Chemistry 102(01) Spring 2008 Instructor: Dr. Upali Siriwardane e-mail: upali@chem.latech.edu Office: CTH 311 Phone 257-4941 Office Hours: M,W 8:00-9:00 & 11:00-12:00 am;                       Tu,Th,F  10:00 - 12:00 a.m.:  Test Dates: March 19,  2008 (Test 1): Chapter 13 April 16, 2008 (Test 2): Chapters 14 & 15 May 12, 2008 Chapters 16 17 & 18 Comprehensive Final Exam: May 14, 2008:Chapters 13, 14, 15, 16, 17 and 18

Chapter 17. Aditional Aqueous Equilibria 17.1 Buffer Solutions 17.2 Acid-Base Titrations 17.3 Acid Rain 17.4 Solubility Equilibria and the Solubility Product Constant, Ksp 17.5 Factors Affecting Solubility / Precipitation: Will It Occur?

CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) Hydrolysis Reaction of a basic anion with water is an ordinary Brønsted-Lowry acid-base reaction. CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq) This type of reaction is given a special name. Hydrolysis The reaction of an anion with water to produce the conjugate acid and OH-. The reaction of a cation with water to produce the conjugate base and H3O+.

How do you calculate pH of a salt solution? Find out the pH, acidic or basic? If acidic it should be a salt of weak base If basic it should be a salt of weak acid if acidic calculate Ka from Ka= Kw/Kb if basic calculate Kb from Kb= Kw/Ka Do a calculation similar to pH of a weak acid or base

What is the pH of 0. 5 M NH4Cl salt solution. (NH 3; Kb = 1 What is the pH of 0.5 M NH4Cl salt solution? (NH 3; Kb = 1.8 x 10-5) Find out the pH, acidic if acidic calculate Ka from Ka= Kw/Kb Ka= Kw/Kb = 1 x 10-14 /1.8 x 10-5) Ka= 5.56. X 10-10 Do a calculation similar to pH of a weak acid

Continued NH4+ + H2O <==> H 3+O + NH3 [NH4+] [H3+O ] [NH3 ] Ini. Con. 0.5 M 0.0 M 0.00 M Eq. Con. 0.5 - x x x [H 3+O ] [NH3 ] Ka(NH4+) = -------------------- = [NH 4+] x2 ---------------- ; appro.:0.5 - x . 0.5 (0.5 - x)

Continued x2 Ka(NH4+) = ----------- = 5.56 x 10 -10 0. 5 x2 = 5.56 x 10 -10 x 0.5 = 2.78 x 10 -10 x= sqrt 2.78 x 10 -10 = 1.66 x 10-5 [H+ ] = x = 1.66 x 10-5 M pH = -log [H+ ] = - log 1.66 x 10-5 pH = 4.77 pH of 0.5 M NH4Cl solution is 4.77 (acidic)

Common Ion Effect This is an example of Le Châtelier’s principle. The shift in equilibrium caused by the addition of an ion formed from the solute. Common ion An ion that is produced by more than one solute in an equilibrium system. Adding the salt of a weak acid to a solution of weak acid is an example of this.

Common Ion Effect Weak acid and salt solutions E.g. HC2H3O2 and NaC2H3O2 Weak base and salt solutions E.g. NH3 and NH4Cl. H2O + C2H3O2- <==> OH- + HC2H3O2 (common ion) H2O + NH4+ <==> H3+O + NH3 (common ion)

HA(aq) + H2O(l) H3O+(aq) + A-(aq) Buffers Solutions that resist pH change when small amounts of acid or base are added. Two types Mixture of weak acid and its salt Mixture of weak base and its salt HA(aq) + H2O(l) H3O+(aq) + A-(aq) Add OH- Add H3O+ shift to right shift to left Based on the common ion effect.

Buffers The pH of a buffer does not depend on the absolute amount of the conjugate acid-base pair. It is based on the ratio of the two. Henderson-Hasselbalch equation. Easily derived from the Ka or Kb expression. Starting with an acid pH = pKa + log Starting with a base pH = 14 - ( pKb + log ) [A-] [HA] [HA] [A-]

Henderson-Hasselbalch Equation HA(aq) + H2O(l) H3O+(aq) + A-(aq) [H3O+] [A-] Ka = ---------------- [HA] [H3O+] = Ka  ([HA]/[A-]) pH = pKa + log([A-]/[HA]) when the [A-] = [HA] pH = pKa

Calcualtion of pH of Buffers Henderson Hesselbach Equation [ACID] pH = pKa - log --------- [BASE] pH = pKa + log ---------

