Acids-Bases Arrhenius: Acid…. A substance that increases the hydrogen ion, H+, concentration when dissolved in H2O. Eg. HCl, H2SO4, HC2H3O2 (CH3COOH), etc. Base…. A substance that increases hydroxide ion, OH-, concentration. Eg. NaOH, NH4OH, etc. Bronsted-Lowry (1923) (not worried about increasing/decreasing H+ or OH- concentrations) Acid … a “proton donor”, p+ = H+, any formula HA. Eg. HCl, H2SO4, HC2H3O2 (CH3COOH), etc. Base …. A “proton acceptor” Eg. OH-, NH3 ; Cl- Note: HOH can do both! Really no different from Arrhenius. This is different from Arrhenius!

Coordinate covalent bond Acid solutions: HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) Coordinate covalent bond .. H .. + .. .. H :Cl: H :O: - + .. .. H :O: + :Cl: .. .. H H Hydrogen ion, H+, or hydronium ion, H3O+ has formed H3O+(aq) = H+(aq) HCl(g) + H2O(l) H+(aq) + Cl-(aq) Arrhenius would say the concentration of H+ has increased. Bronsted and Lowry would say the HCl donated a p+ (H+ion), and the water accepted a p+ (H+ion). Either way the HCl is an acid. The water is a base ….it was a proton acceptor!

HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) Strength of an acid can be determined by conductivity. Good or poor electrolyte? Large or small # of ions? 12 M HCl good Large HCl(g) + H2O(l) H3O+(aq) + Cl-(aq) Initially: 100 0 0 @equil: 0 100 100 Ka = [H3O+] [Cl-] Strong Acid = Large value [HCl] Acids with Ka values greater then “1” are considered strong acids. They 100% ionize. No ions! Note: 100% pure HCl is a poor conductor. Why?

Strength of an acid can be determined by conductivity. Good or poor electrolyte? Large or small # of ions? Pure HC2H3O2 no conduction none Weak acid Diluted HC2H3O2 poor small HC2H3O2 (g) + H2O(l) H3O+(aq) + C2H3O2 -(aq) Initially: 100 0 0 @equil: 97 3 3 < 5% ionization….. Very weak acid [H3O+][C2H3O2-] Ka = = a Very small # = 1.76 X 10-5 [HC2H3O2]

Properties of Acids: 1. Conduct electricity if they ionize. 2. React with metals to form H2 3. Neutralize bases. 4. Turn litmus red 5. Taste sour

More Bronsted-Lowry Acid/Base info: General acid/base reaction: HA + B = A- + HB+ acid1 base2 base1 acid2 conjugate pair Acid/base HCl(aq) + NH3(aq) = NH4+ (aq) + Cl-(aq) acid1 base2 acid2 base1 HCl…. Ka is larger Which is a stronger acid? Which direction is favored? product At equilibrium there is more ________________

Titration problems What is the volume of a 0.325 M NaOH solution needed to just neutralize 65.2 ml of 1.37 M HNO3? Balanced equation: 1 NaOH + 1 HNO3 = 1 HOH + 1NaNO3 65.2ml x x x x x = 274.8 = 275ml NaOH Or using: MaVa#H’s = MbVb#OH’s Since mole ratio of H’s and OH’s is 1:1, (1.37M)(65.2ml)(1) = (0.325M)(Vb)(1) Vb = 275ml

Basic Solutions: Strong Arrhenius bases: Group IA hydroxides. Eg. NaOH, KOH Lower Group IIA hydroxides. Eg. Sr(OH)2, Ba(OH)2 100% “Dissociate”: NaOH(S) Na+(aq) + OH-(aq) 1.0 M Ba(OH)2 = 1.0mol Ba+2 + 2.0mol OH- The Kb for these bases would be very large; > than 1. They would also be good conductors of electricity. Strong Bronsted-Lowry bases: Acid anions of weak acids: eg. C2H3O2-, from HC2H3O2 or OH-, from HOH

More Bronsted-Lowry Acid/Base info: Bases react with water to produce OH- General equation: B(g) + H2O(l) = BH+(aq) + OH-(aq) NH3(g) + H2O(l)  NH4+(aq) + OH-(aq) Base1 acid2 acid1 base2 [NH4+][OH-] = 1.79 x 10-5 Kb = [NH3] Which direction is favored in the reaction?

Hydrolysis Salt + H2O  Acidic solution Neutral solution Basic solution Salt + H2O  In water: H2O + H2O  H3O+ + OH- Kw = 1 x 10-14 Equal amounts of ions, but very little! Because the [H3O+] = [OH-]…. Water is neutral If one of the ions from a salt removes one of the ions from the water….the other ion from the water will be in excess!

