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. 1.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, H 3 0 + ) HA + H 2 O  H 3 0 + + A -

3. -A base ionizes in water to give hydroxyl ions. Weak bases generally ionize as follows: B + H 2 O  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 + H 2 O  H 3 0 + + Cl -

6. WATER IONIZATION CONSTANT, Kw -Pure water ionize slightly, or so called autoprotolysis. H 2 O + H 2 O  H 3 0 + + OH - -The equi. Cons. Express. Can be write as: K = [H 3 0 + ][OH - ] [H 2 O] 2 -Consider no change in water conc., therefore, K = [H 3 0 + ][OH - ] Give special symbol, Kw, Kw = [H 3 0 + ][OH - ] = 1.0 x 10 -14 at 25 o C

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[H 3 0 + ]; pOH = -log[OH - ] -In pure water at 25 o C, the hydronium and hydroxide ion concs. Are both 1.0 x 10 -7 M. pH = -log(1.0 x 10 -7 ) = 7.00 - 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. Equation HOAcH+H+ OAc - Initial (M) Change (M) Equilibrium (M) 1.0 x 10 -3 -x 1.0 x 10 -3 - x 0 +x x 0 +x x x 2 = 1.75 x 10 -5 1.0 x 10 -3 - x x is smaller than C, neglect it, therefore, x 2 = 1.75 x 10 -5 = 1.32 x 10 -4 M = [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 - + H 2 O  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.