CHAPTER 13 Acids and Bases 13.3 Acid-Base Equilibria

HCl(aq) → H+(aq) + Cl–(aq) Strong: Complete dissociation HCl(aq) → H+(aq) + Cl–(aq) 1 M Acids 1 M < 0.02 M H2C6H6O6(aq) → 2H+(aq) + C6H6O6–(aq) Weak: Partial dissociation Strong: Complete dissociation Ca(OH)2(aq) → Ca2+(aq) + 2OH–(aq) 1 M 2 M Bases Weak: Partial dissociation 1 M ~ 0.004 M NH3(aq) + H2O(l) NH4+(aq) + OH–(aq)

Incomplete dissociations are examples of chemical equilibria We will learn how to calculate the pH of a weak acid or a weak base solution 1 M < 0.02 M H2C6H6O6(aq) → 2H+(aq) + C6H6O6–(aq) Weak: Partial dissociation Acids Weak: Partial production of OH- 1 M ~ 0.004 M NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) Bases

Dissociation of water and Kw H2O(l) H+(aq) + OH–(aq) For dilute solutions there is an equilibrium between [H+] and [OH–] ion product constant: in dilute aqueous solutions, [H+] x [OH–] = 1.0 x 10–14 = Kw at 25oC.

A certain solution has a concentration of H+ ions of 0. 0005 M A certain solution has a concentration of H+ ions of 0.0005 M. What is the concentration of OH– ions? Is the solution acidic or basic?

A certain solution has a concentration of H+ ions of 0. 0005 M A certain solution has a concentration of H+ ions of 0.0005 M. What is the concentration of OH– ions? Is the solution acidic or basic? Asked: [OH–] and pH of the new solution Given: [H+] = 0.0005 M Relationships: [H+] x [OH–] = 1.0 x 10–14 pH = –log[H+]

A certain solution has a concentration of H+ ions of 0. 0005 M A certain solution has a concentration of H+ ions of 0.0005 M. What is the concentration of OH– ions? Is the solution acidic or basic? Asked: [OH–] and pH of the new solution Given: [H+] = 0.0005 M Relationships: [H+] x [OH–] = 1.0 x 10–14 pH = –log[H+] Solve:

A certain solution has a concentration of H+ ions of 0. 0005 M A certain solution has a concentration of H+ ions of 0.0005 M. What is the concentration of OH– ions? Is the solution acidic or basic? Asked: [OH–] and pH of the new solution Given: [H+] = 0.0005 M Relationships: [H+] x [OH–] = 1.0 x 10–14 pH = –log[H+] Solve: Answer: The concentration of OH– ions is 2 x 10–11 M. The solution has a pH of 3.3, and is a moderately strong acid.

Ka is the equilibrium constant for an acid Complete dissociation Large Ka (1.3 x 106) Partial dissociation Small Ka (1.8 x 10–5) Ka is the equilibrium constant for an acid

The Ka for acetic acid is a small number(1.8 x 10–5) because [products] < [reactants]

pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). Write down the balanced net ionic equation

No product at the beginning of the reaction pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). No product at the beginning of the reaction Initial molarity Write down the balanced net ionic equation Set up the equilibrium relationship

pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). An unknown amount is consumed (–x) For every x of reactant consumed, x of product is produced (+x), 1:1 ratio Write down the balanced net ionic equation Set up the equilibrium relationship

Add the molarities from Initial and Change pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). + Add the molarities from Initial and Change Write down the balanced net ionic equation Set up the equilibrium relationship

Finding x will help us calculate pH pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). x = [H+] at equilibrium Finding x will help us calculate pH Write down the balanced net ionic equation Set up the equilibrium relationship Use the RICE table to solve for [H+]

pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).

pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5).

pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). Write down the balanced net ionic equation Set up the equilibrium relationship Use the RICE table to solve for [H+] Calculate the pH from [H+]

Same answer as from solving the quadratic equation pH of weak acids Find the pH of a 0.1 M solution of acetic acid (Ka = 1.8 x 10–5). Same answer as from solving the quadratic equation

pH of weak acids This equation only works when MKa > 100

Polyprotic acids H2A HA– + H+ A2– + 2H+ A polyprotic acid can dissociate more than once and donates more than one proton: H2A HA– + H+ A2– + 2H+ Carbonic acid is a polyprotic acid: pH = –log[H+]

Equilibrium constants A weak acid: HC2H3O2 H+ + C2H3O2– Ka NH3 + H2O NH4+ + OH– Kb acetic acid Equilibrium constants A weak base: ammonia

pH of weak bases Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5). Write down the balanced net ionic equation

pH of weak bases Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5). Write down the balanced net ionic equation Set up the equilibrium relationship

pH of weak bases Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5). Write down the balanced net ionic equation Set up the equilibrium relationship Use the RICE table to solve for [OH–]

pH of weak bases Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5).

pH of weak bases Find the pH of a 1.0 M ammonia solution (Kb = 1.8 x 10–5). Write down the balanced net ionic equation Set up the equilibrium relationship Use the RICE table to solve for [OH–] Calculate the pH using pH = 14 + log[OH–]

Use a RICE table to find the pH of a weak acid or a weak base Once [H+] or [OH–] has been determined, the pH can be calculated