pH = - log [H 3 O + ] [H 3 O + ] = 10 - pH mol/L For pure water at 25 o C pH = - log (1.0 x ) = 7.00 For a change in pH by 1, H 3 O + concentration changes by 10 Higher pH, lower H 3 O + concentration pH of pure water is 7 pH of an acidic solution is less than 7 pH of a basic solution is greater than 7

pOH = - log [OH - ] pK w = - log K w pK w = at 298 K [H 3 O + (aq)] [OH - (aq)] = K w - log[H 3 O + (aq)] - log[OH - (aq)] = - log K w pH + pOH = pK w At 298 K pH + pOH = 14.00

Strengths of Acids and Bases The pH of 0.10 M HCl(aq) will be recorded as close to 1 The pH of a 0.10 M solution of CH 3 COOH(aq) solution is recorded as ~ 3. H 3 O + (aq) concentration in 0.10 M HCl(aq) is greater than that in 0.10 M CH 3 COOH(aq) HCl(aq) + H 2 O(l)  H 3 O + (aq) + Cl - (aq) CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq)

K a = [CH 3 COOH(aq)] [H 3 O + (aq)][CH 3 COO - (aq)] At 298 K, K a for CH 3 COOH(aq) = 1.8 x NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) K b = [NH 3 (aq)] [NH 4 + (aq)][OH - (aq)] At 298 K, K b for NH 3 (aq) = 1.8 x K a : acidity constant or acid dissociation constant K b : basicity constant or base dissociation constant

The proton donor strength of an acid is measured by the value of K a ; higher K a, stronger the acid The proton acceptor strength of a base is measured by K b ; higher K b, stronger the base pK a = - log K a pK b = - log K b The larger the pK values, weaker the acid or base

K a K b = [H 3 O + (aq)] [OH - (aq)] = K w Or pK a + pK b = pK w NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) K b = [NH 3 (aq)] [NH 4 + (aq)][OH - (aq)] NH 4 + (aq) + H 2 O(l)  H 3 O + (aq) + NH 3 (aq) K a = [NH 4 + ( aq)] [NH 3 (aq)][H 3 O + (aq)] Relationship between conjugate acid/base pairs

The stronger the acid/base, the weaker its conjugate base/acid pK a - pink pK b - blue HClO 2 (aq)/ ClO 2 - (aq) HOCl(aq)/ OCl - (aq) CH 3 COOH(aq)/ CH 3 COO - (aq) NH 4 + (aq) / NH 3 (aq) CH 3 NH 3 + (aq) / CH 3 NH 2 (aq)

Using tabulated K a and K b values determine which species is stronger as an acid or base 1) as acid HF(aq) or HIO(aq) 2)as base C 6 H 5 NH 2 (aq) or (CH 3 ) 3 N(aq) 3) as acid C 6 H 5 NH 3 + (aq) or (CH 3 ) 3 NH + (aq)

Molecular Structure and Acid Strength The more polar or weaker the H-A bond, the stronger the acid Effect of bond strength HF < HCl < HBr < HI H - I bond is weakest H 2 O < H 2 S < H 2 Se < H 2 Te H-Te bond weakest

For an acid HA, greater the electronegativity of A, stronger the acid electronegativity difference N-H0.8 F-H1.8 HF is an acid in water, NH 3 is a base

Solutions of Weak Acids/Bases For a strong acid and base; assume that deprotonation/protonation reactions go to completion HCl(aq) + H 2 O(l)  H 3 O + (aq) + Cl - (aq) pH = - log [H 3 O + (aq)] Knowing the concentration of HCl, can determine pH For weak acids/bases, set up equilibrium table to determine the H 3 O + (aq) / OH - (aq) concentration at equilibrium, knowing the value of K a /K b.

CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq) Calculate the pH and percentage deprotonation of 0.10 M CH 3 COOH(aq) given that K a is 1.8 x K a = [CH 3 COOH(aq)] [H 3 O + (aq)][CH 3 COO - (aq)] CH 3 COOH(aq)CH 3 COO - (aq)H 3 O + (aq) Initial Change- xx x Equilibrium x x x 1.8 x = x 2 /( x)

Since K a is so small, assume that x << x ≈ x 2 /(0.10) x = 1.3 x M [H 3 O + (aq)] = 1.3 x M pH = 2.89 % deprotonation = 100% x ([CH3COO - (aq)]/[CH 3 COOH] initial ) = 100% x (1.3 x M)/(0.10 M) = 1.3 % Note: x < 5% of 0.10, OK to make this approximation

For a weak base B(aq) + H 2 O(l)  HB + (aq) + OH - (aq) Use a similar approach to determine pOH knowing K b, and then determine pH Determine the pH and percentage protonation of a 0.20 M aqueous solution of methylamine, CH 3 NH 2. The K b for CH 3 NH 2 is 3.6 x pH = 11.9 % protonation = 4.2%

pH of Salt Solutions CH 3 COOH(aq) + NaOH(aq)  CH 3 COONa(aq) + H 2 O(l) “neutralization” reaction If a 0.3M solution of CH 3 COOH(aq) is added to a 0.3M solution of NaOH, pH of resulting solution is not 7.0 but ~ 9.0 Solution of a salt is a solution of an acid (usually the cation) and a base (usually the anion), and the pH depends on their relative strength. CH 3 COO - (aq) determines the pH of the solution

Ni(H 2 O) 6 2+ (aq) + H 2 O(l)  H 3 O + (aq) + Ni(H 2 O) 5 (OH) + (aq)

Estimate the pH of 0.15 M NH 4 Cl(aq). K b (NH 3 (aq)) = 1.8 x NH 4 + (aq) is an acid and Cl - (aq) is neutral; expect pH < 7 NH 4 + (aq) + H 2 O(l)  H 3 O + (aq) + NH 3 (aq) K a = [NH 4 + (aq) ] [H 3 O + (aq)][NH 3 (aq)] NH 4 + (aq)NH 3 (aq)H 3 O + (aq) Initial Change-x x x Equilibrium0.15-x x x

K a (NH 4 + (aq)) = KwKw K b (NH 3 (aq)) 5.6 x = x2x x Assume x << 0.15 x ≈ 9.2 x (agrees with the assumption) pH = - log(9.2 x ) = 5.04

Polyprotic Acids & Bases A polyprotic acid can donate more than one H + Carbonic acid: H 2 CO 3 (aq); dissolved CO 2 in water Sulfuric acid: H 2 SO 4 (aq) Phosphoric acid: H 3 PO 4 (aq) A polyprotic base: can accept more than one proton Carbonate ion: CO 3 2- (aq) Sulfate ion: SO 4 2- (aq) Phophate ion: PO 4 3- (aq) Treat each step of protonation or deprotonation sequentially

H 2 CO 3 (aq) + H 2 O(l)  H 3 O + (aq) + HCO 3 - (aq)K a1 = 4.3 x HCO 3 - (aq) + H 2 O(l)  H 3 O + (aq) + CO 3 2- (aq)K a2 = 4.8 x Typically: K a1 >> K a2 >> K a3 >>… Harder to loose a positively charged proton from a negatively charged ion, because of attraction between opposite charges.