Lesson 2: Acid-Base Equilibria

Part a: Weak Acid and Weak Base Dissociation

Part a: Weak Acid and Weak Base Dissociation
Part b: Percent Dissociation and K_a and K_b
Part c: Dissociation of Polyprotic Acids
Part d: Determination of the pH of Salts
Part e: Common Ion Effects and Buffers
Part f: Titration Analysis

 

The Big Idea

Weak acids and bases only partially dissociate. By using K_a and K_b, ICE tables, and some algebra, we can predict pH and ion concentrations. This page demonstrates the details of how to solve such problems with numerous examples.

 

Acid and Base Equilibrium Systems

We learned in Chapter 15 that weak acids and weak bases only partially dissociate in water. Once they reach an equilibrium, the predominant species is the undissociated acid or base. The analysis of such systems requires the use of an equilibrium constant - K_a for weak acids and K_b for weak bases. These equilibrium constants are indicators of the relative amount of dissociation.
 
By combining a K_a or K_b equation with an ICE table, the equilibrium concentrations and the pH of the equilibrium system can be determined. On this first page of Lesson 2, we will analyze the dissociation of acids and bases to relate the equilibrium constants to the equilibrium concentrations and pH values.
 
As is the case in many areas of Chemistry, success in Lesson 2 will require the use of skills developed in earlier chapters of our Chemistry Tutorial. Such skills (and links to help pages that address the skill) include:

1. Writing equilibrium constant equations for a reaction. (If necessary, review Writing K Expressions.)
2. Using the Bronsted-Lowry model to write proton transfer equations for a chemical reactions involve acids and bases. (If necessary, review Bronsted-Lowry Proton Transfer Reactions.)
3. Mathematically relating pH, pOH, [H₃O⁺], and [OH^-]. (If necessary, review The pH Scale.)

 
 
 

How to Approach an Acid-Base Equilibrium Problem

Lesson 2 will involve a wide collection of problem types. While the approach to each type of problem may vary slightly, nearly every problem will share these strategic steps.

1. Write the balanced equation for the acid-base reaction.
2. Write the equilibrium constant equation for the reaction.
3. Set up and  complete an ICE table that is consistent with the given information. There will typically be a blend of numerical values and unknown variables (i.e., x) in the table cells.
4. Substitute values and expressions from the equilibrium row of the ICE table into the equilibrium constant equation.
5. Use algebra to solve for the unknown quantity.

 
As we proceed through this Lesson, we will use these steps as we approach each problem.  
 

Determining Ion Concentrations

The first type of problem we will analyze involves an acid (or a base) with a known concentration and known K_a (or K_b) value. The problem will involve the determination of the concentrations of ions at equilibirum.
 
 
 

Example 1 - Ion Concentrations of an HCN Solution

Determine the ion concentrations of a 0.10 M solution of hydrocyanic acid, HCN. The K_a is 6.2x10^-10.
 

Solution:

Step 1: The hydrocyanic acid will transfer a proton to water and form its conjugate base (CN^-). The water accepts the proton and becomes its conjugate acid (H₃O⁺). The balanced equation is:
   
Step 2: The equilibrium constant equation is:
   
Step 3: The completed ICE table is shown below. Liquids are not included in the equilibrium constant equation. As such, we do not worry about the water column of the ICE table. The Initial concentration of HCN is 0.10 M (given). The ion concentrations are initially 0 M. The reaction proceeds to the right to reach equilibrium, meaning that the reactant amounts decrease by x mol/L while the product amounts increase by x mol/L. The Equilibrium row of the table is always the Change row added to the Initial row. Presumably, x is a very small number since weak acids only partially dissociate. The result is that the 0.10 - x is approximately equal to 0.10. For instance, if x were 0.00010, then
   
This simplifies the algebra associated with the solution. We regard the assumption that 0.10 - x  ≈ 0.10 to be valid if the value of x is less than 5% the initial concentration of the acid. This is often referred to as the 5% Rule.
   
Step 4: The value of K_a and the value and expressions in the equilibrium row of the ICE table are substituted into equilibrium constant equation from Step 2.
   
Step 5: The unknown quantities in this problem are the concentrations of the two ions - H₃O⁺ and CN^-. As shown in the last row of the ICE table, these concentrations are equal to x. Algebra is performed on the equation in Step 4 to solve for x.
   
