Acid-Base Strength: Ka, Kb, Kw OR any K really…

Relative Strengths Of Binary Acids H –X The greater the tendency for the transfer of a proton from HX to H 2 O, the more the forward reaction is favored and the stronger the acid. in a periodic group: The weaker the bond, the stronger the acid. The larger the resultant anion’s radius, the stronger is the acid. The strengths of binary acids increase from top to bottom in a group of the periodic table.

Relative Strengths Of Binary Acids H –X in a periodic group: Bond dissociation energy: the weaker the bond, the stronger the acid. Bond dissociation energy569> 431 > 368> 297 (kJ/mol)HF HCl HBr HI Acid strength K a 6.6x10 -4 < ~10 6 < ~10 8 < ~10 9 Anion radius: the larger the anion’s radius, the stronger the acid. Anion radius (ppm)136< 181 < 195< 216 (kJ/mol)HF HCl HBr HI Acid strength K a 6.6x10 -4 < ~10 6 < ~10 8 < ~10 9 The strength of binary acids increase from top to bottom in a group of the periodic table.

Relative Strengths Of Binary Acids H –X in a period: The larger the electronegativity difference between H and X, the more easily the proton is removed and the stronger is the acid.  EN 0.4<0.9<1.4< 1.9 Acid strengthCH 4 NH 3 H 2 O HF The strengths of binary acids increase from left to right across a period of the periodic table.

Representative Trends In Strengths of Binary Acids

The Acid dissociation constant, Ka

Ka A weak acid only ionizes to a small extent and comes to a state of chemical equilibrium. We can determine how much it ionizes by calculating the equilibrium constant for this reaction, the ionization constant, Ka. The larger the Ka the more acid ions are found in solution and the stronger the acid because the more easily it donates a proton. The reverse is true for the smaller the Ka

HCOOH (aq) + H2O (l) H3O + (aq) + HCOO - (aq) Ka = [H3O+][HCOO-] [HCOOH] Notice how the Ka ignores the water since we are dealing with dilute solutions of acids, water is considered a constant, and when multiplied by both sides it is cancelled out.

Equilibrium In Solutions Of Weak Acids And Weak Bases weak acid:HA + H 2 O º H 3 O + + A - [H 3 O + ][A - ] K a = [HA] weak base:B + H 2 O º HB + + OH - [HB + ][OH - ] K b = [B] You need to be able to write acid and base ionization equations!!!

Practice: 1.A solution of a weak acid, “HA”, is made up to be 0.15 M. Its pH was found to be Calculate the value of Ka. Steps to follow: 1.Write balanced equation 2.Calculate [H+] using 10 -pH 3.Set up chart for equilibrium (ICE or i Δ f) 4.Solve using Ka expression

Answer 1.HA (aq) + H 2 O (l) H 3 O + (aq) + A - (aq) 2.[H3O+]= = M 3.Set up table: HAH2OH2O H3O+H3O+ A-A i c e r

4.Ka = [H 3 O + ][ A - ] [HA] = [0.0011][0.0011] [0.149] = 8.1 x 10 -6

Percent Dissociation The fraction of acid molecules that dissociate compared with the initial concentration of the acid. Percent Dissociation = [H3O+] x 100% [HA i ] For the previous question: Percent Dissociation = [0.0011] x 100% =0.73 % [0.15]

Practice: The ionization constant, Ka, for a hypothetical weak acid, HA, at 25°C is 2.2 x a) Calculate the [H3O+] of a 0.20 M solution of HA. b) Calculate the percent ionization of HA. c) Calculate the [A-]. d) What initial concentration of HA is needed to produce a [H3O+] of 5.0 x 10-3 M?

Answer a) HA(aq) + H2O(l) H3O+(aq) + A-(aq) Ka = [H3O+][A-] = 2.2 x [HA] 2.2 x = (x)(x) 0.20 M x 2 = (2.2 x ) (0.20 M) x = M The [H3O+] is M. Because it is a 1:1 ratio they are both the same concentration (x)

b) % ionization = M x 100% = 3.3% 0.20 M c) From the stoichiometry of the reaction, [H3O+] = [A-] Therefore, [A-] = M

Found d) from table set up d) 2.2 x = (5.0 x M)(5.0 x M) x – 5.0 x M 2.2 x (x – 5.0 x M) = 2.5 x x x – 1.1 x = 2.5 x x = (2.5 x x ) / 2.2 x x= 0.12 M The initial concentration of HA required is 0.12 M.

Acid And Base Ionization Constants weak acid:CH 3 COOH + H 2 O º H 3 O + + CH 3 COO - [H 3 O + ][CH 3 COO - ] Acid ionization constant: K a = [CH 3 COOH] weak base:NH 3 + H 2 O º NH OH - [NH 4 + ][OH - ] Base ionization constant: K b = [NH 3 ] Acid and base ionization constants are the measure of the strengths of acids and bases.

Kb When using weak bases, the same rules apply as with weak acids, except you are solving for pOH and using [OH-]

Another relationship Useful to know: Ka x Kb = Kw = 1.0 x