Chemistry
Textbooks
Boundless Chemistry
Acid-Base Equilibria
Buffer Solutions
Chemistry Textbooks Boundless Chemistry Acid-Base Equilibria Buffer Solutions
Chemistry Textbooks Boundless Chemistry Acid-Base Equilibria
Chemistry Textbooks Boundless Chemistry
Chemistry Textbooks
Chemistry
Concept Version 7
Created by Boundless

Buffers Containing a Base and Conjugate Acid

An alkaline buffer can be made from a mixture of the base and its conjugate acid, but the formulas for determining pH take a different form.

Learning Objective

  • Calculate the pH of an alkaline buffer system consisting of a weak base and its conjugate acid.


Key Points

    • The pH of bases is usually calculated using the hydroxide ion (OH-) concentration to find the pOH first.
    • The formula for pOH is pOH=-log[OH-]. A base dissociation constant (Kb) indicates the strength of the base.
    • The pH of a basic solution can be calculated by using the equation: pH = 14.00 - pOH.

Terms

  • buffers

    A weak acid or base used to maintain the acidity (pH) of a solution near a chosen value and which prevent a rapid change in pH when acids or bases are added to the solution.

  • alkaline

    Having a pH greater than 7.


Full Text

A base is a substance that decreases the hydrogen ion (H+) concentration of a solution. In the more generalized Brønsted-Lowry definition, the hydroxide ion (OH-) is the base because it is the substance that combines with the proton. Ammonia and some organic nitrogen compounds can combine with protons in solution and act as Brønsted-Lowry bases. These compounds are generally weaker bases than the hydroxide ion because they have less attraction for protons. For example, when ammonia competes with OH- for protons in an aqueous solution, it is only partially successful. It can combine with only a portion of the H+ ions, so it will have a measurable equilibrium constant. Reactions with weak bases result in a relatively low pH compared to strong bases. Bases range from a pH of greater than 7 (7 is neutral like pure water) to 14 (though some bases are greater than 14).

An alkaline buffer can be made from a mixture of a base and its conjugate acid, similar to the way in which weak acids and their conjugate bases can be used to make a buffer.

Ammonia to ammonium ion

Two-dimensional image depicting the association of proton (H+) with the weak base ammonia (NH3) to form its conjugate acid, ammonium ion (NH4+).

Calculating the pH of a Base

The pH of bases is usually calculated using the OH- concentration to find the pOH first. This is done because the H+ concentration is not a part of the reaction, while the OH- concentration is. The formula for pOH is:

pOH=−log([OH−])pOH=-log([O{ H }^{ - }])pOH=−log([OH​−​​])

By multiplying a conjugate acid (such as NH4+) and a conjugate base (such as NH3) the following is given:

Ka×Kb=[H3O+][NH3][NH4+]×[NH4+][OH−][NH3]{ K }_{ a }\times { K }_{ b }=\frac { [{ H }_{ 3 }{ O }^{ + }][N{ H }_{ 3 }] }{ [N{ H }_{ 4 }^{ + }] } \times \frac { [N{ H }_{ 4 }^{ + }][{ OH }^{ - }] }{ [N{ H }_{ 3 }] }K​a​​×K​b​​=​[NH​4​+​​]​​[H​3​​O​+​​][NH​3​​]​​×​[NH​3​​]​​[NH​4​+​​][OH​−​​]​​

Ka×Kb=[H3O+][OH−]=Kw{ K }_{ a }\times { K }_{ b }=[{ H }_{ 3 }{ O }^{ + }][{ OH }^{ - }]={ K }_{ w }K​a​​×K​b​​=[H​3​​O​+​​][OH​−​​]=K​w​​

log(Ka)+log(Kb)=log(Kw){ log(K }_{ a })+{ log(K }_{ b })=log({ K }_{ w })log(K​a​​)+log(K​b​​)=log(K​w​​)

pKa+pKb=pKw=14.00{ pK }_{ a }+{ pK }_{ b }=p{ K }_{ w }=14.00pK​a​​+pK​b​​=pK​w​​=14.00

The pH can be calculated using the formula:

pH=14−pOH{ pH={14}-pOH }pH=14−pOH

Weak bases exist in chemical equilibrium much in the same way as weak acids do. A base dissociation constant (Kb) indicates the strength of the base. For example, when ammonia is put in water, the following equilibrium is set up:

NH3+H2O⇌NH4++OH−N{ H }_{ 3 }+{ H_2O}\rightleftharpoons N{ H }_{ 4 }^{ + } +{OH^-}NH​3​​+H​2​​O⇌NH​4​+​​+OH​−​​

Kb=[NH4+][OH−][NH3]{ K }_{ b }=\frac { [N{ H }_{ 4 }^{ + }][{ OH }^{ - }] }{ [N{ H }_{ 3 }] }K​b​​=​[NH​3​​]​​[NH​4​+​​][OH​−​​]​​

Bases that have a large Kb will ionize more completely, meaning they are stronger bases. NaOH (sodium hydroxide) is a stronger base than (CH3CH2)2NH (diethylamine) which is a stronger base than NH3 (ammonia). As the bases get weaker, the Kb values get smaller.

Example:

Calculate the pH of a buffer solution consisting of 0.051 M NH3 and 0.037 M NH4+. The Kb for NH3 = 1.8 x 10-5.

NH3+H2O⇌NH4++OH−N{ H }_{ 3 }+{ H_2O}\rightleftharpoons N{ H }_{ 4 }^{ + } +{OH^-}NH​3​​+H​2​​O⇌NH​4​+​​+OH​−​​

Kb=[NH4+][OH−][NH3]{ K }_{ b }=\frac { [N{ H }_{ 4 }^{ + }][{ OH }^{ - }] }{ [N{ H }_{ 3 }] }K​b​​=​[NH​3​​]​​[NH​4​+​​][OH​−​​]​​

Assuming the change (x) is negligible to 0.051 M and 0.037 M solutions:

Kb=[0.037][x][0.051]{ K }_{ b }=\frac { [0.037][x] }{ [0.051] }K​b​​=​[0.051]​​[0.037][x]​​

1.8 x 10-5=[0.037][x][0.051]=\frac { [0.037][x] }{ [0.051] }=​[0.051]​​[0.037][x]​​

x = [OH-] = 2.48 x 10-5

pOH = 4.61

pH = 14 - 4.61 = 9.39

[ edit ]
Edit this content
Prev Concept
Calculating Changes in a Buffer Solution
Relative Amounts of Acid and Base
Next Concept
Subjects
  • Accounting
  • Algebra
  • Art History
  • Biology
  • Business
  • Calculus
  • Chemistry
  • Communications
  • Economics
  • Finance
  • Management
  • Marketing
  • Microbiology
  • Physics
  • Physiology
  • Political Science
  • Psychology
  • Sociology
  • Statistics
  • U.S. History
  • World History
  • Writing

Except where noted, content and user contributions on this site are licensed under CC BY-SA 4.0 with attribution required.