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Chapter 9 Aqueous Solutions and Chemical Equilibria

Chapter 9 Aqueous Solutions and Chemical Equilibria. 1) Non electrolytes : Are substances that dissolve in water but do not produce any ions and do not conduct an electric current.

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Chapter 9 Aqueous Solutions and Chemical Equilibria

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  1. http:\\asadipour.kmu.ac.ir...36 slides

  2. Chapter 9Aqueous Solutions and Chemical Equilibria 1) Non electrolytes:Are substances that dissolve in water but do not produce any ions and do not conduct an electric current. 2) Electrolytes: Form ions when dissolved in water or other solvents and produce solutions that conduct electricity. 1) Strong Electrolytes: Ionize essentially completely. Strong conductor of electricity. 2) Weak Electrolytes: Ionize only partially. Poorer conductor than strong electrolyte. http:\\asadipour.kmu.ac.ir...36 slides

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  4. Acids and Bases: An acid is a proton donor and a base is a proton acceptor (Bronsted-Lowry concept). Conjugate Acids and Bases: A conjugate base is the species formed when an acid loses a proton. NH3 + H2O NH4+ + OH- base1 acid2conjugate acid1conjugate base2 http:\\asadipour.kmu.ac.ir...36 slides

  5. Amphiprotic Solvents: A solvent that can act either as an acid or as a base depending on the solute. Water is the classic example. http:\\asadipour.kmu.ac.ir...36 slides

  6. Three general types of solvents : 1) Protic solvent : amphiprotic solvents - possess both acidic and basic properties - undergoes self-ionization (=autoprotolysis) ex. Water, alcohols, acetic acid, ammonia ethylenediamine 2) Aprotic solvents - have no appreciable acidic or basic character - do not undergo autoprotolysis - ex. Benzene, carbon tetrachloride, pentane 3) Basic solvents - have basic properties but essentially no acidic tendencies - do not undergo autoprotolysis ex. ethers, esters, pyridine and amines 2H2O  H 3O+ + OH– 2C2H5OH  C2H5OH2+ + C2H5O– 2HOAc  H2OAc+ + OAc– 2NH3 NH4+ + NH2– http:\\asadipour.kmu.ac.ir...36 slides

  7. Strong acid: Reacts with water so completely that no undissociated solute molecules remain. Weak acid: Reacts incompletely with water to give solution that contain significant amounts of both the parent acid and its conjugate base. Strong base: Completely dissociated in water solution Weak base: Incomplete dissociation in water solution http:\\asadipour.kmu.ac.ir...36 slides

  8. Relative Strength http:\\asadipour.kmu.ac.ir...36 slides

  9. Strengths of Acids and Bases In a leveling solvent, several acids are completely dissociated and show the same strength. In a differentiating solvent, various acids dissociate to different degrees and have different strengths. http:\\asadipour.kmu.ac.ir...36 slides

  10. The tendency of a solvent to accept or donate protons determine the strength of a solute acid or base dissolved in it. H2O + HClO4→ H3O+ + ClO4- H2O + HCl → H3O+ + Cl- CH3COOH + HClO4  CH3COOH2+ + ClO4- CH3COOH + HCl CH3COOH2+ + Cl- base1 acid2 acid1 base2 http:\\asadipour.kmu.ac.ir...36 slides

  11. Chemical Equilibrium & Equilibrium Constants wW + xX yY + zZ Where, the capital letters represent the formulas of participating chemical species and the lowercase letters are the small whole numbers (# of moles) required to balance the equation. The equilibrium-constant expression is Where, the bracketed terms are molar concentration if the species is a dissolved solute or partial pressure (atm) if the species is a gas. If one of the species is a pure liquid, a pure solid, or the solvent in excess (dilute soln.), no term for this species appear in the equilibrium-constant expression. Equilibrium constant K is a temperature dependent quantity. http:\\asadipour.kmu.ac.ir...36 slides

  12. The Ion-Product Constant for Water Aqueous solutions contain small concentrations of hydronium and hydroxide ions as a consequence of the dissociation reaction. 2H2O H3O+ + OH- Equilibrium constants, The concentration of water in dilute aqueous solution is enormous when compared with the concentration of hydrogen and hydroxide ions. [H2O] can be taken as constant, K[H2O]2 = Kw = [H3O+][OH-] where the new constant is the ion-product constant for water. At 25oC, Kw 1.00 x 10-14 -logKw = - log[H3O]+ - log[OH-] pKw = pH + pOH = 14.00 http:\\asadipour.kmu.ac.ir...36 slides

  13. Temperature dependence of pH H+ + OH-  H2O +Q Body temperature = 37 oC Blood pH = 7.35 ~ 7.45 [HCl] in gastric juice = 0.1 ~ 0.02M if [H+] = 0.02M pH = 1.7 pOH = 13.6 –1.7 = 11.9 at 37 oC http:\\asadipour.kmu.ac.ir...36 slides

