AcidbaseequilibriumPDF

897 views

Published on

Pdf file on the powerpoint presentation on acids and bases.
This is a downloadable file.

Published in: Education, Business, Technology
0 Comments
1 Like
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total views
897
On SlideShare
0
From Embeds
0
Number of Embeds
2
Actions
Shares
0
Downloads
24
Comments
0
Likes
1
Embeds 0
No embeds

No notes for slide

AcidbaseequilibriumPDF

  1. 1. Acid Base EquilibriumMr.V
  2. 2. Acid base theoriesArrhenius TheoryBrönsted – Lowry TheoryLewis acid – base Theory
  3. 3. Acid Base behaviourIn this unit when ever we discuss acids andbases we study the behaviour of compoundswhen dissolved in waterThe solvent is all the time Water
  4. 4. Arrhenius TheoryAcid - any substance that increases thehydronium (H3O+) ion concentration inneutral water.  (or H+ ions for simplicity)Base - any substance that increases thehydroxide (OH-) ion concentration in neutralwater.The following pages shows some examplesof acids and bases
  5. 5. Strong Acids = Ka is very largePerchloric acid HClO4Hydroiodic acid HIHydrobromic acid HBrSulfuric acid H2SO4Hydrochloric acid HClNitric acid HNO3Memorise these examples
  6. 6. Weak Acid = Ka is very small at25oCSulfurous acid H2SO3 Ka = 1.2 x 10 -2Hydrogen sulphate ion HSO4- Ka = 1.0 x 10-2Phosphoric acid H3PO4 Ka = 7.1 x 10-4Citric acid H3C6H5O7 Ka = 7.1 x 10-4Nitrous acid HNO2 Ka = 7.1 x 10-4Hydrofluoric acid HF Ka = 6.8 x 10-4Formic Acid HCO2H Ka = 1.8 x 10-4
  7. 7. Weak Acids continued at 25oCAcetic acid HC2H3O3 Ka = 1.8 x 10-5Carbonic acid H2CO3 Ka = 4.5 x 10-7Hydrogen cyanide HCN Ka = 6.2 x 10-10Ammonium ion NH4+ Ka = 5.7 x 10-10Bicarbonate ion HCO3- Ka = 4.7 x 10-11Water H2O Ka = 1.8 x 10-16
  8. 8. Arrhenius BasesSodium hydroxide NaOHAmmonium hydroxide NH4OHCalcium hydroxide Ca(OH)2 (Lime water)
  9. 9. Kw and pHWater is weakly dissociated and theequilibrium can be shown like thisH2O + H2O = H3O+ + OHH2O = H+ + OH-Kw = [H+][OH-]Kw = 10-14 at 250C for waterH2O = H+ + OH--x +x +x
  10. 10. pHpH = -log[H+]pOH = -log[OH-]For water pH = -log[10-7] = - [-7log 1o] =7Similarly pOH = 7pH + pOH = 14
  11. 11. Brönsted Lowry TheoryAn acid is a proton donorA base is a proto acceptorAn amphoteric substance is one that can actboth as an acid and a base.Amphoteric substances are proton acceptorsor donors eg. H2O, HCO3-, etc.
  12. 12. Brönsted acids bases examplesNH3 + H2O       NH4+     +     OH-base     acid           conjugate   conjugate                                acid             baseHCl + H2O       H3O+    +        Cl-acid    base             conjugate      conjugate                                       acid                base
  13. 13. Ka – Dissociation constant ofan acidThe equilibrium constant for the dissociationfor a weak acid that is an equilibriummixtureEg. HA + H2O H3O+ + A-
  14. 14. Kb = Dissociation constant of abaseNH3 (aq) + H2O NH4+ (aq) +OH-(aq)CO32-(aq) + H2O HCO3-(aq) +OH-(aq)B (aq) + H2O BH+(aq) + OH-(aq)
  15. 15. pKa and pKbThese expressions are analogous to theexpression for pH. We derived it to expressthe concentrations in a more convenientway using whole numbers.pKa = -log KapKb = -log KbSmaller value of pKa , stronger the acidSmaller value of pKb , stronger the base.
  16. 16. Ka and Kb for Conjugate acid –base pairFor a conjugate acid base pair this isapplicable.Ka x Kb = KwHA H+ + A-A- + H2O HA + OH-
  17. 17. Conjugate acid base pairsKa x Kb
  18. 18. Relationships between Ka,Kb,Kw.
  19. 19. Calculations involving KaCalculate the value of Ka for a specific acidfrom the [H+] or pH of a solution for whichwe also know the initial concentration of theacidCalculate the equilibrium concentrations ofH+ and A- from the initial concentration of aspecific weak acid and its Ka value
  20. 20. Calculations involving KbCalculate the value of Ka for a specific acidfrom the [OH-] or pOH of a solution for whichwe also know the initial concentration of thebase or from pH as pOH = 14 - pHCalculate the equilibrium concentrations of[H+ ]or [OH-]from the initial concentration ofa specific weak base and its Kb value (Orwhat is almost the same is calculating thepH or pOH from the values of Kb and initialconcentration of base)
  21. 21. Hydrolysis of IonsThere are many ionic salt’s that make theaqueous solution acidic or basic.The reaction of ionic salts with water iscalled “Hydrolysis of salts”Hydrolysis can be explained using BrönstedLowry theory.If the conjugate acid or base is strong thenthere will be hydrolysis.Weak acids and bases produce strong
  22. 22. Illustration of HydrolysisA- is a strong conjugate baseA- (aq)+ H2O = HA(aq) + OH-The dissociation constant for the hydrolysisin this case is Kb as one of the products is abase.
  23. 23. Hydrolysis of strong conjugateacidNH4+ + H2O = H3O+ + NH3NH4+ is a strong conjugate acid thereforeundergoes hydrolysis and the solutionbecomes more acidic or pH is lowered.Other positive metal ions that can undergohydrolysis is Fe3+, Al3+ etc.………………
  24. 24. Titration of weak acid Vs baseAcetic acid Vs NaOH
  25. 25. Polyprotic acids examplesH2SO4H2CO3H3PO4
  26. 26. Polyprotic acidFor any diprotic acid the value of Ka1 > Ka2It is easier to pull a H+ from a neutral moleculethan an ionKa1 is usually 104 to 105 times larger than Ka2.Therefore the contribution of acidity due to thesecond or third proton is negligible and can beignored. 
  27. 27. Calculating Ka from pH #1Formic acid HCHO2 is a monoprotic acid. In a0.100 M solution of formic acid, the pH is2.38 at 25oC. Calculate Ka for formic acid atthis temperature.Ans: Ka = 1.8x10-4
  28. 28. #1Determine the [H+] ion concentration frompHMake Ice table sand substitute theconcentration of H+ in the table andcomplete it to get the equilibriumconcentration of HA, H+ and A-Use the values to determine the Ka for theacid.
  29. 29. Calculating [H+], pH from Kafor weak acid #2The concentration of vinegar HC2H3O2 wasfound to be 0.75 M acetic acid, Calculate thevalue for the [H+] ion concentration and pHfor this acid solution if Ka for acetic acid is1.8 x 10-5
  30. 30. #2Create an ice table initial concentration is0.75 M for acetic acid. Change inconcentration is –x,x and x respectively forHA, H+ and A-Since Ka is very small 0.75 –x can be takenas 0.75 and solve for x.X is the concentration of H+ use this valueto calculate pH

×