Weak base

In chemistry, a weak base is a chemical base that does not ionize fully in an aqueous solution. As Bronsted-Lowry bases are proton acceptors, a weak base may also be defined as a chemical base in which protonation is incomplete. This results in a relatively low pH level 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). The pH level has the formula: :$mbox\{pH\}\; =\; -log\_\{10\}\; left\; [\; mbox\{H\}^+\; ight]$Since bases are proton acceptors, the base receives a hydrogen ion from water, H₂O, and the remaining H⁺ concentration in the solution determines the pH level. Weak bases will have a higher H⁺ concentration because they are less completely protonated than stronger bases and, therefore, more hydrogen ions remain in the solution. If you plug in a higher H⁺ concentration into the formula, a low pH level results. However, the pH level of bases is usually calculated using the OH^- concentration to find the pOH level first. This is done because the H⁺ concentration is not a part of the reaction, while the OH^- concentration is.:$mbox\{pOH\}\; =\; -log\_\{10\}\; left\; [\; mbox\{OH\}^-\; ight]$

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

:$K\_a\; imes\; K\_b\; =\; \{\; [H\_3O^+]\; [NH\_3]\; over\; [NH\_4^+]\; \}\; imes\; \{\; [NH\_4^+]\; [OH^-]\; over\; [NH\_3]\; \}\; =\; [H\_3O^+]\; [OH^-]$

Since $\{K\_w\}\; =\; [H\_3O^+]\; [OH^-]$ then, "$K\_a\; imes\; K\_b\; =\; K\_w$"

By taking logarithms of both sides of the equation, the following is reached:

:$logK\_a\; +\; logK\_b\; =\; logK\_w$

Finally, multipying throughout the equation by -1, the equation turns into:

:$pK\_a\; +\; pK\_b\; =\; pK\_w\; =\; 14.00$

After acquiring pOH from the previous pOH formula, pH can be calculated using the formula pH = pK_w - pOH where pK_w = 14.00.

Weak bases exist in chemical equilibrium much in the same way as weak acids do, with a Base Ionization Constant (K_b) (or the Base Dissociation Constant) indicating the strength of the base. For example, when ammonia is put in water, the following equilibrium is set up:

:$mathrm\{K\_b=\{\; [NH\_4^+]\; [OH^-]\; over\; [NH\_3]$

Bases that have a large K_b will ionize more completely and are thus stronger bases. As stated above, the pH of the solution depends on the H⁺ concentration, which is related to the OH^- concentration by the Ionic Constant of water (K_w = 1.0x10^-14) (See article Self-ionization of water.) A strong base has a lower H⁺ concentration because they are fully protonated and less hydrogen ions remain in the solution. A lower H⁺ concentration also means a higher OH^- concentration and therefore, a larger K_b.

NaOH (s) (sodium hydroxide) is a stronger base than (CH₃CH₂)₂NH (l) (diethylamine) which is a stronger base than NH₃ (g) (ammonia). As the bases get weaker, the smaller the K_b values become. The pie-chart representation is as follows:
* purple areas represent the fraction of OH- ions formed
* red areas represent the cation remaining after ionization
* yellow areas represent dissolved but non-ionized molecules.

Percentage protonated

As seen above, the strength of a base depends primarily on the pH level. To help describe the strengths of weak bases, it is helpful to know the percentage protonated-the percentage of base molecules that have been protonated. A lower percentage will correspond with a lower pH level because both numbers result from the amount of protonation. A weak base is less protonated, leading to a lower pH and a lower percentage protonated.

The typical proton transfer equilibrium appears as such:

:$B(aq)\; +\; H\_2O(l)\; leftrightarrow\; HB^+(aq)\; +\; OH^-(aq)$

B represents the base.

:$Percentage\; protonated\; =\; \{molarity\; of\; HB^+\; over\; initial\; molarity\; of\; B\}\; imes\; 100\%\; =\; \{\; [\{HB\}^+]\; over\; [B]\; \_\{initial\; \{\; imes\; 100\%\}$

In this formula, [B] _initial is the initial molar concentration of the base, assuming that no protonation has occurred.

A typical pH problem

Calculate the pH and percentage protonation of a .20 M aqueous solution of pyridine, C₅H₅N. The K_b for C₅H₅N is 1.8 x 10^-9.

First, write the proton transfer equilibrium:

:$mathrm\{H\_2O(l)\; +\; C\_5H\_5N(aq)\; leftrightarrow\; C\_5H\_5NH^+\; (aq)\; +\; OH^-\; (aq)\}$

:$K\_b=mathrm\{\; [C\_5H\_5NH^+]\; [OH^-]\; over\; [C\_5H\_5N]\; \}$

The equilibrium table, with all concentrations in moles per liter, is

This means .0095% of the pyridine is in the protonated form of C₅H₆N⁺.

Examples

*Alanine, C₃H₅O₂NH₂
*Ammonia, NH₃
*Methylamine, CH₃NH₂
*Pyridine, C₅H₅N

Other weak bases are essentially any bases not on the list of strong bases.

ee also

* Strong base
* Weak acid

References

*Atkins, Peter, and Loretta Jones. Chemical Principles: The Quest for Insight, 3rd Ed., New York: W.H. Freeman, 2005.

External links

*http://wine1.sb.fsu.edu/chm1046/notes/AcidBase/WeakBase/WeakBase.htm
*http://www.chemguide.co.uk/physical/acidbaseeqia/bases.html
*http://bouman.chem.georgetown.edu/S02/lect16/lect16.htm
*http://www.intute.ac.uk/sciences/reference/plambeck/chem1/p01154.htm