  14. Solubility-Products Ba(IO3)2(s) Ba2+(aq) + 2IO3-(aq) K[Ba(IO3)2(s)] = Ksp = [Ba2+][IO3-]2 Where, the new constant, Ksp, is called the solubility-product. The position of this equilibrium is independent of the amount of Ba(IO3)2 as long as some solid is present. Common Ion Affect The common-ion effect is responsible for the reduction in solubility of an ionic precipitate when a soluble compound combining one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. http:\\asadipour.kmu.ac.ir...36 slides

  15. Dissociation Constants for Acids and Bases When a weak acid or a weak base is dissolved in water, partial dissociation occurs, HNO2 + H2O H3O+ + NO2- , NH3 + H2O NH4+ + OH- , Where Ka and Kb are acid dissociation constant for nitrous acid and base dissociation constant for ammonia respectively. Dissociation Constants for Conjugate Acid/Base Pairs: NH3 + H2O NH4+ + OH- , NH4+ + H2O NH3 + H3O+ , Ka. Kb = [H3O+ ][OH- ] = Kw Kw = Ka. Kb http:\\asadipour.kmu.ac.ir...36 slides

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  18. Calculation of pH : strong acids and bases 1. Strong acid or base Cacid or base >> 10–6 M 2. Both 10–6 < Cacid or base < 10–8 M 3. Water Cacid or base << 10–8 M Calculated pH as a function of the concentration of a strong acid or strong base dissolved in water. http:\\asadipour.kmu.ac.ir...36 slides

  19. A buffered solution is one that resists change of pH on adding acids or bases, or on diluting it with solvent. The buffer consists of mixture of an acid and its conjugated base. HA  A– + H+ Ka = [A–][H+] / [HA] [H+] = Ka [HA] / [A–] – log [H+] = – log Ka [HA] / [A–] = – log Ka – log [HA] / [A–] = – log Ka + log [A–]/[HA] pH = pKa+ log [A–] / [HA] pKa is constant [A–]/[HA]  pH 100:1 or 1:100 pKa 2 10:1 or 1:10 pKa 1 1:1 pKa 0 Buffer solutions Henderson-Hasselbalch equation http:\\asadipour.kmu.ac.ir...36 slides

  20. preparation • 1) HA + A- B + HB+ • 2) HA + OH-B + H+ • 3) A- + H+HB+ + OH- http:\\asadipour.kmu.ac.ir...36 slides

  21. Ex. Tris buffer NH2 N+H3 C C HOH2C HOH2C + H+  CH2OH CH2OH HOH2C HOH2C Tris hydrochloride(FW=157.597) : BH+Cl– BH+ + Cl– BH+ =B + H+ Find the pH of a solution prepared by dissolving 12.43g of tris plus 4.67g of tris hydrochloride in 1.00 L of water. pKa = 8.075 Tris = B : 121.136 g/L = 1.00M Tris hydrochloride = BH+ : 157.597 g/L = 1.00 M 12.43 g/L = x M 4.67 g/L = y M x = 0.1026 M y = 0.0296 M pH = pKa+ log [B] / [BH+] = 8.075 + log (0.1026 / 0.0296) = 8.61 http:\\asadipour.kmu.ac.ir...36 slides

  22. Effect of dilution of a buffer solution Ex. 1) Calculate the pH of a buffer prepared by mixing 0.250 mol of acetic acid with 0.100 mol of sodium acetate and diluting to 1.00 L. pKa = – log 1.75×10–5 = 4.75 2) Calculate the pH when 10.0 ml of this buffer is diluted to 250 mL with water. HOAC H+ + OAC– Ka = [OAC–][H+] / [HOAC] [H+] = Ka [HOAC] / [OAC–] pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36 HOAC 0.250 M×10.0 ml = x M×250 ml x = 0.0100 M NaOAC 0.100 M ×10.0 ml = x M×250 ml x = 0.00400 M pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.00400M)/(0.0100M) = 4.36 http:\\asadipour.kmu.ac.ir...36 slides

  23. http:\\asadipour.kmu.ac.ir...36 slides Fig 9-4, p.255

  24. Effect of addition of acids or bases to buffer. Buffer HA A-+ H+ NaOH + HA A- HCl + A- HA pH of 100 ml of buffer consisting of 0.20M HA and 0.10M NaA pH = pKa+ log [NaA] / [HA] = pKa+ log (0.10/0.20) = pKa+ 0.30 If 0.01 mole of strong base is added HA =0.10M. A- =0.2M pH = pKa+ log [NaA] / [HA] = pKa+ log (0.20/0.10) = pKa– 0.30 pH = 0.60 Not very large http:\\asadipour.kmu.ac.ir...36 slides

