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Strong Bases (examples , pH calc. ) Acid-Base Theories Weak Electrolites

Dr. M. Sasvári: Medical Chemistry Lectures 3. Strong Bases (examples , pH calc. ) Acid-Base Theories Weak Electrolites. Strong bases Complete ionization in water. BOH. B + + OH -. I. II. III. IV. V. VI. VII. VIII. Li. Na. K. Remarks. Strong bases:.

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Strong Bases (examples , pH calc. ) Acid-Base Theories Weak Electrolites

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  1. Dr. M. Sasvári: Medical Chemistry Lectures 3. • Strong Bases (examples, pH calc.) • Acid-Base Theories • Weak Electrolites Chem Eq 3

  2. Strong bases Complete ionization in water BOH B+ + OH- Chem Eq 3

  3. I. II. III. IV. V. VI. VII. VIII. Li Na K Remarks Strong bases: LiOH (Lithium hydroxide) removes CO2 in spacecraft's NaOH (Sodium hydroxide) oven cleaner, drain opener KOH (Potassium hydroxide) manufacturing liquid soaps Li+ + OH- K+ + OH- Na+ + OH- Strong bases Complete ionization in water Oxidation No.: +1 Chem Eq 3

  4. Example: 2K + 2H2O → 2K+ + 2OH - + H2 KOH is a strong base its formation: exergonic reaction heat → H2 gas ignites Chem Eq 3

  5. Strong bases I. II. III. IV. V. VI. VII. VIII. Ca Sr Ba Strong bases: Remarks Ca(OH)2(Calcium hydroxide) slaked lime (used in mortar) Sr(OH)2(Strontium hydroxide) Ba(OH)2(Barium hydroxide) X-ray of the stomach Ca2+ + 2OH- Sr2+ + 2OH- Ba2+ + 2OH- Complete ionization in water Oxidation No.: +2 Chem Eq 3

  6. NaOH Na+ + OH- Ca(OH)2 Ca2+ + 2 OH- Normality of a strong base = the [OH-]of the solution 1 N strong base contains 1 mol of OH-per liter Normality = Molarity * n (number of dissociable OH-) n = 1 n = 2 Chem Eq 3

  7. K+ KOH + OH- H2O H+ + OH- • pH of a strong base [OH-] = [ strong base] in Normality H+ from water [OH-] from water is negligible Chem Eq 3

  8. Examples KOH (N) pOH pH 13 1 [OH-]from H2O: 10-7 M [OH-]from KOH: 10-8 M 0.01 =10-2 2 12 11 0.001 =10-3 3 Total [H+] > 10-7 M • pH of a strong base pOH = - lg [OH-]= -lg c strong base (N) pH = 14 - pOH 14 1 =10 0 0 0.1 =10-1 =10-8 pOH < 7 pH > 7 The pH of a base is always above 7! Chem Eq 3

  9. g 4 mol 0.1 = Molarity = = mol = = 0.2 M pOH= -log10 (0.2)= 0.7 BASE! Mw 40 L 0.5 What is the M and N of the NaOH solution with pH 11.7? 4 g NaOH (Mw=40) is dissolved in water upto a total volume of 500 ml. What is the pH of the solution? = 0.1 0.2 N pH= 14-0.7=13.3 BASE! pOH= 14 – 11.7 = 2.3 10-2.3 =0.005 N 0.005 M Chem Eq 3

  10. Chem Eq 3

  11. EQ H2O H+ + OH- INI H2O H+ + OH- + OH- INI H2O H+ ACID added BASE added The Arrhenius concept (1884) • An acid, when dissolved in water, • increases the [ H+](decreases pH) • decreases [OH-] (increases pOH) • A base, when dissolved in water, • decreases the [ H+](increases pH) • increases [OH-] (decreases pOH) [H+] [OH-] [H+]  [OH-] Chem Eq 3

  12. The Bronsted-Lowry concept Acid = donates a proton in a proton-transfer reaction Conjugated base ACID Base = accepts a proton in a proton-transfer reaction Conjugated acid BASE Chem Eq 3

  13. - H A A + H H B B + H Protonated forms Deprotonated forms Bases Conjugated bases Acids Conjugated acids Chem Eq 3

  14. Conjugated BASE ACID CH3COO- CH3COOH + H+ Acetic acid Acetate ion CONJUGATED ACID BASE NH4+ ammonium ion ammonia NH3 + H+ Protonated forms Deprotonated forms Examples Chem Eq 3

  15. example: NH3 + H+ NH4 + H + H : : : H N : : H N H : : : + H+ H H The Lewis concept ammonia ammonium ion H+ accepts an electron pair Lewis acid NH3 donates an electron pair Lewis base Chem Eq 3

  16. example: 4 NH3 + Cu2+ [Cu(NH3)4]2+ H : : H N : 2+ : NH3 4 + Cu2+ H : H3N: Cu :NH3 : NH3 Tetraamminecopper(II) ion ammonia copper(II)ion The central metal ionaccepts an electron pair Lewis acid The ligand donates an electron pair Lewis base Complex ion(coordination compound) Chem Eq 3

  17. Weak electrolites • Degree of ionization (a) • - pKa, pKb • - Strength of Acidity/Basicity Chem Eq 3

  18. CH3COO-+ H3O+ CH3COOH + H2O hydroxonium ion acetic acid acetate ion INI 1M (excess) 1M 1M 0.006 M 0.006 M 0.006  = 0.003 = 2 moles in ionized form  = total number of moles Degree of Ionization 0 < < 1 Endergonic process, reaction goes backwards acetate ions will be protonated by the hydroxonium ion 0.006 M of the ionized forms remain in the new EQ EQ (2-0.006) M Degree of ionization (a) in the EQ: Chem Eq 3

  19. [H+] 10-4 0.01 or 1% a = = = c 0.01 [OH-] 10-3 0.05 or 5% a = = = c 0.02 The [H+] concentration of a 0.01 M monoprotic acid solution is 10-4. Calculate the degree of ionization! The [OH-] concentration of a 0.02M base solution is 10-3. Calculate the degree of ionization! Chem Eq 3

  20. (0.006)2 Ka = =1.8 10-5 (2-0.006) CH3COOH CH3COO - + H+  Ka =  2 c Ka = Ka = 1 -  Acidic Ionization Constant (pKa) acid Ka is an equilibrium constant of acidic dissociation (concentration of H2O included) Degree of Ionization (a) and the Ionization constant (Ka): c)c) c - c) c = total concentration (HA + A-) Chem Eq 3

  21. A-+ H3O+ AH + H2O 0.1 M 0.01 M 2 c Ka = 1 -  Will the a change if we dilute aweak acid? higher dilution  more complete ionization increase awill Ka = 1.7 10-5 a = 1.3 % a = 4.2 % Ka change! does not Chem Eq 3

  22. NH3 + H2O KOH NH4++ OH - K++ OH - Kb is an equilibrium constant of basic dissociation (concentration of H2O included) KOH Kb = KOH base 2 c Kb = 1 -  Basic Ionization Constant (pKb) in general: Chem Eq 3

  23. 7 4 2 pKa pK< 2 K> 10 -2 Strength of Acidity/Basicity the higher is the pKa/pKb the weaker is the acidity/basicity Strong acids and bases conjugated acids conjugated bases weak acids weak bases Chem Eq 3

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