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ACID-BASE EQUILIBRIA

ACID-BASE EQUILIBRIA. INTRODUCTION Acidity and basicity of a solution is important factor in chemical reactions. In this topic you will review acid-base theories and the basic pH concept. Others topic involve various calculation on acid-base equil., including weak acids and bases.

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ACID-BASE EQUILIBRIA

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  1. ACID-BASE EQUILIBRIA • INTRODUCTION • Acidity and basicity of a solution is important factor in chemical reactions. • In this topic you will review acid-base theories and the basic pH concept. • Others topic involve various calculation on acid-base equil., including weak acids and bases.

  2. Acid-base theories • -to explain or clasify acidic and basic properties of substances. • Arrhenius Theory - Restrict to aqueous solutions only • An acid is any substances that ionizes (partially or completely) in water to give hydrogen ions (which associate with the solvent to give hydronium ions, H30+) HA + H2O  H30+ + A-

  3. A base ionizes in water to give hydroxyl ions. Weak bases generally ionize as follows: B + H2O  BH+ + OH- 2. Theory of Solvent systems Accepted for solvent that ionize to give a cation and an anion. An acid is defined as a solute that yieldsthe cation of the solvent while a ase is a solute that yields the anion of the solvent. 3. Bronsted-Lowry theory - Applicable to acid-base reactions in nonionizable solvents such as benzene or dioxane. An acid is any substance that can donate a proton, and base is any substance that can accept a proton.

  4. Introduce half reaction equations. 4. Lewis Theory • Introduced the electronic theor of acids and bases. • An acid is a substance tat can accept an electron pair and a base is a substance that can donate an electron pair.

  5. ACID BASE EQUILIBRIA IN WATER INTRODUCTION • For strong acid, ionization is complete, therefore an equi. cons. would have a value of infinity. • For weak acid, will partially ionizes, therefore can calculate the equi. Cons. Value. • For example: an ionize of acetic acid HOAc + H2O  H30+ + Cl-

  6. WATER IONIZATION CONSTANT, Kw • Pure water ionize slightly, or so called autoprotolysis. H2O + H2O  H30+ + OH- • The equi. Cons. Express. Can be write as: K = [H30+][OH-] [H2O]2 • Consider no change in water conc., therefore, K = [H30+][OH-] Give special symbol, Kw, Kw = [H30+][OH-] = 1.0 x 10-14 at 25oC

  7. WATER AND THE Ph SCALE • The pH of a solution is defined as the negative of the base-10 logarithm (log) of the hydronium (hydroxide) ion concentration. pH = -log[H30+ ]; pOH = -log[OH-] -In pure water at 25oC, the hydronium and hydroxide ion concs. Are both 1.0 x 10-7M. pH = -log(1.0 x 10-7) = 7.00 - Solution with pH < 7.00 are acidic and solution with pH > 7.00 are basic.

  8. Example : Calculate the pH of a 2.0 x 10-3 M solution of HCl. Solution: HCl is completely ionized, [H+] = 2.0 x 10-3 pH = log (2.0 x 10-3) = 3 – log 2.0 = 3 – 0.3 = 2.70 Exercise:Calculate the pOH and the pH of a 5.0 x 10-2 M solution of NaOH

  9. Solution: [OH-] = 5.0 x 10-2 M pOH = -log (5.0 x 10-2 M) = 2 – log 5.0 = 2 – 0.70 = 1.30 pH + 1.30 = 14.00 ; pH = 12.70 Or [OH+]= 1.0 x 10-14 = 2.0 x 10-13 M 5.0 x 10-2 pH = -log (2.0 x 10-13 ) = 13 – 0.30 = 2.70

  10. Calculation for weak acids and bases • Ionization cons. for some common acids have been calculated. • Follow the rules for equi. cons. calculation. Example: Calculate the pH and pOH of a 1.0 x 10-3 solution of acetic acid Solution : HOAc  H+ + OAc- Ka= [H+][OAc-] = 1.75 x 10-5 [HOAc]

  11. x2 = 1.75 x 10-5 1.0 x 10-3 - x x is smaller than C, neglect it, therefore, x2 = 1.75 x 10-5 = 1.32 x 10-4M = [H+ ] 1.0 x 10-3 pH = -log 1.32 x 10-4 = 3.88 pOH = 14.00 – 3.88 = 10.12

  12. SALTS OF WEAK ACIDS AND BASES • Salt of a weak acid is a strong electrolyte and will completely ionize.(NaOAc) • Anion of the salt of basic acid is a Bronsted base, which will accept protons. • It partially hydrolizes in water to form hydroxide ion and the corresponding undissociated acid. • Hydrolizes the conjugate base • The ionization cons. is equal to the basicity cons. • Example: OAc- + H2O  HOAc + OH-

  13. Kh = Kb = [HOAc][OH-] = hydrolisis cons. [OAc-] Kb = Kw = 1.0 x 10-14 = 5.7 x 10-10 Ka 1.75 x 10-5 - We can derive that KaKb = Kw • This equation can also be applied for weak bases • Concentration of products is consider small and can be neglected compared to C of reactants.

  14. BUFFERS • Solution that resists change in pH when a small amount of an acid or base is added or when the solution is diluted. • Consists of a mixtur of a weak acid and its conjugate base or a weak base and its conjugate acid at a predetermined cons. or ratios. • Example: a system of acetic acid-acetate buffer. HOAc H+ + OAc- - When acetate ions is supplied, there is changes in H+ conc.

  15. [H+] = Ka[HOAc] [OAc-] -log [H+] = -log Ka – log[HOAc] [OAc-] pH = pKa – log[HOAc] [OAc-] pH = pKa + log [OAc-](conjugate base/proton acceptor) HOAc] (acid) - Note that at any situation products (numerator) always above and reagents (denominator) always below.

  16. Try this example: Calculate the pH of buffer prepared by adding 10 ml of 0.10 M acetic acid to 20 ml of 0.10 M sodium acetate.

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