Buffers and blood Control of blood pH Oxygen is transported primarily by hemoglobin in the red blood cells. CO2 is transported both in plasma and the red blood cells. CO2 (aq) + H2O H2CO3 (aq) The bicarbonate buffer is essential for controlling blood pH H+(aq) + HCO3-(aq)

Buffer Capacity Refers to the ability of the buffer to retard changes in pH when small amounts of acid or base are added The ratio of [A-]/[HA] determines the pH of the buffer whereas the magnitude of [A-] and [HA] determine the buffer capacity

Adding an Acid or Base to a Buffer

Buffer Systems

Titrations of Acids and Bases Analyte Titrant analyte + titrant => products

Indicators Acid-base indicators are highly colored weak acids or bases. HIn In- + H+ color 1 color 2 They may have more than one color transition. Example. Thymol blue Red - Yellow - Blue One of the forms may be colorless - phenolphthalein (colorless to pink)

Acid-Base Indicator Weak acid that changes color with changes in pH HIn + H2O  H3O+ + In- acid base color color [H3O+][In-] Ka = [HIn] They may have more than one color transition. Example. Thymol blue Red - Yellow - Blue

What is an Indicator? Indicator is an weak acid with different Ka, colors to the acid and its conjugate base. E.g. phenolphthalein HIn <===> H+ + In- colorless pink Acidic colorless Basic pink

Selection of an indicator for a titration a) strong acid/strong base b) weak acid/strong base c) strong acid/weak base d) weak acid/weak base Calculate the pH of the solution at he equivalence point or end point

pH and Color of Indicators

Red Cabbage as Indicator

Indicator examples Acid-base indicators are weak acids that undergo a color change at a known pH. pH phenolphthalein

Titration Apparatus Buret delivering base to a flask containing an acid. The pink color in the flask is due to the phenolphthalein indicator.

Endpoint vs. Equivalence Point point where there is a physical change, such as color change, with the indicator Equivalence Point # moles titrant = # moles analyte #molestitrant=(V  M)titrant #molesanalyte=(V  M)analyte

Acid-Base Indicator Behavior acid color shows when [In-] 1 £ [HIn] 10 [H3O+][In-] 1 =  [H3O+] = Ka [HIn] 10 base color shows when [H3O+][In-] = 10  [H3O+] = Ka [HIn] [In-] 1 ³ [HIn] 10

Indicator pH Range acid color shows when pH + 1 = pKa and base color shows when pH - 1 = pKa Color change range is pKa = pH  1 or pH = pKa  1

Titration curves Acid-base titration curve A plot of the pH against the amount of acid or base added during a titration. Plots of this type are useful for visualizing a titration. It also can be used to show where an indicator undergoes its color change.

Indicator and Titration Curve 0.1000 M HCl vs 0.1000 M NaOH

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 0.00 mL of NaOH added, initial point [H3O+] = CHCl = 0.1000 M pH = 1.0000

((35.00mL)  (0.1000M)) - ((15.00mL)  (0.1000M)) EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 15.00 mL of NaOH added Va  Ma > Vb  Mb thus (Va  Ma) - (Vb  Mb) [H3O+] = (Va + Vb) ((35.00mL)  (0.1000M)) - ((15.00mL)  (0.1000M)) = (35.00 + 15.00)mL = 4.000  10-2 M pH = 1.3979

at equivalence point of a strong acid - strong base titration EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 35.00 mL of NaOH added Va  Ma = Vb  Mb , equivalence point at equivalence point of a strong acid - strong base titration pH º 7.0000

EXAMPLE: Derive the titration curve for the titration of 35.00 mL of 0.1000 M HCl with 0.00, 15.00, 35.00, and 50.00 mL of 0.1000 M NaOH. at 50.00 mL of NaOH added Vb  Mb > Va  Ma , post equvalence point (Vb  Mb) - (Va  Ma) [OH-] = (Va + Vb) ((50.00mL)  (0.1000M)) - ((35.00mL)  (0.1000M)) = (35.00 + 50.00)mL = 1.765  10-2 M pOH = 1.7533 pH = 14.00 - 1.7533 = 12.25

Titration of Strong Acid with Strong Base equivalence point x

Titration of Weak Acid with Strong Base

Effect of Acid Strength on Titration Curve

Titration of Weak Base with Strong Acid

Titration of Diprotic Weak Acid with Strong Base

pH range of Indicators litmus (5.0-8.0) bromothymole blue (6.0-7.6) methyl red (4.8-6.0) thymol blue (8.0-9.6) phenolphthalein (8.2-10.0) thymolphthalein (9.4-10.6)

Acid Rain acid rain is defined as rain with a pH < 5.6 pH = 5.6 for rain in equilibrium with atmospheric carbon dioxide

Sulfuric Acid from Sulfur burning SO2 S + O2 => SO2 SO3 2 SO2 + O2 => 2 SO3 Sulfuric Acid SO3 + H2O => H2SO4

2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq) Nitric Acid 2 NO2(g) + H2O(l) => HNO3(aq) + HNO2(aq)

How Acid Precipitation Forms

Acid Precipitation in U.S.

