Buffered Solutions (Acid/Salt) and pH Changes

Basic Concept

A buffered solution is one that resists change in pH when either hydroxide ions or protons are added. A buffered solution may contain a weak acid and its salt or a weak base and its salt. 

This module calculates the pH of buffered solutions and the pH changes in the solution when a strong base is added.

1. pH of a buffered solution

In the buffered solution containing 0.50M acetic acid (HC₂H₃O₂, K_a = 1.8 x 10^-5) and 0.50M sodium acetate (NaC₂H₃O₂), the dissociation of the acetic acid will control the pH of the solution:

        HC₂H₃O₂(aq)  = H⁺ (aq) + C₂H₃O₂^- (aq)

        K_a = [H⁺][C₂H₃O₂^-] / [HC₂H₃O₂] = 1.8 x 10^-5

The concentrations are as follows: 

               HC₂H₃O₂ (aq)  =     H⁺ (aq) +     C₂H₃O₂^- (aq)

Initial:         0.50                        0                    0.50

Change:       -x                           x                   x

Equilibrium:    0.50 - x                 x                   0.50 + x 

             K_a = 1.8 x 10^-5 = [H⁺][C₂H₃O₂^-] / [HC₂H₃O₂] = (x) (0.50 + x) / (0.50 - x) (x) (0.50) / 0.50 

              x 1.8 x 10^-5

                 pH = 4.74

    2. pH changes in buffered solutions

When added 0.010 mol solid NaOH is added to 1.0L of the buffered solution, the stoichiometry for the reaction is as follows:

                               HC₂H₃O₂     +     OH^-              ==    C₂H₃O₂^-  + H₂O

Before reaction:        0.50 mol              0.010 mol            0.50 mol

After reaction:           0.49 mol              0                         0.51 mol

The equilibrium of the solution can be shown in a shorthand notation:

               HC₂H₃O₂ (aq) ==     H⁺ (aq) +     C₂H₃O₂^- (aq)

Initial:         0.49                        0                    0.51

Change:       -x                           x                   x

Equilibrium:    0.49 - x                 x                   0.51 + x 

                K_a = 1.8 x 10^-5 = [H⁺][C₂H₃O₂^-] / [HC₂H₃O₂] = (x) (0.51 + x) / (0.49 - x) (x)(0.51) / 0.48

Thus

                x 1.7 x 10^-5

                [H⁺] = x =  1.7 x 10^-5 M

                pH = 4.76

The change in pH from adding 0.010 mol OH^- to the buffered solution is only

                4.76 - 4.74 = +0.02

If 0.01 mol solid NaOH is added to 1.0L water, the pH will change from 7.00 to 12.00 (+5.00).

 User Instructions

This is two step processes, enter the known data and press Calculate to output the unknowns.

1. Select Buffered Solution Acid/Salt link from the front page or Buffered Solution Acid/Salt tab from the Acid, Baseand Salt module. The Input and Output screen appears.

2. In the Input area, enter the known quantities with a proper significant figure with the above example.

3. Click Calculate to output the answer.

4. The Show Work area on the right shows you step-by-step how your problem has been solved.

To start a new problem, click Reset. All Input fields will be cleared. Follow Step 1-3 again.