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Ch 18 buffers
 

Ch 18 buffers

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    Ch 18 buffers Ch 18 buffers Presentation Transcript

    • The Color Change of the Indicator Bromthymol BlueFig. 19.5
    • Acid-Base Indicators Usually dyes that are weak acids and display different colours in protonated/deprotonated forms. HIn(aq.) H+ (aq.) +In- (aq.) Ka = [H ][In ] + - [ HIn]In general we seek an indicator whose transition range(±1pH unit from the indicator pKa) overlaps the steepest partof the titration curve as closely as possible
    • pKa H Cl Cl + H -7 O OCH3 C CH3 C + H 4.74 O H O H H H C H H C + H ~50 H HStrong acids give weak bases.Weak acids give strong bases.
    • H Cl Cl + H basicity O O CH3 C CH3 C + H O H Oacidity H H H C H H C + H H H Strong acids give weak bases. Weak acids give strong bases.
    • Colors and Approximate pH Range ofSome Common Acid-Base Indicators
    • Which molecule is the stronger acid, ethanolor acetic acid? Ka 10-16ethanol 10-4.74 more stable anionacetic acid because ofthe stronger O resonance andacid CH3 C inductive effects O
    • Predict whether trifluoroacetic acid will be astronger or weaker acid than acetic acid. Ka H O H O H C C H C C + H 10-4.74 H OH H O acetic acid F O F O F C C F C C + H 10-0.23 F OH F O Fluorine is more electro-trifluoroacetic acid negative than hydrogen. Anion more acidic acid is more stable.
    • It is a general principle that the more stablethe anion the more acidic is the acid.The principle is also successful across a row ofthe periodic table.increasingelectronegativity pKa H H H C H H C + H 48 H H H H H N H H N + H 38 H H O H O + H 15.7
    • The relative strengths of acids and bases isgiven by the Ka of the acid. Ka H Cl Cl + H 107 O OCH3 C CH3 C + H 10-4.74 O H O CH3 CH3CH3 N H CH3 N + H 10-9.81 CH3 CH3
    • Identify the strongest acid. Ka H Cl Cl + H 107 O OCH3 C CH3 C + H 10-4.74 O H O CH3 CH3CH3 N H CH3 N + H 10-9.81 CH3 CH3
    • Identify the weakest acid. Ka H Cl Cl + H 107 O OCH3 C CH3 C + H 10-4.74 O H O CH3 CH3CH3 N H CH3 N + H 10-9.81 CH3 CH3
    • Identify the weakest base. Ka H Cl Cl + H 107 O OCH3 C CH3 C + H 10-4.74 O H O CH3 CH3CH3 N H CH3 N + H 10-9.81 CH3 CH3
    • Identify the strongest base. Ka H Cl Cl + H 107 O OCH3 C CH3 C + H 10-4.74 O H O CH3 CH3CH3 N H CH3 N + H 10-9.81 CH3 CH3
    • In chemistry, particularly biology, a largenumber of compounds are acids and bases.HO NH2 HO NH3 + HHO HO dopamine HO CO2H HO CO2 + H CO2H CO2H CO2H CO2H citric acidBiological fluids are often buffered (constant pH) anit is useful to know the predominant species presentat a given pH.
    • Consider acetic acid with a Ka = 10-4.74 at pH = 4.74 [CH3CO2H] = [CH3CO2] H at lower pH, more acidic than 4.74, acetic acid is the major species present
    • Consider acetic acid with a Ka = 10-4.74 at pH = 4.74 [CH3CO2H] > [CH3CO2] H at lower pH, more acidic than 4.74, acetic acid is the [CH3CO2H] major species present
    • Consider acetic acid with a Ka = 10-4.74 at pH = 4.74 [CH3CO2H] < [CH3CO2] OH at lower pH, more acidic than 4.74, acetic acid is the [CH3CO2H] major species present at higher pH, less acidic than 4.74, acetate ion is the major species present [CH3CO2]
    • If acetic acid is introduced into the bloodwhat will be the predominant species present?Will it be acetate ion or acetic acid? The pH of blood is maintained at ∼ 7.4 If the pH of blood was 4.74 then the acetate ion would be equal to the acetic acid ion concentration. O O [HOCCH3] = [ OCCH3] at pH = 4.74 If the pH is raised to 7.4 will the concentration of acetate ion increase or decrease?
