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{{Acids and bases}}
In chemistry, a '''weak base''' is a [[chemical]] [[base (chemistry)|base]] that does not [[ionize]] fully in an [[aqueous solution]]. As [[Brønsted–Lowry base]]s 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 [[Base (chemistry)#Strong bases|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:
:<math>\mbox{pH} = -\log_{10} \left[ \mbox{H}^+ \right]</math>
Since bases are [[proton]] acceptors, the base receives a hydrogen ion from water, H<sub>2</sub>O, and the remaining H<sup>+</sup> [[concentration]] in the solution determines pH. Weak bases will have a higher H<sup>+</sup> 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<sup>+</sup> concentration into the formula, a low pH results. However, pH of bases is usually calculated using the OH<sup>-</sup> concentration to find the pOH first. This is done because the H<sup>+</sup> concentration is not a part of the reaction, while the OH<sup>-</sup> concentration is.
:<math>\mbox{pOH} = -\log_{10} \left[ \mbox{OH}^- \right]</math>


By multiplying a conjugate acid (such as NH<sub>4</sub><sup>+</sup>) and a conjugate base (such as NH<sub>3</sub>) the following is given:
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:<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, multiplying 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<sub>w</sub> - pOH''' where pK<sub>w</sub> = 14.00.
 
Weak bases exist in [[chemical equilibrium]] much in the same way as [[weak acid]]s do, with a '''[[base dissociation constant]] (K<sub>b</sub>)''' 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<sub>b</sub> will ionize more completely and are thus stronger bases. As stated above, pH of the solution depends on the H<sup>+</sup> concentration, which is related to the OH<sup>-</sup> concentration by the '''[[self-ionization constant]] (K<sub>w</sub> = 1.0x10<sup>−14</sup>)'''. A strong base has a lower H<sup>+</sup> concentration because they are fully protonated and less hydrogen ions remain in the solution. A lower H<sup>+</sup> concentration also means a higher OH<sup>-</sup> concentration and therefore, a larger K<sub>b</sub>.
 
<!-- Image with unknown copyright status removed: [[Image: basestrength.jpg]] -->
 
NaOH (s) (sodium hydroxide) is a stronger base than (CH<sub>3</sub>CH<sub>2</sub>)<sub>2</sub>NH (l) ([[diethylamine]]) which is a stronger base than NH<sub>3</sub> (g) (ammonia). As the bases get weaker, the smaller the K<sub>b</sub> 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 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:
 
:<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]<sub>initial</sub> 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<sub>5</sub>H<sub>5</sub>N. The K<sub>b</sub> for C<sub>5</sub>H<sub>5</sub>N is 1.8 x 10<sup>−9</sup>.
 
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
 
{| width:75%; height:200px border="1"
|+
|-style="height:40px"
! !! C<sub>5</sub>H<sub>5</sub>N !! C<sub>5</sub>H<sub>6</sub>N<sup>+</sup> !! OH<sup>-</sup>
|-
! initial normality
| .20 || 0 || 0
|-
! change in normality
| -x || +x || +x
|-
! equilibrium normality
| .20 -x || x || x
|}
 
{| width:75%; height:200px border="1"
|-
| 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<sup>-</sup>] with [OH<sup>-</sup>]=x
| <math>\mathrm pOH \approx -log(1.9 \times 10^{-5}) = 4.7 </math>
|-
| From pH = pK<sub>w</sub> - pOH,
| <math>\mathrm pH \approx 14.00 - 4.7 = 9.3</math>
|-
| From the equation for percentage protonated with [HB<sup>+</sup>] = x and [B]<sub>initial</sub> = .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<sub>5</sub>H<sub>5</sub>NH<sup>+</sup>.
 
==Examples==
* [[Alanine]],
* [[Ammonia]], NH<sub>3</sub>
* [[Methylamine]], CH<sub>3</sub>NH<sub>2</sub>], C<sub>5</sub>H<sub>8</sub>O<sub>2</sub>
 
Other weak bases are essentially any bases not on the list of [[strong base]]s.
 
==See also==
* [[Strong base]]
* [[Weak acid]]
 
==References==
{{reflist}}
* Atkins, Peter, and Loretta Jones. Chemical Principles: The Quest for Insight, 3rd Ed., New York: W.H. Freeman, 2005.
 
==External links==
* [http://www.chemguide.co.uk/physical/acidbaseeqia/bases.html Explanation of strong and weak bases] from ChemGuide
* [http://bouman.chem.georgetown.edu/S02/lect16/lect16.htm Guide to Weak Bases from Georgetown course notes]
* [http://www.intute.ac.uk/sciences/reference/plambeck/chem1/p01154.htm Article on Acidity of Solutions of Weak Bases] from Intute
 
[[Category:Bases]]

Latest revision as of 23:34, 7 January 2015


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