Acid/base problems may fall into 4 categories: strong acid/base, weak acid/base, buffers and hydrolysis.

Strong Acids and Strong Bases

Remember: The strength of the acid is determined by how far the equilibrium lies to the right. Qualitatively, this may be judged by the K_a of the acid. A large K_a indicates a strong acid; a small K_a indicates a weak acid.

Weak Acids and Bases

Weak acids and weak bases do not dissociate completely. Equilibrium exists between the weak acid, water, H₃O⁺, and the anion of the weak acid. The equilibrium lies to the left hand side of the equation, indicating that not much H₃O⁺ is being produced. The fact that very little H₃O⁺ is being produced is the definition of a weak acid. The K_a for a weak acid is small, usually a number less than 1.

Dissociation of a Weak Acid

In this type of problem, you will be asked to find the hydronium ion concentration and/or the pH of a weak acid whose initial concentration is known. A typical problem may be:

What is the hydronium ion concentration and pH of a 0.010 M solution of hypochlorous acid, K_a = 3.5 x 10^-8?

1^st write the balanced equation for the dissociation:

HClO(aq) + H₂O(l)ó H₃O⁺(aq) + ClO^­¯(aq)

2^nd write the Equilibrium expression:    Note: H₂O is not in the equation.  Why not?

3^rd set up your ICE table:       

HClO(aq)         ó       H₃O⁺(aq) +      ClO^­¯(aq)

                        Initial               0.010M                        0                      0

                        Change            -x                                 +x                    +x

                        Equilibrium     0.010-x                        x                      x

4^th put the value in the equilibrium expression and solve for x:

The solution for x becomes simplified because the x shown in the denominator can be neglected. This x can be neglected because it will be negligibly small compared to the concentration, 0.10 M. To determine whether x is negligible, compare the magnitude of the last decimal place of the concentration of the acid to the magnitude of the equilibrium constant. If the difference in magnitude is greater than 100, the x may be neglected. In this case, the concentration is known to the 10^-3 place and the equilibrium constant is the magnitude of 10^-8. The difference in magnitude is 10⁵, therefore, x may easily be neglected.

5^th calculate x:             x² = (0.010)(3.5x10^-8) = 3.5 x 10^-10  ;  x =1.87 x 10^-5M

6^th Calculate the pH using the newly found value of H₃O⁺ : [H₃O⁺] = 1.87 x 10^-5 M

pH = -log(.87 x 10^-5 M) = 4.73

The same process shown above works for the dissociation of a base as well, you simply need to calculate the [OH^-], calculate pOH and then convert to pH.

Problems:

1. What is the hydroxide ion concentration and pH of a 0.10 M solution of NH₃, K_b = 1.8 x 10^-5?

NH₃     + H₂O              ó        NH₄⁺   +          OH^-

Initial                           0.10                                         0                      0

Change                        -x                                             +x                    +x

Equilibrium                 0.10 – x                                   x                      x

Assume x small compared to concentration (Ka is ~10^-5 concentration is 10^-1, difference is 10,000).

                        1.8 x 10^-5 = x²/0.10                  x = 1.34 x 10^-3  so [OH] = 1.34 x 10^-3 M

pOH = -log(1.34 x 10^-3 M) = 2.9        SO pH = 14-2.9 = 11.1

1. Suppose you have a 1.0 L solution that is initially 0.20 M in acetic acid (HAc) and 0.10 M in sodium acetate (NaAc). What will be the pH of this solution?  What will the pH be if you add 2 mL of 1.0M HCl to the system? K_a = 1.8 x 10^-5

Original reaction:      HAc + H₂O ó H₃O⁺ + Ac^-

Initial                          0.20                 0          0.10    

Change                       -x                     +x        +x

Equilibrium               0.20 – x           x          0.10 + x

Again, assume x is small in comparison to the concentration of acid and base, simplify and calculate x:

K_a = 0.10x/0.20 = 1.8 x 10^-5            x = 3.6 x 10^-5               [H₃O] = 3.6 x 10^-5 M 

so pH = –log(3.6 x 10^-5M)  = 4.44

Adding Strong acid adds H⁺  : 0.002L x 1.0 mol/L = 0.002 moles added.  So the equilibrium is shifted to the left, back towards the weak acid.  We add the acid concentration to the acid and subtract it from the base: 

After the acid addition, the new concentrations are: [Hac] = ~0.202M ; [Ac^-] = ~0.098M.  These are our new initial concentrations.  The ~ is there because the concentrations include the 2 mL volume of acid.

Reaction:                   HAc    + H₂O   ó      H₃O⁺               + Ac^-

Initial                          ~0.202                         0                      ~0.098

Change                       -x                                 +x                    +x

Equilibrium               0.202 –x                      x                      0.098 + x

Assume x is small, then 1.8 x 10^-5 = 0.098x/0.202            x = 3.64 x 10^-5 M

New pH = 4.43

This is basically a buffer system so the change should be small!

Salt Hydrolysis –

When a strong acid and a strong base is neutralized in the reaction:

H⁺ + OH^- ó H₂O

Usually, a neutral salt is formed.