    • If acetic acid is introduced into the bloodwhat will be the predominant species present?Will it be acetate ion or acetic acid? pH 4.74 ⇒ 7.4 O H O O [HOCCH3] = [ OCCH3] If the pH is raised to 7.4 (more basic) will the concentration of acetate ion increase or decrease?
    • If acetic acid is introduced into the bloodwhat will be the predominant species present?Will it be acetate ion or acetic acid? Acetate ion is the major species present if acetic acid is introduced into blood. O O [HOCCH3] = [ OCCH3] at pH = 4.74 If the pH is raised to 7.0 will the concentration of acetate ion increase or decrease?
    • If cocaine is introduced in the blood what willbe the major species present? KaCH3 H CH3 10-8.81 N N CO2CH3 CO2CH3 + H O O O O protonated form neutral form
    • If cocaine is introduced in the blood what willbe the major species present? Ka CH3 10-8.81 N at pH = 8.81 = CO2CH3 + H O O protonated form neutral form pH 8.81 ⇒ 7.4 (more acidic)
    • The Henderson-Hasselbalch Equation Take the equilibrium ionization of a weak acid: HA(aq) + H2O(aq) = H3O+(aq) + A-(aq) [H3O+] [A-] Ka = [HA]Solving for the hydronium ion concentration gives: [HA] [H3O+] = Ka x [A-]Taking the negative logarithm of both sides: -log[H3O +] = -log Ka - log ( ) [HA] [A-] ( ) pH = -log Ka - log [HA] [A-] Generalizing for any conjugate acid-base pair : [base] pH = -log Ka + log ( [acid] ) Henderson-Hasselbalch equation
    • H-A H + A p H = -log [H⊕ ] [A-] pH = pKa + log [HA] a useful concept: when [H-A] = [ A ]Biological fluids are often buffered (constant pH) anit is useful to know the predominant species presentat a given pH.
    • HENDERSON-HASSELBALCH EQUATION For acids:  [A − ]  When [A-] = [HA], pH = pK a + log   [HA]     pH = pKa [B] pKa appliesFor bases: pH = pK a + log [BH+ ] to this acid Kb → B + H2O ← BH+ + OH- Ka base acid acid baseDerivation: HA  H+ + A- [H+ ][A − ]  [H+ ][A − ]   [A − ]   [A − ] Ka = - log K a = − log   + = − log [H ] − log   pK a = pH − log   [HA]   [HA]  [HA]   [HA]         [A −]  pH =pK a + log  [HA]    
    • BUFFERS Mixture of an acid and its conjugate base.Buffer solution → resists change in pH when acids or bases are added or when dilution occurs.Mix:A moles of weak acid + B moles of conjugate baseFind:• moles of acid remains close to A, and• moles of base remains close to B⇒ Very little reaction HA  H+ + A- Le Chatelier’s principle
    • Why does a buffer resist change in pH when small? amounts of strong acid or bases is added? The acid or base is consumed by A- or HA respectivelyA buffer has a maximum capacity to resist change to pH. Buffer capacity, β: → Measure of how well solution resists change in pH when strong acid/base is added. dCb − dCa β= = dpH dpH Larger β ⇒ more resistance to pH change
    • How a Buffer Works Consider adding H3O+ or OH- to water and also to a buffer For 0.01 mol H3O+ to 1 L water: [H3O+] = 0.01 mol/1.0 L = 0.01 M pH = -log([H3O+]) = 2.0So, change in pH from pure water: ∆pH = 7.00 – 2.00 = 5.0For the H2CO3- / HCO3- system: pH of buffer = 7.38 Addition of 0.01 mol H3O+ changes pH to 7.46So change in pH from buffer: ∆pH = 7.46 – 7.38 = 0.08 !!!
