Applications of Aqueous Equilibria

Reaction of Weak Bases with Water The generic reaction for a base reacting with water, producing its conjugate acid and hydroxide ion: B + H 2 O  BH + + OH - (Yes, all weak bases do this – DO NOT endeavor to make this complicated!)

Reaction of Weak Bases with Water The base reacts with water, producing its conjugate acid and hydroxide ion: CH 3 NH 2 + H 2 O  CH 3 NH OH - K b = 4.38 x 10 -4

K b for Some Common Weak Bases BaseFormula Conjugate Acid KbKb Ammonia NH 3 NH x Methylamine CH 3 NH 2 CH 3 NH x Ethylamine C 2 H 5 NH 2 C 2 H 5 NH x Diethylamine (C 2 H 5 ) 2 NH (C 2 H 5 ) 2 NH x Triethylamine (C 2 H 5 ) 3 N (C 2 H 5 ) 3 NH x Hydroxylamine HONH 2 HONH x HydrazineH 2 NNH 2 H 2 NNH x Aniline C 6 H 5 NH 2 C 6 H 5 NH x Pyridine C 5 H 5 N C 5 H 5 NH x Many students struggle with identifying weak bases and their conjugate acids.What patterns do you see that may help you?

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 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 C 6 H 5 COOH NaC 6 H 5 COO – Sodium benzoate Acetic CH 3 COOH NaH 3 COO – Sodium acetate Carbonic H 2 CO 3 NaHCO 3 - Sodium bicarbonate Propanoic HC 3 H 5 O 2 NaC 3 H 5 O 2 - Sodium propanoate Hydrocyanic HCN KCN - potassium cyanide The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)

Base/Salt Buffering Pairs The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO 3 ) Base Formula of the base Example of a salt of the weak acid Ammonia NH 3 NH 4 Cl - ammonium chloride Methylamine CH 3 NH 2 CH 3 NH 2 Cl – methylammonium chloride Ethylamine C 2 H 5 NH 2 C 2 H 5 NH 3 NO 3 - ethylammonium nitrate Aniline C 6 H 5 NH 2 C 6 H 5 NH 3 Cl – aniline hydrochloride Pyridine C 5 H 5 N C 5 H 5 NHCl – pyridine hydrochloride

Common Ion Effect A solution contains HF (ka = 7.2 x ) and its salt (NaF). What effect does the presence of the dissolved salt have on the dissociation equilibrium? NaF(s) Na + (aq) + F - (aq) H 2 O(l)

Step 1: Identify the major species HF, Na +, F - and H 2 O HF(aq) H + (aq) + F - (aq) HF(aq) H + (aq) + F - (aq) + F - (aq) + Na + A solution contains HF (ka = 7.2 x ) and its salt (NaF).

Give this a try! A buffered solution contains 0.50 M acetic acid (HC 2 H 3 O 2, Ka = 1.8 x ) and 0.50 M sodium acetate (NaC 2 H 3 O 2 ). Calculate the pH of this solution. pH = 4.74 Now….try this! Calculate the change in pH that occurs when 0.01 M solid NaOH is added to 1.0 L of the buffered solution above. Compare this pH change with that which occurs when 0.01 M NaOH is added to 1.0 L of water.

Titration of an Unbuffered Solution A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH

Titration of a Buffered Solution A solution that is 0.10 M CH 3 COOH and 0.10 M NaCH 3 COO is titrated with 0.10 M NaOH

Comparing Results Buffered Unbuffered

Unbuffered Buffered  In what ways are the graphs different?  In what ways are the graphs similar?

Henderson-Hasselbalch Equation

Try This! Calculate the pH of a solution containing 0.75 M lactic acid (Ka = 1.4 x ) and 0.25 M sodium lactate. Lactic acid (HC 3 H 5 O 3 ). pH = 3.38

Weak Acid/Strong Base Titration A solution that is 0.10 M CH 3 COOH is titrated with 0.10 M NaOH Endpoint is above pH 7

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

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 NH 3 Endpoint is below pH 7

Selection of Indicators

Some Acid-Base Indicators Indicator pH Range in which Color Change Occurs Color Change as pH Increases Crystal violet Thymol blue Orange IV Methyl orange Bromcresol green Methyl red Chlorophenol red Bromthymol blue Phenol red Neutral red Thymol blue Phenolphthalein Thymolphthalein Alizarin yellow Indigo carmine yellow to blue red to yellow red to yellow red to yellow yellow to blue red to yellow yellow to red yellow to blue yellow to red red to amber yellow to blue colourless to pink colourless to blue yellow to blue blue to yellow

pH Indicators and their ranges

K sp Values for Some Salts at 25  C NameFormulaK sp Barium carbonate BaCO x Barium chromate BaCrO x Barium sulfate BaSO x Calcium carbonate CaCO x Calcium oxalate CaC 2 O x Calcium sulfate CaSO x Copper(I) iodide Cu I 1.3 x Copper(II) iodate Cu( I O 3 ) x Copper(II) sulfide CuS 6.0 x Iron(II) hydroxide Fe(OH) x Iron(II) sulfide FeS 6.0 x Iron(III) hydroxide Fe(OH) x Lead(II) bromide PbBr x Lead(II) chloride PbCl x Lead(II) iodate Pb( I O 3 ) x Lead(II) iodide Pb I x Lead(II) sulfate PbSO x NameFormulaK sp Lead(II) bromide PbBr x Lead(II) chloride PbCl x Lead(II) iodate Pb( I O 3 ) x Lead(II) iodide Pb I x Lead(II) sulfate PbSO x Magnesium carbonate MgCO x Magnesium hydroxide Mg(OH) x Silver bromate AgBrO x Silver bromide AgBr 5.4 x Silver carbonate Ag 2 CO x Silver chloride AgCl 1.8 x Silver chromate Ag 2 CrO x Silver iodate Ag I O x Silver iodide Ag I 8.5 x Strontium carbonate SrCO x Strontium fluoride SrF x Strontium sulfate SrSO x Zinc sulfide ZnS 2.0 x

Solving Solubility Problems For the salt AgI at 25  C, K sp = 1.5 x AgI(s)  Ag + (aq) + I - (aq) I C E O O +x x x 1.5 x = x 2 x = solubility of AgI in mol/L = 1.2 x M

Solving Solubility Problems For the salt PbCl 2 at 25  C, K sp = 1.6 x PbCl 2 (s)  Pb 2+ (aq) + 2Cl - (aq) I C E O O +x +2x x 2x 1.6 x = (x)(2x) 2 = 4x 3 x = solubility of PbCl 2 in mol/L = 1.6 x M

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

Precipitation and Qualitative Analysis

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

NH 3, CN -, and H 2 O are Common Ligands

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 numberExample(s) 2Ag(NH 3 ) 2 + 4CoCl 4 2- Cu(NH 3 ) Co(H 2 O) 6 2+ Ni(NH 3 ) 6 2+

Complex Ions and Solubility AgCl(s)  Ag + + Cl - K sp = 1.6 x Ag + + NH 3  Ag(NH 3 ) + K 1 = 2.1 x 10 3 Ag(NH 3 ) + NH 3  Ag(NH 3 ) 2 + K 2 = 8.2 x 10 3 AgCl + 2NH 3  Ag(NH 3 ) Cl - K = K sp  K 1  K 2