  25. Effect of addition of acids or bases to buffer. Ex. 1) Calculate the pH of a buffer prepared by mixing 0.250 mol of acetic acid with 0.100 mol of sodium acetate and diluting to 1.00 L. pKa = – log 1.75×10–5 = 4.75 2) Calculate the pH when 10.0 ml of 0.1M NaOH is added. HOAC H+ + OAC– Ka = [OAC–][H+] / [HOAC] [H+] = Ka [HOAC] / [OAC–] pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36 HOAC N1V1-N2V2= 1000 ×0.250 -10 ×0.1=0.240 mole NaOAC N1V1+N2V2= 1000 ×0.100 +10 ×0.1=0.110 mole pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.110)/(0.240) = 4.41 Not very different http:\\asadipour.kmu.ac.ir...36 slides

  26. Buffer capacity The relative ability of a buffer solution to resist pH change upon addition of an acid or a base.  = dCb / dpH = – dCa / dpH A buffer is most effective in resisting changes in pH when pH= pKa (i.e., when [HA] = [A–]). Choose a buffer for an experiment whose pKa is as close as possible to the desired pH. The useful pH range of a buffer is usually considered to be pKa 1 pH unit. http:\\asadipour.kmu.ac.ir...36 slides

  27. Buffer Capacity • [Buffer]=[Acid]+[Base] • [Acid]↑ & [Base]↑Capacity↑ • In equimolar buffersis is important • Capacity↑ http:\\asadipour.kmu.ac.ir...36 slides

  28. Buffer capacity as a function of the logarithm of the ratio CNaA/CHA. The maximum buffer capacity occurs when the concentration of acid and conjugated base are equal; that is, when 0= 1 = 0.5. http:\\asadipour.kmu.ac.ir...36 slides

  29. Buffer capacity Ex. ) Calculate the pH and capacity of a buffer prepared by mixing 0.250 mol of acetic acid with 0.100 mol of sodium acetate and diluting to 1.00 L. pKa = – log 1.75×10–5 = 4.75 pH = pKa+ log [OAC–] / [HOAC] = 4.75 + log ( 0.100M)/(0.250M) = 4.36 [OAC–] is less than [HOAC]and reduce sooner,by adding H+ , so it is determinant of capacity. 4.36 -1= 3.36= 4.75 + log ( 0.100M) - X/(0.250M) +X -1.39 = log ( 0.100M) - X/(0.250M) +X 0.041=0.1-X/0.250+X X=0.064M http:\\asadipour.kmu.ac.ir...36 slides

  30. Cb versus pH for a solution containing 0.100F HF with pKa =5.00. • Buffer capacity versus pH for the same system reaches a maximum when pH = pKa. The lower curve is the derivative of the upper curve. http:\\asadipour.kmu.ac.ir...36 slides

  31. The composition of buffer solutions as a function of pH:  value CT = CHOAC + CNaOAC 0 : fraction of the total concentration of acid that is undissociated 0= [HOAC] / CT 1 : the fraction dissociated 1= [OAC–] / CT Ka = [OAC–][H3O+] / [HOAC] [OAC–] = Ka [HOAC] /[H3O+] CT = CHOAC + CNaOAC = [HOAC] + [OAC–] = [HOAC] + Ka [HOAC] /[H3O+] ={[HOAC][H3O+]+[HOAC] Ka)]} / [H3O+]=[HOAC](Ka + [H3O+])/ [H3O+] 0= [HOAC] / CT = [H3O+] / ([H3O+]+ Ka) 1= [OAC–] / CT = Ka / ([H3O+]+ Ka) http:\\asadipour.kmu.ac.ir...36 slides

  32. Variation in  of HOAC with pH http:\\asadipour.kmu.ac.ir...36 slides

  33. The fraction of dissociation of weak electrolyte increases as the electrolyte is diluted. Stronger acid is more dissociated than the weaker acid at all concentrations. http:\\asadipour.kmu.ac.ir...36 slides

  34. 1 value : Fraction of dissociation of a weak acid Ex. 0.0500 M benzoic acid Ka = [A–][H+] / [HA] = (x)(x) / (F –x)  x = [H+] = [A–] = 1.77 ×10–3   = x / F = 1.77 ×10–3 / 0.05 = 0.0354 = 3.54 % The fraction of dissociation of weak acids increases as the acid is diluted. http:\\asadipour.kmu.ac.ir...36 slides

  35. 1 value: Fraction of association of a weak base Ex. 0.0372 M cocaine solution Kb = 2.6×10–6x = [BH+] =[OH–] = KbF = 3.10 ×10–4  1 = x / F = 3.10 ×10–4 /0.0372 = 0.0083 = 0.83 % http:\\asadipour.kmu.ac.ir...36 slides

  36. The End!!!!!!! http:\\asadipour.kmu.ac.ir...36 slides

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