Acids and Bases Chapter 20 Lesson 2

Definitions Acids – produce H + Bases - produce OH - Acids – donate H + Bases – accept H + Acids – accept e - pair Bases – donate e - pair Arrehenius Bronsted-Lowry Lewis only in water any solvent used in organic chemistry, wider range of substances

HA Let’s examine the behavior of an acid, HA, in aqueous solution. What happens to the HA molecules in solution?

HA H+H+ A-A- Strong Acid 100% dissociation of HA Would the solution be conductive?

HA H+H+ A-A- Weak Acid Partial dissociation of HA Would the solution be conductive?

HA H+H+ A-A- Weak Acid HA  H + + A - At any one time, only a fraction of the molecules are dissociated.

Ka (Dissociation Constant) 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 + (aq) + A - (aq) 2.[H+]= = M 3.Set up table: HAH2OH2O H+H+ A-A i Δ f

4.Ka = [H + ][ A - ] [HA] = [0.0011][0.0011] [0.139] = 8.7 x 10 -6

Percent Dissociation The fraction of acid molecules that dissociate compared with the initial concentration of the acid. Percent Dissociation = [H+] 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 [H+] 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 [H+] of 5.0 x M?

Answer a) HA(aq) + H2O(l) H+(aq) + A-(aq) Ka = [H+][A-] = 2.2 x [HA] 2.2 x = (x)(x) 0.20 M x 2 = (2.2 x ) (0.20 M) x = M The [H+] 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, [H+] = [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

Buffer Solutions A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added. A buffer contains a weak acid with its salt (conjugate base) or a weak base with its salt (conjugate acid) CH 3 COOH/CH 3 COONa NH 3 /NH 4 Cl

Depicting Buffer Action

How A Buffer Solution Works The acid component of the buffer can neutralize small added amounts of OH -, and the basic component can neutralize small added amounts of H 3 O +. Pure water does not buffer at all.