Hydrolysis (continue) NaClO + H2O : NaClO is a salt of a base NaOH and an acid HClO Na+ + OH- vs H+ + ClO- HClO…. Weak acid, H+ ion removed from solution NaOH…strong base, OH- ion still present in solution basic Solution will be ___________

Lewis Acid/Base + Lewis acid: electron pair acceptor Lewis base: electron pair donor + H H .. .. H+ + :N: H .. : N : H H .. H Lewis acid H Lewis base “adduct” … the product of a L.acid/base rxn

Acid-Base Equilibria Review: In pure water & all aqueous solutions: H2O + H2O = H3O+ + OH- Kw = [H3O+][OH-] = 1 x 10-14 1 x 10-7 [H3O+] = [OH-] = or = [H3O+] > [OH-] = Acidic solution [H3O+] < [OH-] = Basic solution [H3O+] = [OH-] = Neutral solution

Solutions of Strong Acids/ Bases: 0.10 M HNO3 actually consists of 0.10 M H3O+ or [H+] = 0.10 M 0.10 M NaOH…. Adds 0.10 OH- to the water solution Ionization & dissociation are 100% for strong acids & bases HNO3 + H2O  H3O+ + NO3- NaOH + H2O  Na+ (aq) + OH-(aq) GONE GONE In 0.10M Ba(OH-)2 [OH-] = 0.20M In 0.20 M H2SO4 [H+] = 0.40 M

Solutions of Weak acids/bases 0.10 M HC2H3O2 : a weak acid - when it reacts with water it does not ionize 100%. What is the [H3O+] for this solution? Ka = 1.8 x 10-5 Equation for the ionization of HC2H3O2 in water: H3O+ + C2H3O2- HC2H3O2 + H2O = Write the MAE, equilibrium expression of this reaction: [H3O+] [C2H3O2-] Ka = = 1.8 x 10-5 [HC2H3O2]

Weak acids (continue) [H3O+] [C2H3O2-] Ka = = 1.8 x 10-5 [HC2H3O2] “rule of thumb”… if %ionization < 5%, the value for x is so small it can be ignored. (otherwise you would need to use quadratic equation to solve). ignore % ionization < 5% when: [HA] / Ka > 100 0.10  1.8 x 10-5 = 5,556 !

Weak acids (continue) So…. 1.8 x 10-5 x = 1.3 x 10-3 = [H3O+] 2 = Also = [C2H3O2- ] 1.3 x 10-3 % ionization = x 100 = 1.3% .10 < 5% as predicted Weak acids ionize very little…. there is very little H3O+ present.

Weak bases…. x = 2.5 x 10-3 = [OH-] B + H2O = BH+ + OH- NH4+ + OH- NH3 + H2O = Given: 0.35M NH3 solution; Kb = 1.8 x 10-5 What is the concentration of OH- ? 2 [NH4+][OH] 1.8 x 10-5 = = [NH3] x = 2.5 x 10-3 = [OH-] ignore…<5%

Weak bases (continue) If the [OH-] = 2.5 x 10 -3 , what is the [H3O+]? Given: [H3O+][OH-] = 1 x 10-14 [H3O+][2.5 x 10 -3] = 1 x 10 -14 4.0 x 10-12 [H3O+] = Yes Are we saying that this solution has both H3O+ & OH - in it? All acid or base solutions have them….Why? They have H2O ! Why is this solution Basic? [OH-] > [H3O+]

pH A simple scale for ranking the H3O+ concentrations of dilute acid/base solutions. (Sorenson). pH = - Log [H3O+ ] The logarithm of a number is that number expressed as an exponent of the base 10. For example, the logarithm of 1 is 0, 1 x 100. If [H3O+] = 0.001 = pH = -log 0.001 = 1x10-3 This is the pH 3 If this is 1

pH Scale 1M BOH 1M HA Acidity increases Basicity increases pH neutral pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 [H3O+] 100 10-1 10-2 10-310-410-5 10-6 1x10-7 10-8 10-9 10-10 10-11 10-12 10-13 10-14 [OH-] 10-12 10-8 1x10-7 10-6 10-3 100 8 7 6 3 pOH 12 [H3O+][OH-]= 1x10-14 pH = - log[H3O] pH + pOH = 14 Formulas to remember

More About the pH Scale Given a 3.25 x 10-4 M HNO3 , solve for: [H3O+] [OH-] = HNO3 is a strong acid. Thus, 3.25 x 10-4M HNO3 = prediction [H3O+] = 3.25 x 10- 4 -log [3.25 x 10-4] = -(-3.488) = 3.49 1 x 10-14 = 3.08 x 10-11 3.25 x 10-4 d. pOH = -log [OH-] = -log [3.08 x 10-11] = 10.5 Or pOH = 14.00 –3.49 = 10.51

x = 1.3 x 10-3 Sample Problem 1.8 x 10-5 0.10 - x 1.8 x 10-5 = [H3O+] What is the pH of a 0.10 M HC2H3O2 solution? [H3O+] [C2H3O2-] = 1.8 x 10-5 Ka = [HC2H3O2 ] = x [H3O+] = [C2H3O2-] 0.10 - x [HC2H3O2 ] = 2 = 1.8 x 10-5 x = 1.3 x 10-3 = [H3O+] ignore pH = - Log [1.3 x 10-3] = 2.9