Chemistry: The Central Science Acid-Base Equilibria Chemistry: The Central Science Chapter 16 Stacey Dobrosky Cool and Ray Guest

Arrhenius Definition Acids produce hydrogen ions when dissolved in water. HCl → H+ + Cl- Bases produce hydroxide ions when dissolved in water. NaOH → Na+ + OH- Problem: NH3 (ammonia) could not be an Arrhenius base.

A Brønsted–Lowry acid… …must have a removable (acidic) proton. A Brønsted–Lowry base… …must have a pair of nonbonding electrons.

Brønsted-Lowry Definitions According to this theory, an acid is a proton donor and a base is a proton acceptor EOS

If it can be either an acid or base… ...it is amphiprotic (amphoteric). HCO3− HSO4− H2O

What Happens When an Acid Dissolves in Water? Water acts as a Brønsted–Lowry base and extracts a proton (H+) from the acid. As a result, the conjugate base of the acid and a hydronium ion are formed. NH3 works under this definition as a base

Conjugate Acids and Bases: From the Latin word conjugare, meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids.

The conjugate acid of a base is the base plus the attached proton and the conjugate base of an acid is the acid minus the proton EOS

Identify the acid, base, conjugate acid, conjugate base and pairs HCN + NH3  CN- +NH4+ Acid – HCN Base – NH3 Conjugate Base - CN- Conjugate Acid - NH4+

Lewis Acids Lewis acids are defined as electron-pair acceptors. Includes cations and incomplete octets. Atoms with an empty valence orbital can be Lewis acids.

Lewis Bases Lewis bases are defined as electron-pair donors. Should have a lone pair of electrons. Anything that could be a Brønsted–Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however.

Lewis Acids and Bases Boron triflouride wants more electrons, because it has an incomplete octet. F F H H B F F :N H B N H F H F H

Acid and Base Strength Strong acids are completely dissociated in water. Their conjugate bases are quite weak. Weak acids only dissociate partially in water. Their conjugate bases are weak bases.

Acid and Base Strength Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong.

Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl(aq) + H2O(l)  H3O+(aq) + Cl−(aq) H2O is a much stronger base than Cl−, so the equilibrium lies so far to the right K is not measured (K>>1).

Acid and Base Strength HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq) Acetate ion is a stronger base than H2O, so the equilibrium favors the left side (K<1).

Autoionization of Water As we have seen, water is amphoteric. In pure water, a few molecules act as bases and a few act as acids. This is referred to as autoionization. H2O(l) + H2O(l) H3O+(aq) + OH−(aq)

pH Many of the concentration measurements in acid-base problems are given to us in terms of pH and pOH. p (anything) = -log (anything) pH = -log [H+] pOH = -log [OH-] pKa = -log Ka

pH pH= -log[H+] Used because [H+] is usually very small As pH decreases, [H+] increases exponentially

pH – cont. When [H+] = [OH-], the solution is neutral, and pH=7 When [H+] is greater than [OH-], the solution is acidic and pH is less than 7 When [H+] is less than [OH-], the solution is basic, and pH is greater than 7

pH – cont. Since pH is based on a log scale, every 1 unit change in pH is 10 times the concentration. pH of 3 = 10 times more concentrated than pH of 4, and 100 times more concentrated than pH of 5

pH and pOH pH is a measure of the strength of an acid; pH = –log[H3O+] and [H3O+] = 10–pH pH is a measure of the strength of an acid; low pH = stronger acid pOH = –log[OH–] and [OH–] = 10–pOH pOH is a measure of the strength of a base; low pOH = stronger base EOS

pH of strong acids Finding the pH of strong acids is easy since there is almost complete dissociation Common strong acids include: HCl, HBr, HI, HNO3, H2SO4, HClO4

pH of weak acids Weak acids do not completely dissociate in water. Equilibrate!! Problems solved similar to equilibrium problems.

Autoionization of Water S Autoionization of Water Under the Brønsted-Lowry definition of acids and bases, water can be considered both an acid and a base A substance that can act as both acid and base is called amphoteric or amphiprotic An amphoteric substance reacts as a base when combined with something more acidic than it (and vice versa)

Autoionization of Water Water autoionizes, by donating a proton to another water molecule: 2H2O  H3O+ + OH- or H2O  H+ + OH- No individual ion remains ionized for long At room temperature only 2 out of every 109 molecules are ionized at any given moment

Autoionization of Water H2O  H+ + OH- This autoionization helps to see the amphoteric nature of water Demo

Autoionization of Water H2O  H+ + OH- Since autoionzation of water is an equilibrium process, we can write the equilibrium constant for it: Kc = [H+][OH-] The concentration of water is excluded, because the concentration of any pure liquid is one

