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Weak base

Acid-base extraction
Acid-base reaction
Acid-base physiology
Acid-base homeostasis
Acid dissociation constant
Acidity function
Buffer solutions
pH
Proton affinity
Self-ionization of water
Acids:
- Lewis acids
- Mineral acids
- Organic acids
- Strong acids
- Superacids
- Weak acids
Bases:
- Lewis bases
- Organic bases
- Strong bases
- Superbases
- Non-nucleophilic bases
- Weak bases

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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 $K w = [H 3 O +][O H -]$ then, $K_a \times K_b = K_w$

By taking logarithms of both sides of the equation, the following is reached:

l o g K a + l o g K b = l o g K w

Finally, multipying throughout the equation by -1, the equation turns into:

p K a + p K b = p K w = 14.00

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:

$\mathrm{K_b={[NH_4^+][OH^-]\over[NH_3]}}$

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 (CH₃CH₂)₂NH (l) (diethylamine) which is a stronger base than NH₃ (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.

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1 Percentage protonated
2 A typical pH problem
3 Examples
4 See also
5 References

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 K_b 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 = pK_w - pOH,	$\mathrm pH \approx 14.00 - 4.7 = 9.3$
From the equation for percentage protonated with [HB⁺] = x and [B]_initial = .20,	$\mathrm percentage \ protonated = {1.9 \times 10^{-5} \over .20} \times 100\% = .0095\%$