Weak base

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Acids and bases

Acid dissociation constant Acid-base extraction Acid–base reaction Acid–base titration Dissociation constant Acidity function Buffer solutions pH Proton affinity Amphoterism Self-ionization of water Acid strength
Acid types
Brønsted Lewis Mineral Organic Strong Superacids Weak
Base types
Brønsted Lewis Organic Strong Superbases Non-nucleophilic Weak
v t e

In chemistry, a weak base is a chemical base that does not ionize fully in an aqueous solution. 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:

$K_{a}\times K_{b}={[H_{3}O^{+}][NH_{3}] \over [NH_{4}^{+}]}\times {[NH_{4}^{+}][OH^{-}] \over [NH_{3}]}=[H_{3}O^{+}][OH^{-}]$

Since ${K_{w}}=[H_{3}O^{+}][OH^{-}]$ 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 = 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 dissociation constant (K_b) 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, pH of the solution depends on the H⁺ concentration, which is related to the OH^- concentration by the self-ionization constant (K_w = 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 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.

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:

$B(aq)+H_{2}O(l)\leftrightarrow HB^{+}(aq)+OH^{-}(aq)$

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:

${\mathrm {H_{2}O(l)+C_{5}H_{5}N(aq)\leftrightarrow C_{5}H_{5}NH^{+}(aq)+OH^{-}(aq)}}$

$K_{b}={\mathrm {[C_{5}H_{5}NH^{+}][OH^{-}] \over [C_{5}H_{5}N]}}$

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

This means .0095% of the pyridine is in the protonated form of C₅H₅NH⁺.

Examples

Alanine,
Ammonia, NH₃
Methylamine, CH₃NH₂], C₅H₈O₂

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.

External links

Explanation of strong and weak bases from ChemGuide
Guide to Weak Bases from Georgetown course notes
Article on Acidity of Solutions of Weak Bases from Intute

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