Autoionization of Water H2O  H+ + OH- Since this is specifically for the autoionization of water we use Kw as the equilibrium constant. At 25oC: Kw = [H+][OH-] = 1.0 x 10-14 [H+] > [OH-]  acid [H+] < [OH-]  base [H+] = [OH-]  neutral

Autoionization of Water H2O  H+ + OH- Kw = [H+][OH-] = 1.0 x 10-14 At this point it is very useful to remember that pH = - log [H+] Since Kw is a constant, if we are given [OH-] it is a very simple matter to calculate pH

Acid dissociation constant = Ka R Acid dissociation constant = Ka Acid dissociation constant is the equilibrium constant for an acid. HA(aq) + H2O(l) H3O+(aq) + A-(aq) Ka = [H3O+][A-] [HA] H3O+ is often written H+ , ignore the water in equation.

Acid dissociation constant Ka HA(aq) H+(aq) + A-(aq) Ka = [H+][A-] [HA] We can write the expression for any acid. Strong acids dissociate completely. Equilibrium far to right. Conjugate base is weak.

Write the acid dissociation/ionization reaction and its Ka (omit water) HC2H3O2 for Acetic Acid

Kb = Base dissociation constant B(aq) + H2O(l)  BH+(aq) + OH-(aq) Kb =[BH+][OH-] [B]

Strong Acids Equilibrium lies far to the right Almost 100% dissociation HA  H+ + A- 1% 99% Gives a weak conjugate base (weaker than water) Large Ka

Weak acids Equilibrium lies to the left Very little H+ HA  H+ + A- 99% 1% Strong conjugate base (stronger than water) Small Ka

Back to Pairs Strong acids Ka is large [H+] is >>> to [HA] A- is a weaker base than water Weak acids Ka is small [H+] <<< [HA] A- is a stronger base than water

Weak Acids and Bases For weak acids and bases, equations can be written to describe equilibrium conditions EOS

Acid–Base Strength (cont’d) Ka values are used to compare the strengths of weak acids;  K,  strength EOS For strong acids, water has a leveling effect; that is, when the strong acids are dissolved in water, they all completely ionize to the hydronium ion

Strong Bases Large Kb Almost completely dissociated Conjugate acid is weak Alkali metal hydroxides NaOH, LiOH, KOH Ba(OH)2 Ca(OH)2 or Sr(OH)2

Kw, Ka, Kb, pH, Oh my! How does this all relate and tie together??? The symbols are actually very basic: Kw is the equilibrium constant for water Ka is the acid-dissociation constant Kb is the base-dissociation constant pH (power of hydrogen) is the –log of the concentration of hydrogen

Kw, Ka, Kb, Oh my! Given an acid reaction: HA(aq) + H2O(l)  H+(aq) + A-(aq) Ka = [H+] [A-] [HA] Given a base reaction: B(aq) + H2O(l)  HB+(aq) + OH-(aq) Kb = [HB+] [OH-] [B]

Kw, Ka, Kb, Oh my! Here we have a conjugate acid-base pair: NH4+(aq)  NH3 (aq) + H+ (aq) NH3(aq) + H2O(l)  NH4+ (aq) + OH- (aq) Ka = [NH3] [H+] [NH4+] Kb = [NH4+] [OH-] [NH3]

Mathematical Approach We can add the two equations together: NH4+(aq)  NH3 (aq) + H+ (aq) NH3(aq) + H2O(l)  NH4+ (aq) + OH- (aq) NH3(aq) + H2O(l) + NH4+(aq)  NH3 (aq) + H+(aq) + NH4+(aq) + OH- (aq) Cancel out the duplicate terms on opposite sides of the arrows… You are now left with: H2O  H+ + OH-

Mathematical Approach If we try to multiply Ka and Kb. this is what we get: Ka x Kb [NH3] [H+] x [NH4+] [OH-] = [NH4+] [NH3] Therefore… Ka x Kb = [H+] x [OH-] Ka x Kb = 1.0 x 10-14 = Kw

−log [H3O+] + −log [OH−] = −log Kw = 14.00 Watch This! Because [H3O+] [OH−] = Kw = 1.0  10−14, we know that −log [H3O+] + −log [OH−] = −log Kw = 14.00 or, in other words, pH + pOH = pKw = 14.00

Mathematical Approach KW = [H+][OH-] = 1.0 x 10-14 Ka × Kb = Kw In many books you will find a table for Ka they rarely are given for Kb since it is such an easy calculation

Relationships Kw = [H+][OH-] -log Kw = -log([H+][OH-]) -log Kw = -log[H+]+ -log[OH-] pKw = pH + pOH Kw = 1.0 x10-14 14.00 = pH + pOH Given any one of these we can find the other three: [H+], [OH-], pH and pOH