You may want to review the Acids and Bases unit before proceeding.

Strong acid - Dissociates more or less completely when it dissolves in water.  In other words, it is a very good H⁺ donor.

Reactions involving the dissociation of strong acids can be written with a regular forward arrow, because the equilibrium lies very far to the right, i.e. favoring the products (the dissociated ions).

Weak acid - Dissociates only slightly when it dissolves in water.  In other words, it is a poor H⁺ donor.

Reactions involving the dissociation of weak acids are written with a forward/backward arrow to indicate that the products are not heavily favored.

Acid-dissociation equilibrium constant (K_a) - A measure of the relative strength of an acid.  For the generic acid dissociation reaction with water,

The acid-dissociation equilibrium constant is the mathematical product of the equilibrium concentrations of the products of this reaction divided by the equilibrium concentration of the original acid,  In other words, K_a is the product of the concentrations of the products over the concentration of the reactants (not including water):

Water is not included in the acid-dissociation equilibrium expression because the [H₂O] has no effect on the equilibrium.

As the K_a value of an acid increases, so does the strength of the acid.  By definition:

• strong acid: K_a > 1
• weak acid: K_a < 1

When a strong acid reacts with water, water acts as a strong base.  The strong acid and the strong base react to form a weaker acid and a weaker base.  However, when a weak acid reacts with water, water acts as a weak base.  The reaction attempts to convert the weak acid and the weak base into a stronger acid and a stronger base, but the reaction cannot naturally proceed in this direction.  This is why strong acids dissociate nearly completely whereas weak acids dissociate only slightly.

Example Equilibrium Calculations

• Strong acid (or base) calculations
  • Strong acid calculation #1
  • Strong acid calculation #2
• Weak acid calculations
  • Weak acid calculation #1
    • with detailed step-by-step instructions
  • Summary of how to solve weak acid calculations
  • Weak acid calculation #2
  • Weak acid calculation #3 - When the Assumption Fails!!
• Mixtures (salt solutions)

Strong Acid (or Base) Calculations

These calculation problems are the easiest type of equilibrium problems.  Since strong acids (and bases) dissociate nearly 100% in water,

Therefore, it can be assumed that [H₃O⁺] at equilibrium is equal to the initial concentration of the acid.

So let's try a problem.

1) Calculate the pH of a 0.05 M HBr solution.

First, thing we should always do in any equilibrium problem is write the equilibrium reaction.  That way we never forget what is happening.

Note that the arrow is a regular forward arrow since this HBr a strong acid, and it dissociates almost 100% in water.  

As stated above, the [H₃O⁺] at equilibrium is equal to the initial concentration of the acid.

The pH is by definition the negative logarithm of the [H₃O⁺].

Let's try one more strong acid calculation.

2) Calculate the pH of a solution with 0.325 g HBr in 500. mL H₂O.

First, set up the reaction:

Since we are given the mass of the HBr and the volume of the solution, we have to calculate the molarity of the HBr solution:

• NOTE:  Molarity is in units of moles/Liter so the grams must be converted to moles (using the periodic table) and the milliliters must be converted to Liters.

Again, since HBr is a strong base, the initial concentration of the HBr is essentially equal to the [H₃O⁺] at equilibrium:

Now, used the pH formula to calculate the pH:

Here is a list of common strong acids and common strong bases to help determine which acid/base equilibrium problems can be solved using the above method.  It can be assumed that the folloiwng compounds dissociate nearly 100% in water, and therefore [H₃O⁺] (for acids) or [OH^-] (for bases) equals the initial acid or base concentration.

Weak Acid Calculations

These calculations are a bit more complicated because weak acids dissociate only slightly in water.

It can NOT be assumed that the [H₃O⁺] at equilibrium is equal to the initial concentration of the acid.

Let's try a sample problem.

1) Calculate the [H₃O⁺] and the pH for 0.462 M acetic acid.  (K_a for acetic acid = 1.8 x 10^-5)

First, set up the REACTION:

Note that the arrow in the above reaction is a forward/backward arrow since acetic acid is a weak acid.  The arrow indicates that the reaction does not proceed completely to the right.

In weak acid problems, it is helpful to set up a CHART for the initial and equilibrium concentrations of the reactants and the products.  When doing so, however, the reactant water (H₂O) is ignored because the [H₂O] does not effect the equilibrium.  The initial concentration of the acid is given in the problem.  The initial concentrations of the products are zero since the reaction has yet to happen.  At this point, it is unknown how much of the acid will dissociate (i.e. how far the reaction will proceed to reach equilibrium), so we assume that some unknown amount represented by the symbol delta () dissociates.  Therefore, the the equilibrium concentration of the acid is written as the initial concentration minus the unknown amount, .  The equilibrium concentration of the products are determined from the mole ratios of the reactants to the products.  According to the reaction equation above, one mole of H₃O⁺ is formed for every one mole of HC₂H₃O₂ consumed.  Likewise, one mole of C₂H₃O₂^- is formed for every one mole of HC₂H₃O₂ consumed.  Therefore the amount of H₃O⁺ and C₂H₃O₂^- formed (the equilibrium concentration) is equal to the unknown amount of the acid that dissociates, .

