Buffered Solutions (Base/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 acid is added.

1. pH of a buffered solution

In a buffered solution example that contains 0.25M NH3 (Kb = 1.8 x 10-5) and 0.40M NH4Cl, the equilibrium is:

        NH3(aq)  + H2O = NH4+ (aq) + OH- (aq)

        Kb = [NH4+][OH-] / [NH3] = 1.8 x 10-5

The concentrations are as follows: 

               NH3(aq)   =  N H4+ (aq) + OH- (aq)

Initial:         0.25           0.40                0

Change:       -x               x                   x

Equilibrium:    0.50 - x     0.40 + x         x 

           Kb = 1.8 x 10-5 = [N H4+][OH-] / [NH3] = (0.40 + x) (x) / (0.25 - x) (0.40) (x) / 0.25 

            x 1.1 x 10-5

               pOH = 4.95

            pH = 14.00 - 4.95 = 9.05

 2. pH changes in buffered solutions

When 0.10 mol gaseous HCl is added to 1.0L the buffered solution, the stoichiometry is shown as follows:

                               NH3     +             H+              ==    NH4+ 

Before reaction:         0.25 mol            0.10 mol             0.40 mol

After reaction:           0.15 mol            0                         0.50 mol

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

                   NH3     +             H+              ==    NH4+ 

Initial:         0.15                     0                        0.50

Change:       -x                        x                        x

Equilibrium:  0.15 - x                x                       0.50 + x 

             Ka = Kw / Kb = 5.6 x 10-10 = [H+][NH3] / [NH4+] = (x) (0.15 -  x) / (0.50 + x) (x)(0.15) / 0.50

Thus

                x 1.9 x 10-9

                [H+] = x =  1.9 x 10-9 M

                pH = 8.73

The change in pH by adding 0.10 mol H+ to this buffered solution is only

                8.73 - 9.05 = -0.32

If 0.10 mol HCl is added to 1.0L water, the pH will from 7.00 to 1.00 (-6.00).

 User Instructions

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

1. Select Buffered Solution Base/Salt link from the front page or Buffered Solution Base/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.