A salt formed between a strong acid and a weak base is an acid salt. Ammonia is a weak base, and its salt with any strong acid gives a solution with a pH lower than 7.

For example, let us consider the reaction:

HCl + NH₄OH ó NH₄⁺ + Cl^- + H₂O

In the solution, the NH₄⁺ ion reacts with water (called hydrolysis) according to the equation:

NH₄⁺ + H₂O ó NH₃ + H₃O⁺.

A basic salt is formed between a weak acid and a strong base. The basicity is due to the hydrolysis of the conjugate base of the (weak) acid used in the neutralization reaction. For example, sodium acetate formed between the weak acetic acid and the strong base NaOH is a basic salt. When the salt is dissolved, ionization takes place:

NaCH₃COO = Na⁺ + CH₃COO^-

In the presence of water, CH₃COO^- undergoes hydrolysis:

H₂O + CH₃COO^- = CH₃COOH + OH^-

Salts of weak acids and weak bases

A salt formed between a weak acid and a weak base can be neutral, acidic, or basic depending on the relative strengths of the acid and base.

If K_a(cation) > K_b(anion) the solution of the salt is acidic.

If K_a(cation) = K_b(anion) the solution of the salt is neutral.

If K_a(cation) < K_b(anion) the solution of the salt is basic.

How do you get K_b if you are only given K_a or vice versa? 

K_a*K_b = K_w = 1 x 10^-14                       or                     pK_a + pK_b = pK_w

Problems:

1. Is a solution of sodium acetate acidic, neutral or basic? Basic 2) Are solutions of ammonium chloride acidic, basic or neutral? Acidic 3) Calculate the pH of a 0.100 M KCN solution. 
   K_a(HCN) = 6.2E-10, K_b(CN-) = 1.6E-5.

CN^- + H₂O ó HCN  + OH^-

                I           0.1                   0          0

            C         -x                     +x        +x

            E          0.1-x                x          x

1.6 x 10^-5 = x²/0.1        assuming x small  x = 0.0013 = [OH^-]  pOH = 2.9 so pH = 11.1

Types of Acid-Base Reactions - Strong Acid and Strong Base: 

HA + MOH    à H₂O + MA             where MA is a salt. This is a neutralization reaction. The stoichiometry of the reaction determines the amount of acid or base remaining unreacted in the solution and thus the pH of the solution.  (Remember limiting and excess reagents?)

Problem:  What is the pH of a solution containing 10 mL of 1.0 M HCl and 20 mL of 1.0 M NaOH?  10 mL of base remains after reaction.  So 0.01 moles of base = pOH of 2, so the pH will be 14-2 = 12.

Strong Acid and Weak Base:

When you add together a strong acid with a weak base you still get a salt:

HCl(aq) + NH₃(aq) à NH₄⁺Cl(aq)

But like most salts, it is dissociated in water.  The NH₄⁺ as you know is an acid, so it further reacts with the water in the solution as:

NH₄⁺ + H₂O ó NH₃ + H₃O⁺

The pH of the solution will be determined by the K_a of  NH₄⁺.  (K_a = 5.6 x 10^-10)

Problem: What will be the pH of a solution containing 20 mL of 1M HCl and 20 mL of 1M NH₃?  HCl(aq) + NH₃(aq) à NH₄⁺Cl(aq) ; 0.02 moles of NH₄⁺Cl(aq) produced in the reaction so start with that:

                        NH₄⁺ + H₂O ó NH₃ + H₃O⁺

            I           0.02                             0          0

            C         -x                                 +x        +x

            E          0.02 – x                       x          x

5.6 x 10^-10 = x²/0.02    assuming small x.      x = 3.35 x 10^-6 = [H₃O⁺]  pH = 5.5

Strong Base and Weak Acid:

When you add together a strong base with a weak acid you still get a salt:

CH₃COOH (aq) + NaOH(aq) ó NaCH₃COO^¯(aq) + H₂O(l)

But like most salts, it is dissociated in water.  The CH₃COO^¯ as you know is the conjugate base of the weak acid, so it further reacts with the water in the solution as:

CH₃COO^¯(aq) + H₂O(l) ó CH₃COOH(aq) + OH^-(aq)

The pH of the solution is thus dependent on the K_b of the anion of the salt. K_b = 5.6x10^-10

Problem: What will be the pH of a solution containing 20 mL of 1M CH₃COOH and 20 mL of 1M NaOH?  CH₃COOH (aq) + NaOH(aq) ó NaCH₃COO^¯(aq) + H₂O(l) ; 0.02 mol of  NaCH₃COO^¯ are made so starting with that:

                        CH₃COO^¯(aq) + H₂O(l) ó CH₃COOH(aq) + OH^-(aq)

            I           0.02                                         0                      0

            C         -x                                             +x                    +x

            E         0.02-x                                      x                      x

5.6 x 10^-10 = x²/0.02    assuming small x.      x = 3.35 x 10^-6 = [OH^-]  pOH = 5.5 so pH = 8.5

Weak Base and Weak Acid:               K_Rxn = K_w/K_aK_b  ; [H₃O] = (K_wK_a/K_b)^1/2

This produces a salt composed of the conjugate acid and base.  The pH depends on the K_a and K_b of those species.