# Weak base

Jump to: navigation, search

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

C5H5N C5H6N+ OH-
initial normality .20 0 0
change in normality -x +x +x
equilibrium normality .20 -x x x

 Substitute the equilibrium molarities into the basicity constant $K_b=\mathrm {1.8 \times 10^{-9}} = {x \times x \over .20-x}$ We can assume that x is so small that it will be meaningless by the time we use significant figures. $\mathrm {1.8 \times 10^{-9}} \approx {x^2 \over .20}$ Solve for x. $\mathrm x \approx \sqrt{.20 \times (1.8 \times 10^{-9})} = 1.9 \times 10^{-5}$ Check the assumption that x << .20 $\mathrm 1.9 \times 10^{-5} \ll .20$; so the approximation is valid Find pOH from pOH = -log [OH-] with [OH-]=x $\mathrm pOH \approx -log(1.9 \times 10^{-5}) = 4.7$ From pH = pKw - pOH, $\mathrm pH \approx 14.00 - 4.7 = 9.3$ From the equation for percentage protonated with [HB+] = x and [B]initial = .20, $\mathrm percentage \ protonated = {1.9 \times 10^{-5} \over .20} \times 100\% = .0095\%$

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

## Examples

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

## References

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