solubility is an equilibrium position for how much can dissolve. A common ion can change this.

Ksp Values for Some Salts at 25C Name Formula Ksp  Barium carbonate   BaCO3   2.6 x 10-9   Barium chromate   BaCrO4   1.2 x 10-10   Barium sulfate   BaSO4   1.1 x 10-10   Calcium carbonate   CaCO3   5.0 x 10-9   Calcium oxalate   CaC2O4   2.3 x 10-9   Calcium sulfate   CaSO4   7.1 x 10-5   Copper(I) iodide   CuI   1.3 x 10-12   Copper(II) iodate   Cu(IO3)2   6.9 x 10-8   Copper(II) sulfide   CuS   6.0 x 10-37   Iron(II) hydroxide   Fe(OH)2   4.9 x 10-17   Iron(II) sulfide   FeS   6.0 x 10-19   Iron(III) hydroxide   Fe(OH)3   2.6 x 10-39   Lead(II) bromide   PbBr2   6.6 x 10-6   Lead(II) chloride   PbCl2   1.2 x 10-5   Lead(II) iodate   Pb(IO3)2   3.7 x 10-13   Lead(II) iodide   PbI2   8.5 x 10-9   Lead(II) sulfate   PbSO4   1.8 x 10-8  Name Formula Ksp  Lead(II) bromide   PbBr2   6.6 x 10-6   Lead(II) chloride   PbCl2   1.2 x 10-5   Lead(II) iodate   Pb(IO3)2   3.7 x 10-13   Lead(II) iodide   PbI2   8.5 x 10-9   Lead(II) sulfate   PbSO4   1.8 x 10-8   Magnesium carbonate   MgCO3   6.8 x 10-6   Magnesium hydroxide   Mg(OH)2   5.6 x 10-12   Silver bromate   AgBrO3   5.3 x 10-5   Silver bromide   AgBr   5.4 x 10-13   Silver carbonate   Ag2CO3   8.5 x 10-12   Silver chloride   AgCl   1.8 x 10-10   Silver chromate   Ag2CrO4   1.1 x 10-12   Silver iodate   AgIO3   3.2 x 10-8   Silver iodide   AgI   8.5 x 10-17   Strontium carbonate   SrCO3   5.6 x 10-10   Strontium fluoride   SrF2   4.3 x 10-9   Strontium sulfate   SrSO4   3.4 x 10-7   Zinc sulfide   ZnS   2.0 x 10-25 

Solving Solubility Problems For the salt AgI at 25C, Ksp = 1.5 x 10-16 AgI(s)  Ag+(aq) + I-(aq) I C E O O +x +x x x 1.5 x 10-16 = x2 x = solubility of AgI in mol/L = 1.2 x 10-8 M

Solving Solubility Problems For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5 PbCl2(s)  Pb2+(aq) + 2Cl-(aq) I C E O O +x +2x x 2x 1.6 x 10-5 = (x)(2x)2 = 4x3 x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

Relative solubilities Ksp will only allow us to compare the solubility of solids the 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 pieces you have to do the math.

The Common Ion Effect When the salt with the anion of a weak acid is added to that acid, It reverses the dissociation of the acid. Lowers the percent dissociation of the acid. The same principle applies to salts with the cation of a weak base.. The calculations are the same as last chapter.

Solving Solubility with a Common Ion For the salt AgI at 25C, Ksp = 1.5 x 10-16 What is its solubility in 0.05 M NaI? AgI(s)  Ag+(aq) + I-(aq) I C E O 0.05 +x 0.05+x x 0.05+x 1.5 x 10-16 = (x)(0.05+x)  (x)(0.05) x = solubility of AgI in mol/L = 3.0 x 10-15 M

Precipitation and Qualitative Analysis

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 a acidic solution than in water. CaC2O4 Ca+2 + C2O4-2 H+ + C2O4-2 HC2O4- Reduces C2O4-2 in acidic solution.

Precipitation Ion Product, Q =[M+]a[Nm-]b 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 300.0 mL of 2.00 x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M)precipitate and if so, what is the concentration of the ions?

Selective Precipitations Used to separate mixtures of metal ions in solutions. Add anions that will only precipitate certain metals at a time. Used to purify mixtures. Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.

Selective Precipitation In Basic adding 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 first with insoluble chlorides (Ag, Pb, Ba) Then sulfides in Acid. Then sulfides in base. Then insoluble carbonate (Ca, Ba, Mg) Alkali metals and NH4+ remain in solution.

Complex Ions A Complex ion is a charged species composed of: 1. A metallic cation 2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

NH3, CN-, and H2O are Common Ligands

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 equilibrium. The complex ion will be the biggest ion in solution.

Coordination Number Coordination number refers to the number of ligands attached to the cation 2, 4, and 6 are the most common coordination numbers Coordination number Example(s) 2 Ag(NH3)2+ 4 CoCl42- Cu(NH3)42+ 6 Co(H2O)62+ Ni(NH3)62+

Complex Ions and Solubility AgCl(s)  Ag+ + Cl- Ksp = 1.6 x 10-10 Ag+ + NH3  Ag(NH3)+ K1 = 1.6 x 10-10 Ag(NH3)+ NH3  Ag(NH3)2+ K2 = 1.6 x 10-10 AgCl + 2NH3  Ag(NH3)2+ + Cl- K = KspK1K2