 
 

Example 2 - Ion Concentrations of an NH₃ Solution

Determine the ion concentrations of a 0.50 M solution of NH₃. The K_b is 1.8x10^-5.
 

Solution:

Step 1: The ammonia is a base and will accept a proton from water and form its conjugate acid (NH₄⁺). The water donates the proton and becomes its conjugate base (OH^-). The balanced equation is:
   
Step 2: The equilibrium constant equation is:
   
Step 3: The completed ICE table is shown below. Liquids are not included in the equilibrium constant equation. As such, we do not worry about the water column of the ICE table. The Initial concentration of NH₃ is 0.50 M (given). The ion concentrations are initially 0 M. The reaction proceeds to the right to reach equilibrium, meaning that the reactant amounts decrease by x mol/L while the product amounts increase by x mol/L. The Equilibrium row of the table is always the Change row added to the Initial row. Presumably, x is a very small number since weak bases only partially dissociate. We will make the simplifying assumption that 0.50 - x is approximately equal to 0.50 and check that the 5% rule is satisfied. This simplifies the algebra associated with the solution.
   
Step 4: The value of K_a and the value and expressions in the equilibrium row of the ICE table are substituted into equilibrium constant equation from Step 2.
   
Step 5: The unknown quantities in this problem are the concentrations of the two ions - NH₄⁺ and OH^-. These concentrations are equal to x. Algebra is performed on the equation in Step 4 to solve for x.
   
The equilibrium concentrations are [NH₄⁺] = 3.0x10^-3 M and [OH^-] = 3.0x10^-3 M.
 
 
 

Determining the pH of a Weak Acid or a Weak Base

We learned in Chapter 15 of this Chemistry Tutorial that hydronium ion concentrations ([H₃O⁺]), hydroxide ion concentrations ([OH^_-]), pH, and pOH values are all related. The mathematical relationships between these quantities was carefully presented and numerous examples were shown to demonstrate the use of the relationships. The graphic organizer below lists the four quantities and shows how to determine one quantity from another quantity.  Review our Chapter 15 page on The pH Scale if necessary.
   
In this chapter, we will use these relationships to relate equilibrium concentrations of ions to the pH and pOH values. The following two examples demonstrate these relationships.
   
 

Example 3 - pH of an HC₂H₃O₂ Solution

Determine the pH of a 1.0 M solution of acetic acid, HC₂H₃O₂. The K_a is 1.8x10^-5.
 

Solution:

Step 1: The acetic acid will transfer a proton to water and form its conjugate base (C₂H₃O₂^-). The water accepts the proton and becomes its conjugate acid (H₃O⁺). The balanced equation is:
   
Step 2: The equilibrium constant equation is:
   
Step 3: The completed ICE table is shown below. It is completed in the same manner as it was for Example 1. To simplify the algebra, we assume that ...
     
Step 4: The value of K_a and the value and expressions in the equilibrium row of the ICE table are substituted into equilibrium constant equation from Step 2.
   
Step 5: The unknown quantity in this problem is the pH. We will need to determine the H₃O⁺ concentration in order to determine the pH. This concentration is equal to x. Algebra is performed on the equation in Step 4 to solve for x.
   
The pH can be determined using the equation from Chapter 15 of our Chemistry Tutorial: pH = -log([H₃O⁺]) where [H₃O⁺] is 4.2x10^-3 M.
   
 
 

Example 4 - pH of a C₆H₅NH₂ Solution

Determine the pH of a 0.10 M solution of aniline (C₆H₅NH₂), a weak base. The K_b is 7.1x10^-10.
 

Solution:

Step 1: The aniline is a base and will accept a proton from water to form its conjugate acid (C₆H₅NH₃⁺). The water donates the proton and becomes its conjugate base (OH^-). The balanced equation is:
   
Step 2: The equilibrium constant equation is:
   
Step 3: The completed ICE table is shown below. It is completed in the same manner as it was for Example 2. And to simplify the algebra, we assume that ...
     
Step 4: The value of K_b and the value and expressions in the equilibrium row of the ICE table are substituted into equilibrium constant equation from Step 2.
   
