Weak base
- Acid-base extraction
- Acid-base reaction
- Acid-base physiology
- Acid-base homeostasis
- Dissociation constant
- Acidity function
- Buffer solutions
- pH
- Proton affinity
- Self-ionization of water
- Acids:
- Bases:
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:
- <math>\mbox{pH} = -\log_{10} \left[ \mbox{H}^+ \right]</math>
Since bases are proton acceptors, the base receives a hydrogen ion from water, H_{2}O, 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.
- <math>\mbox{pOH} = -\log_{10} \left[ \mbox{OH}^- \right]</math>
By multiplying a conjugate acid (such as NH_{4}^{+}) and a conjugate base (such as NH_{3}) the following is given:
- <math> K_a \times K_b = {[H_3O^+][NH_3]\over[NH_4^+]} \times {[NH_4^+][OH^-]\over[NH_3]} = [H_3O^+][OH^-]</math>
Since <math>{K_w} = [H_3O^+][OH^-]</math> then, <math>K_a \times K_b = K_w</math>
By taking logarithms of both sides of the equation, the following is reached:
- <math>logK_a + logK_b = logK_w</math>
Finally, multipying throughout the equation by -1, the equation turns into:
- <math>pK_a + pK_b = pK_w = 14.00</math>
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:
- <math>\mathrm{K_b={[NH_4^+][OH^-]\over[NH_3]}}</math>
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 (CH_{3}CH_{2})_{2}NH (l) (diethylamine) which is a stronger base than NH_{3} (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:
- <math>B(aq) + H_2O(l) \leftrightarrow HB^+(aq) + OH^-(aq)</math>
B represents the base.
- <math>Percentage\ protonated = {molarity\ of\ HB^+ \over\ initial\ molarity\ of\ B} \times 100\% = {[{HB}^+]\over [B]_{initial}} {\times 100\%}</math>
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, C_{5}H_{5}N. The K_{b} for C_{5}H_{5}N is 1.8 x 10^{-9}.
First, write the proton transfer equilibrium:
- <math>\mathrm{H_2O(l) + C_5H_5N(aq) \leftrightarrow C_5H_5NH^+ (aq) + OH^- (aq)}</math>
- <math>K_b=\mathrm{[C_5H_5NH^+][OH^-]\over [C_5H_5N]}</math>
The equilibrium table, with all concentrations in moles per liter, is
C_{5}H_{5}N | C_{5}H_{6}N^{+} | 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 | <math>K_b=\mathrm {1.8 \times 10^{-9}} = {x \times x \over .20-x}</math> |
We can assume that x is so small that it will be meaningless by the time we use significant figures. | <math>\mathrm {1.8 \times 10^{-9}} \approx {x^2 \over .20}</math> |
Solve for x. | <math>\mathrm x \approx \sqrt{.20 \times (1.8 \times 10^{-9})} = 1.9 \times 10^{-5}</math> |
Check the assumption that x << .20 | <math>\mathrm 1.9 \times 10^{-5} \ll .20</math>; so the approximation is valid |
Find pOH from pOH = -log [OH^{-}] with [OH^{-}]=x | <math>\mathrm pOH \approx -log(1.9 \times 10^{-5}) = 4.7 </math> |
From pH = pK_{w} - pOH, | <math>\mathrm pH \approx 14.00 - 4.7 = 9.3</math> |
From the equation for percentage protonated with [HB^{+}] = x and [B]_{initial} = .20, | <math>\mathrm percentage \ protonated = {1.9 \times 10^{-5} \over .20} \times 100\% = .0095\% </math> |
This means .0095% of the pyridine is in the protonated form of C_{5}H_{6}N^{+}.
Examples
- Alanine, C_{3}H_{5}O_{2}NH_{2}
- Ammonia, NH_{3}
- Methylamine, CH_{3}NH_{2}
- Pyridine, C_{5}H_{5}N
Other weak bases are essentially any bases not on the list of strong bases.
See also
References
- Atkins, Peter, and Loretta Jones. Chemical Principles: The Quest for Insight, 3rd Ed., New York: W.H. Freeman, 2005.