Weak base

From Wikipedia, the free encyclopedia
Jump to: navigation, search

In chemistry, a weak base is a chemical base that does not ionize fully in an aqueous solution. [1] As Brønsted–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 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). pH 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 pH. 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 results. However, 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.

\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, multiplying 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 dissociation constant (Kb) 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, pH of the solution depends on the H+ concentration, which is related to the OH- concentration by the self-ionization constant (Kw = 1.0x10−14). 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.

Percentage protonated[edit]

As seen above, the strength of a base depends primarily on pH. 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 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[edit]

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 C5H5NH+.

Examples[edit]

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

Simple Facts[edit]

  • An example of a weak base is ammonium. It does not contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions.[2]
  • The position of equilibrium varies from base to base when a weak base reacts with water. The further to the left it is, the weaker the base.[3]

See also[edit]

References[edit]

  1. ^ Brown, Larry N.p, N.p. N.p: Content Technologies , Inc Cram101, 2013. Print.
  2. ^ Atkins, Peter, and Loretta Jones. Chemical Principles: The Quest for Insight, 3rd Ed., New York: W.H. Freeman, 2005.
  3. ^ Clark, Jim. "Strong and Weak Bases."N.p.,2002. Web.

External links[edit]