# 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, Ka = 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)

Ka = [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

Ka = 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

Ka = 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.