    • How a Buffer Works Consider a buffer made from acetic acid and sodium acetate: CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq) [CH3COO-] [H3O+]Ka = or [CH3COOH] + [CH3COOH][H3O ] = Ka x [CH3COO-]
    • How a Buffer WorksLet’s consider a buffer made by placing 0.25 mol of acetic acid and0.25 mol of sodium acetate per liter of solution.What is the pH of the buffer?And what will be the pH of 100.00 mL of the buffer before andafter 1.00 mL of concentrated HCl (12.0 M) is added to the buffer?What will be the pH of 300.00 mL of pure water if the same acidis added? + [CH3COOH] -5 (0.25) [H3O ] = Ka x = 1.8 x 10 x = 1.8 x 10-5 [CH3COO-] (0.25) pH = -log[H3O+] = -log(1.8 x 10-5) = pH = 4.74 Before acid added!
    • How a Buffer WorksWhat is pH if added to pure water?1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H3O+Added to 300.00 mL of water :0.012 mol H3O+ = 0.0399 M H3O+ pH = -log(0.0399 M)301.00 mL soln. pH = 1.40 Without buffer!
    • How a Buffer WorksAfter acid is added to buffer:Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+Initial 0.250 ---- 0.250 0Change +0.012 ---- -0.012 0.012Equilibrium 0.262 ---- 0.238 0.012Solving for the quantity ionized:Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+Initial 0.262 ---- 0.238 0Change -x ---- +x +xEquilibrium 0.262 - x ---- 0.238 + x xAssuming: 0.262 - x = 0.262 & 0.238 + x = 0.238 [CH3COOH] +[H3O ] = Ka x =1.8 x 10-5 x (0.262) = 1.982 x 10-5 [CH3COO-] (0.238)pH = -log(1.982 x 10-5) = 5.000 - 0.297 = 4.70 After the acid is added!
    • How a Buffer WorksSuppose we add 1.0 mL of a concentrated base instead of an acid. Add1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see whatthe impact is: 1.00 mL x 12.0 mol OH-/1000mL = 0.012 mol OH-This will reduce the quantity of acid present and force the equilibriumto produce more hydronium ion to replace that neutralized by theaddition of the base!Conc. (M) CH3COOH(aq) + H2O(aq) CH3COO- + H3O+Initial 0.250 ---- 0.250 0Change - 0.012 ---- +0.012 +0.012Equilibrium 0.238 ---- 0.262 +0.012Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238 0.238 = 1.635 x 10-5[H3O+] = 1.8 x 10-5 x 0.262 pH = -log(1.635 x 10-5) = 5.000 - 0.214 = 4.79 After base is added!
    • How a Buffer WorksBy adding the 1.00mL base to 300.00 mL of pure water we would get ahydroxide ion concentration of: - 0.012 mol OH- = 3.99 x 10-5 M OH- [OH ] = 301.00 mLThe hydrogen ion concentration is: + Kw 1 x 10-14 [H3O ] = - = = 2.506 x 10-10 [OH ] 3.99 x 10-5 MThis calculates out to give a pH of:pH = -log(2.5 6 x 10-10) = 10.000 - 0.408 = 9.59 With 1.0 mL of the base in pure water!In summary: Buffer alone pH = 4.74 Buffer plus 1.0 mL base pH = 4.79 Base alone, pH = 9.59 Buffer plus 1.0 mL acid pH = 4.70 Acid alone, pH = 1.40
    • Problem:Calculate the pH of a solution containing 0.200 M NH3and 0.300 M NH4Cl given that the acid dissociationconstant for NH4+ is 5.7x10-10.NH3 + H2O  NH4+ + OH- pKa = 9.244 Kabase acid [B] pKa applies pH = pK a + log [BH+ ] to this acid (0.200) pH = 9.244 + log (0.300) pH = 9.07