Chemistry 102(01) Spring 2008 Instructor: Dr. Upali Siriwardane e-mail: firstname.lastname@example.org 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, 2008Chapters 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?
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 = 1.8 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. Common ion effect 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)
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 [A-] [HA] [HA] [A-] 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 )
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 BuffersHenderson HesselbachEquation [ACID] pH = pKa - log --------- [BASE] [BASE] pH = pKa + log --------- [ACID]
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
Titrations ofAcids and Bases • Titration • Analyte • Titrant analyte + titrant => products
Indicators Acid-base indicators are highly colored weak acids or bases. HIn In- + H+ color 1color 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
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 Endpoint point where there is a physical change, such as color change, with theindicator Equivalence Point # moles titrant = # moles analyte #molestitrant=(V M)titrant #molesanalyte=(V M)analyte
Acid-Base Indicator Behavior acid color shows when [H3O+][In-] 1 = [H3O+] = Ka [HIn] 10 [In-] 1 £ [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.
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
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 VaMa > VbMbthus (VaMa) - (VbMb) [H3O+] = (Va + Vb) ((35.00mL)(0.1000M)) - ((15.00mL)(0.1000M)) = (35.00 + 15.00)mL = 4.00010-2 M pH = 1.3979
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 VaMa = VbMb , 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 VbMb > VaMa , post equvalence point (VbMb) - (VaMa) [OH-] = (Va + Vb) ((50.00mL)(0.1000M)) - ((35.00mL)(0.1000M)) = (35.00 + 50.00)mL = 1.76510-2 M pOH = 1.7533 pH = 14.00 - 1.7533 = 12.25
Titration of Strong Acid with Strong Base equivalence point x
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
Nitric Acid 2 NO2(g) + H2O(l)=> HNO3(aq) + HNO2(aq)
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: AxByxA+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 Ag2SO42 Ag+ + SO4-2 Ksp = [Ag+]2[SO4-2]
Dissolving Slightly Soluble Salts Using Acids Insoluble salts containing anions of Bronsted-Lowry bases can be dissolved in solutions of low pH