		[HC2H3O2]		[H3O+]		[C2H3O2-]
Initial:		0.462 M		0		0
Equilibrium:		0.462 M -

Next, write the ACID-DISSOCIATION EQUILIBRIUM EXPRESSION (K_a) for acetic acid.  This is the mathematical product of the concentrations of the products over the concentration of the acid:

Now SUBSTITUTE in the values.  The K_a should be provided in the problem.  The concentrations are obtained from the chart created above.  Remember to use the equilibrium values:

At this point, we could solve for by using the quadratic formula, however, this would be a lot of extra work.  Instead, we are going to:

ASSUME that is negligible when subtracted from the initial concentration of the acid

By doing this, we simplify the equilibrium expression:

Now, it is much easier to SOLVE for by simply multiplying both sides of the equation by the initial concentration of the acid and taking the square root of both sides of the equation:

After we have made the assumption that is negligible when subtracted and have solved for , we must CHECK our assumption to make sure that it was correct.  We check our assumption by calculating what percent of the original acid concentration actually dissociated:

• NOTE:  Remember that was the variable assigned to the unknown amount of acid that dissociated.

How do we determine if our assumption was valid or not??

As a general guideline, it is safe to assume that is negligible when subtracted if the percent of the original acid concentration that dissociates is less than 5%.

Since in this problem is much less than 5%, our assumption is valid.

Now, we can't forget what the purpose of the entire problem was.  Our task was to calculate the [H₃O⁺] and pH.  Well, according to our chart, = [H₃O⁺] at equilibrium, so we have completed the first part of the problem:

To find the pH of the solution, we simply use the pH formula:

Let's try another weak acid problem.

2) Calculate the [H₃O⁺] and pH for a 0.10 M solution of acetic acid.  (K_a = 1.8 x 10^-5)

1. Set up the reaction:

2. Set up the chart:

		[HC2H3O2]		[H3O+]		[C2H3O2-]
Initial:		0.10 M		0		0
Equilibrium:		0.10 M -

3. Write the K_a expression:

4. Substitute:

5. Assume that is negligible when subtracted.

6. Solve for .

7. Check the assumption.

Since the acid dissociates less than 5%, is negligible.

Weak Acid Calculation - When the Assumption Fails

3) Calculate the equilibrium concentrations of H₃O⁺, HClO₂, and ClO₂^- for a 0.10 M HClO₂ solution (K_a = 1.1 x 10^-2)

1. Set up the reaction:

2. Set up the chart:

		[HClO2]		[H3O+]		[ClO2-]
Initial:		0.10 M		0		0
Equilibrium:		0.10 M -

3. Write the K_a expression:

4. Substitute:

5. Assume that is negligible when subtracted.

6. Solve for .

7. Check the assumption.

The assumption fails so we have to go back to the expression before the assumption:

We have to now solve for by using the quadratic formula.  In order to use the quadratic formula, our equation has to be in the standard form:

So we have to rearrange our equation to fit the standard form:

The quadratic formula is:

Substitute in the values we found for a, b, and c and solve for :

According to our chart, = [H₃O⁺] = [ClO₂^-] and [HClO₂] = 0.10 - :

Mixtures (salt solutions)

What happens when an acid is added to a salt solution instead of just pure water.  How does the presence of a salt affect the dissociation of the acid?  To help answer these questions, consider the following problem.

Calculate the pH of a solution that is 0.10 M acetic acid AND 0.10 M sodium acetate.

First, let's write the reaction for the dissociation of the salt, sodium acetate and identify the concentrations of each of the components:

• NOTE: The concentrations of the ions are the same as the concentration of the salt because the mole ratio is 1:1.

Now, let's set up the REACTION for the dissociation of the acetic acid.

Now, we set up our concentration CHART.  Remember, however, that the initial concentrations for the conjugate base, C₂H₃O₂^- is no longer 0 because of the salt:

		[HC2H3O2]		[H3O+]		[C2H3O2-]
Initial:		0.10 M		0		0.10 M
Equilibrium:		0.10 M -				0.10 M +

Next, we write the ACID-DISSOCIATION EQUILIBRIUM EXPRESSION (K_a):

SUBSTITUTE in the values from the chart:

ASSUME is negligible when subtracted and added:

SOLVE for :

CHECK the assumption:

Since the acid dissociates less than 5%, is negligible when added and subtracted.

Monoprotic versus Polyprotic acids

All the acids discussed thusfar have been MONOPROTIC with a single H⁺ ion to donate.

In general, acids with more than one H⁺ ion availible to be donated are called POLYPROTIC acids

• DIPROTIC acids have two H⁺ ions which it can donate.
  • e.g.  H₂SO₄ and H₂CO₃

• TRIPROTIC acids have three H⁺ ions which it can donate.
  • e.g.  H₃PO₄

Polyprotic acids undergo a stepwise-dissociation in water, in which one H⁺ ion is lost at a time.

Next:  "Base-Ionization Equilibrium Constant"