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:
 * $$\mbox{pH} = -\log_{10} \left[ \mbox{H}^+ \right]$$

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.
 * $$\mbox{pOH} = -\log_{10} \left[ \mbox{OH}^- \right]$$

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


 * $$ K_a \times K_b = {[H_3O^+][NH_3]\over[NH_4^+]} \times {[NH_4^+][OH^-]\over[NH_3]} = [H_3O^+][OH^-]$$

Since $${K_w} = [H_3O^+][OH^-]$$ then, $$K_a \times K_b = K_w$$

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 = pKw - pOH where pKw = 14.00.

Weak bases exist in chemical equilibrium much in the same way as weak acids do, with a Base Ionization Constant (Kb) (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 Kb 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 (Kw = 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 Kb.

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 Kb 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.


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

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 Kb 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.