Step 5: The unknown quantity in this problem is the pH. We will need to determine the OH^- concentration in order to determine the pH. This concentration is equal to x. Algebra is performed on the equation in Step 4 to solve for x.
   
The value of X is the hydroxide ion concentration, [OH^-]. It can be used to determine the pOH. Once determined, the pH can be calculated. We will use the equation from Chapter 15 of our Chemistry Tutorial: pOH = -log([OH^-]) where [OH^-] is 8.42614 ...x10^-6 M.
   
The pH and pOH value must add to 14.00. Thus, the pH is ...
   
 

Determining the K_a or K_b of a Weak Acid or Base

A pH value and acid or base concentration can be used to determine the unknown K_a or K_b of a weak acid or a weak base. Since we know the pH, we can determine the equilibrium concentrations of the two ions. The K_a or K_b value can then be calculated. The step-by-step solution proceeds as it did in the previous examples.
 
 
 

Example 5 - From pH to K_a for a Weak Acid

A 2.5 M solution of an unknown acid has a pH of 4.12. Determine its K_a value.
 

Solution:

Step 1: We will represent the acid by the generic formula HA. Its dissociation equation is:
   
Step 2: The equilibrium constant equation is:
   
Step 3: The completed ICE table is shown below. It is completed in the same manner as it was for Example 1.
   
Step 4: The equilibrium concentration expressions in the equilibrium row of the ICE table are substituted into equilibrium constant equation from Step 2.
   
Step 5: This problem is a bit different than the others. We are looking to solve for K_a. We know the pH of the solution at equilibrium. We will use an equation from Chapter 15 of our Chemistry Tutorial to calculate the hydronium ion concentration from the pH. The [H₃O⁺] is equal to the x in the ICE table.
   
This is the value of x. We can substitute x into the K_a equation from Step 4 to solve for K_a:
   
 
 

Example 6 - From pH to K_b for a Weak Base

A 1.5 M solution of an unknown base has a pH of 11.42. Determine its K_b value.
 

Solution:

Step 1: We will represent the base by the generic formula B. Its dissociation equation is shows the proton (H⁺ ion) being transferred from water to the base:
   
Step 2: The equilibrium constant equation is:
   
Step 3: The completed ICE table is shown below. It is completed in the same manner as all other examples.
   
Step 4: The algebraic expressions in the equilibrium row of the ICE table are substituted into equilibrium constant equation from Step 2.
   
Step 5: We are looking to solve for K_b.  We need to determine x to solve for K_b. We know the pH of the solution at equilibrium. The pH is related to the pOH and the pOH is related to the [OH^-] in the ICE table. The pOH is 14.00 - pH:
   
We will use an equation from Chapter 15 of our Chemistry Tutorial to calculate the hydroxide ion concentration ([OH^-]) from the pOH. The [OH^-] is equal to the x in the ICE table.
   
This is the value of x. We can substitute x into the K_b equation from Step 4 to solve for K_b:
   
 

 

Before You Leave - Practice and Reinforcement

Now that you've done the reading, take some time to strengthen your understanding and to put the ideas into practice. Here's some suggestions.
 

• Our Calculator Pad section is the go-to location to practice solving problems. You’ll find plenty of practice problems on our Solution Equilibria page. Check out the following problem sets:  Set SAB5: K_a, K_b, and pH 1 || Set SAB6: K_a, K_b, and pH 2
• The Check Your Understanding section below includes questions with answers and explanations. It provides a great chance to self-assess your understanding.
• Download our Study Card on Mathematics of Weak Acid and Base Solutions. Save it to a safe location and use it as a review tool. (Coming Soon.)

 
 
 

Check Your Understanding of K_a and K_b Problems

Use the following questions to practice your skill at solving K_a and K_b problems similar to those on this page. Tap the Check Answer buttons when ready.
 
1. Benzoic acid, HC₆H₅CO₂, has a K_a value of 6.4 × 10^-5. Determine the pH of a 1.56 M benzoic acid solution.
 

 
2. A 2.5 M solution of a weak base, B, undergoes dissociation to form an aqueous solution with a pH value of 11.25. The dissociation equation is:
 
B_(aq)   +   H₂O_(l)    ⇄    BH⁺_(aq)   +   OH^-_(aq)
 
Determine the K_b of the weak base.
  
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