Solubility Product solubility-product - the product of the solubilities solubility-product constant => Ksp constant that is equal to the solubilities of the ions produced when a substance dissolves

Solubility Product Constant In General: AxBy  xA+y + yB-x [A+y]x [B-x]y K = [AxBy] [AxBy] K = Ksp = [A+y]x [B-x]y Ksp = [A+y]x [B-x]y For silver sulfate Ag2SO4  2 Ag+ + SO4-2 Ksp = [Ag+]2[SO4-2]

Solubility Product Constant Values

Dissolving Slightly Soluble Salts Using Acids Insoluble salts containing anions of Bronsted-Lowry bases can be dissolved in solutions of low pH

Calcium Carbonate Dissolved in Acid Limestone Dissolving in Ground Water CaCO3(S) + H2O + CO2 => Ca+2(aq) + 2 HCO3-(aq) Stalactite and Stalagmite Formation Ca+2(aq) + 2 HCO3-(aq) => CaCO3(S) + H2O + CO2

The Common Ion Effect common ion second source which is completely dissociated In the presence of a second source of the ion, there will be less dissolved than in its absence common ion effect a salt will be less soluble if one of its constitutent ions is already present in the solution

(x)(x) = Ksp = [Ag+][Cl-] = 1.82  10-10M2 EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl? AgCl  Ag+ + Cl- Ksp = [Ag+][Cl-] = 1.82  10-10M2 let x = molar solubility = [Ag+] = [Cl-] (x)(x) = Ksp = [Ag+][Cl-] = 1.82  10-10M2 x = 1.35  10-5M

EXAMPLE: What is the molar solubility of AgCl in pure water and in 1 EXAMPLE: What is the molar solubility of AgCl in pure water and in 1.0 M NaCl? AgCl  Ag+ + Cl- Ksp = [Ag+][Cl-] = 1.82  10-10M2 let x = molar solubility = [Ag+] [Cl-] = 1.0 M Ksp = [Ag+][Cl-] = (x)(1.0M) = 1.82  10-10M2 x = 1.82  10-10M

Formation of Complexes ligand - Lewis base complexes - product of Lewis acid-base reaction Ag+(aq) + 2 NH3(aq)  [Ag(NH3)2(aq)]+ Ag+(aq) + Cl-(aq)  AgCl(s) AgCl(s) + 2 NH3(aq)  [Ag(NH3)2(aq)]+ + Cl-(aq)

Sodium Thiosulfate Dissolves Silver Bromide

Formation Constants

Amphoterism

Reactant Quotient, Q Reactant Quotient, Q ion product of the initial concentration same form as solubility product constant Q < Ksp - no precipitate forms an unsaturate solution Q > Ksp - precipitate may form to restore condition of saturated solution Q = Ksp - no precipitate forms, saturated

Will Precipitation Occur?

magnesium ammonium phsphate Kidney Stones Kidney stones are normally composed of: calcium oxalate calcium phosphate magnesium ammonium phsphate

Calcium Oxalate Precipitate formed from calcium ions from food rich in calcium, dairy products, and oxalate ions from fruits and vegetables Ca+2 + C2O4-2  CaC2O4

PRECIPITATION REACTIONS

Analysis of Silver Group These salts formed are insoluble, they do dissolve to some SLIGHT extent. AgCl(s) Ag+(aq) + Cl-(aq) When equilibrium has been established, no more AgCl dissolves and the solution is SATURATED.

Analysis of Silver Group AgCl(s) Ag+(aq) + Cl-(aq) When solution is SATURATED, expt. shows that [Ag+] = 1.67 x 10-5 M. (SOLUBILITY) of AgCl. What is [Cl-]?

Analysis of Silver Group AgCl(s) Ag+(aq) + Cl-(aq) Saturated solution has [Ag+] = [Cl-] = 1.67 x 10-5 M Use this to calculate Ksp Ksp = [Ag+] [Cl-] = (1.67 x 10-5)(1.67 x 10-5) = 2.79 x 10-10

Ksp = solubility product Because this is the product of “solubilities”, we call it Ksp = solubility product constant See Table in the Text

Lead(II) Chloride PbCl2(s) Pb2+(aq) + 2 Cl-(aq) Ksp = 1.9 x 10-5

Solubility of Lead(II) Iodide Consider PbI2 dissolving in water PbI2(s) Pb2+(aq) + 2 I-(aq) Calculate Ksp if solubility = 0.00130 M Solution 1. Solubility = [Pb2+] = 1.30 x 10-3 M [I-] = 2 x [Pb2+] = 2.60 x 10-3 M 2. Ksp = [Pb2+] [I-]2 = [Pb2+] {2 • [Pb2+]}2 = 4 [Pb2+]3 = 4 (solubility)3 Ksp = 4 (1.30 x 10-3)3 = 8.8 x 10-9

Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 If [Hg22+] = 0.010 M, what [Cl-] is req’d to just begin the precipitation of Hg2Cl2? (maximum [Cl-] in 0.010 M Hg22+ without forming Hg2Cl2?)

Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Recognize that Ksp = product of maximum ion concs. Precip. begins when product of ion concs. EXCEEDS the Ksp.

Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 = [Hg22+] [Cl-]2 Solution [Cl-] that can exist when [Hg22+] = 0.010 M, If this conc. of Cl- is just exceeded, Hg2Cl2 begins to precipitate.

Precipitating an Insoluble Salt Hg2Cl2(s) Hg22+(aq) + 2 Cl-(aq) Ksp = 1.1 x 10-18 Now raise [Cl-] to 1.0 M. What is the value of [Hg22+] at this point? Solution [Hg22+] = Ksp / [Cl-]2 = Ksp / (1.0)2 = 1.1 x 10-18 M The concentration of Hg22+ has been reduced by 1016 !

Separating Metal Ions Cu2+, Ag+, Pb2+ Cl- Insoluble PbCl2 AgCl Soluble CuCl2 Heat Soluble PbCl2 Insoluble AgCl Ksp Values AgCl 1.8 x 10-10 PbCl2 1.7 x 10-5 PbCrO4 1.8 x 10-14 CrO4-2 Insoluble PbCrO4

Separating Salts by Differences in Ksp A solution contains 0.020 M Ag+ and Pb2+. Add CrO42- to precipitate Ag2CrO4 (red) and PbCrO4 (yellow). Which precipitates first? Ksp for Ag2CrO4 = 9.0 x 10-12 Ksp for PbCrO4 = 1.8 x 10-14 Solution The substance whose Ksp is first exceeded precipitates first. The ion requiring the smaller amount of CrO42- ppts. first.

Separating Salts by Differences in Ksp Solution Calculate [CrO42-] required by each ion. [CrO42-] to ppt. PbCrO4 = Ksp / [Pb2+] = 1.8 x 10-14 / 0.020 = 9.0 x 10-13 M [CrO42-] to ppt. Ag2CrO4 = Ksp / [Ag+]2 = 9.0 x 10-12 / (0.020)2 = 2.3 x 10-8 M PbCrO4 precipitates first.

Separating Salts by Differences in Ksp How much Pb2+ remains in solution when Ag+ begins to precipitate? Solution We know that [CrO42-] = 2.3 x 10-8 M to begin to ppt. Ag2CrO4. What is the Pb2+ conc. at this point? [Pb2+] = Ksp / [CrO42-] = 1.8 x 10-14 / 2.3 x 10-8 M = 7.8 x 10-7 M Lead ion has dropped from 0.020 M to < 10-6 M

Common Ion Effect Adding an Ion “Common” to an Equilibrium PbCl2(s) Pb2+(aq) + 2Cl-(aq) NaCl Na+(aq) + Cl- (aq)

The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) Solution a) Solubility in pure water = [Ba2+] = [SO42-] = x Ksp = [Ba2+] [SO42-] = x2 x = (Ksp)1/2 = 1.1 x 10-5 M

The Common Ion Effect BaSO4(s) Ba2+(aq) + SO42-(aq) Ksp = 1.1 x 10-10 Solution b) Now dissolve BaSO4 in water already containing 0.010 M Ba2+. Which way will the “common ion” shift the equilibrium? ___ Will solubility of BaSO4 be less than or greater than in pure water?___

The Common Ion Effect BaSO4(s) Ba2+(aq) + SO42-(aq) Solution initial change equilib.

The Common Ion Effect Calculate the solubility of BaSO4 in (a) pure water and (b) in 0.010 M Ba(NO3)2. Ksp for BaSO4 = 1.1 x 10-10 BaSO4(s) Ba2+(aq) + SO42-(aq) Solution [Ba2+] [SO42-] initial 0.010 0 change + y + y equilib. 0.010 + y y

The Common Ion Effect Ksp = [Ba2+] [SO42-] = (0.010 + y) (y) Because y << x (1.1 x 10-5 M) 0.010 + y  0.010. Therefore, Ksp = 1.1 x 10-10 = (0.010)(y) y = 1.1 x 10-8 M = solubility in presence of added Ba2+ ion. Le Chatelier’s Principle is followed!