1 / 22

pH of weak electrolites Indicators The Henderson-Hasselbalch equation Titration

Dr. M. Sasvári: Medical Chemistry Lectures 4. pH of weak electrolites Indicators The Henderson-Hasselbalch equation Titration E xamples for strong/weak electrolites. Assumptions:. [H + ] = [A - ]. 1. No common ion:.     . K a =. . [ H +  ][ H +  ]. K a =.

quade
Download Presentation

pH of weak electrolites Indicators The Henderson-Hasselbalch equation Titration

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Dr. M. Sasvári: Medical Chemistry Lectures 4. • pH of weak electrolites • Indicators • The Henderson-Hasselbalch equation • Titration • Examples for strong/weak electrolites chem Eq 4

  2. Assumptions: [H+] = [A-] 1. No common ion:  Ka =  [H+][H+] Ka = [H+]= Kac pKa-logc pH= 2 pH of Weak Acids: Calculations Calculation of pH from a: a [H+]= c Calculation of pH from Ka c 2. a is small: [HA]  c Calculation of pH from pKa pKa = -log Ka The same assumptions as above chem Eq 4

  3. Calculation of pH from a: [OH-]= c a + OH- Calculation of pH from Kb NH4+ [OH-] Kb = NH3 Kb is an equilibrium constant of basic dissociation (concentration of H2O included) NH3 + H2O NH4+ pOH = -lg [OH-] pH = 14-pOH [OH-]= Assumptions: 1. No common ion: [OH-] = [NH4+] Kbc 2. a is small: [NH3]  c pKb-logc Calculation of pOH from pKb pOH= 2 pH = 14-pOH The same assumptions as above pH of Weak Bases: Calculations chem Eq 4

  4. Acid-Base Indicators chem Eq 4

  5. Strong acid H+ HA + A1- HA H+ + A2- Weak acid • Effect of strong acid on the ionization of a weak acid Common ion: [H+] Mainly from the strong acid Ionization of weak acid is inhibited by the strong acid [HA] c weak acid chem Eq 4

  6. HCl H+ HA + A- Indicator: a weak acid, e.g. methyl orange (HA) HA A- H+ + Indicator Color in acidic range Color in basic range chem Eq 4

  7. HA H+ + A- If a common ion is present: [H+] = [A-]  [H+] = Ka   [H+][A-]  pH = pKa + lg Ka = [HA] • The Henderson-Hasselbalch equation deprotonated form protonated form The pH will determine for a compound the ratio of the two forms If pH = pKa, the two forms are equal chem Eq 4

  8. HInd H+ + Ind- Intermediate color pH = pKa  1 = Color change  10    = pH = pKa + lg   10 =  1 • Acid-Base Indicators pH < pKa pH > pKa pH = pKa-1 pH = pKa pH = pKa+1 chem Eq 4

  9. Methyl orange Phenolphtalein Bromophenol blue pH 1 2 3 4 5 6 7 8 9 10 11 chem Eq 4

  10. Acid-Base Titrations chem Eq 4

  11. 14 12 10 Methyl orange HCl neutralized NaOH present Eqv. Point pH=7 8 6 4 HCl present 2 0 0 20 40 60 80 Titration of strong acid with a strong base Unknown:(0.1 N)HCl (50 ml)titrated with 0.1 N NaOH: chem Eq 4

  12. Calculations-Titration 0.1 N NaOH (0 ml) 0.1 N HCl (50ml) Start: c*v= meqv c= meqv/ml pH = -lg c 50 ml 0.1 N HCl 50*0.1 = 5 meqv 5/50 = 0.1 -lg 0.1 = 1 chem Eq 4

  13. Calculations-Titration 50 ml 0.1 N HCl 50*0.1 = 5 meqv - 10*0.1 = 1 meqv + 10 ml 0.1 N NaOH 0.1 N NaOH (10 ml) 0.1 N HCl (50ml) HCl excess c*v= meqv c= meqv/ml 4/60 = 0.067 N 4 meqv 60 ml pH = -lg 0.067 = 1.2 HW: +20 ml, +40 ml, +49 ml 0.1 N NaOH pH = ? chem Eq 4

  14. Calculations-Titration 0.1 N NaOH ( 50 ml) 0.1 N HCl (50ml) Neutralization 50 ml 0.1 N HCl 50*0.1 = 5 meqv - 50*0.1 = 5 meqv + 50 ml 0.1 N NaOH 0 meqv 100 ml pH = 7 chem Eq 4

  15. Calculations-Titration 0.1 N NaOH ( 51 ml) 0.1 N HCl (50ml) Base excess 50 ml 0.1 N HCl 50*0.1 = 5 meqv - 51*0.1 = 5.1 meqv + 51 ml 0.1 N NaOH 0.1 meqv 101 ml 0.1/101=0.001 pOH = 3 pH = 11 chem Eq 4

  16. pH Start: pH =13 NaOH excess Eqv. point: pH=7 neutralization End: pH below 2 HCl excess ml of HCl Titration of strong base with a strong acid Unknown:(0.1 N) NaOH (50 ml)titrated with 0.1 N HCl: chem Eq 4

  17. EN increases NH3 < H2O < HF V. VI. VII. VIII. ammonia water hydrogenfluoride F N O F H Acidic strenght of HYDRIDES High electronnegativity atom → electron withdrawing effect Electron density decreases around H → enhanced acidity chem Eq 4

  18. H2O < H2S HF < HCl < HBr < HI F H Cl H Br H I H Acidic strength of HYDRIDES V. VI. VII. VIII. F O Cl S Size increases EN increases Br I Acidic strength increases by atomic size chem Eq 4

  19. Acidic strength of OXYACIDS Oxidation No. HClO (hypochlorous acid) +1 HClO2(chlorous acid) +3 HClO3(chloric acid) +5 HClO4 (perchloric acid) +7 Acidity increases by oxidation number chem Eq 4

  20. Weak acids are common acetic acid (vinegar) nicotinic acid (vitamin) ascorbic acid (VitC) citric acid (lemon juice) formic acid(ants) oxalic acid (sorrel) chem Eq 4

  21. Conjugated bases (salts of weak acids) antacids (NaHCO3) baking soda (NaHCO3) washing powder (Na2CO3) All the weak acids are in their conjugated base forms in our body chem Eq 4

  22. Weak bases are rear ammonia alkaloids quinine nicotine cocaine caffeine chem Eq 4

More Related