Weak base

Bases yield solutions in which the hydrogen ion activity is lower than it is in pure water, i.e., the solution is said to have a pH greater than 7.0 at standard conditions, potentially as high as 14 (and even greater than 14 for some bases). The formula for pH is:

${\displaystyle {\mbox{pH}}=-\log _{10}\left[{\mbox{H}}^{+}\right]}$

Bases are proton acceptors; a base will receive a hydrogen ion from water, H₂O, and the remaining H⁺ concentration in the solution determines pH. A weak base will have a higher H⁺ concentration than a stronger base because it is less completely protonated than a stronger base and, therefore, more hydrogen ions remain in its solution. Given its greater H⁺ concentration, the formula yields a lower pH value for the weak base. However, pH of bases is usually calculated in terms of the OH⁻ concentration. This is done because the H⁺ concentration is not a part of the reaction, whereas the OH⁻ concentration is. The pOH is defined as:

${\displaystyle {\mbox{pOH}}=-\log _{10}\left[{\mbox{OH}}^{-}\right]}$

If we multiply the equilibrium constants of a conjugate acid (such as NH₄⁺) and a conjugate base (such as NH₃) we obtain:

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

As ${\displaystyle {K_{w}}=[H_{3}O^{+}][OH^{-}]}$ is just the self-ionization constant of water, we have ${\displaystyle K_{a}\times K_{b}=K_{w}}$

Taking the logarithm of both sides of the equation yields:

${\displaystyle logK_{a}+logK_{b}=logK_{w}}$

Finally, multiplying both sides by -1, we obtain:

${\displaystyle pK_{a}+pK_{b}=pK_{w}=14.00}$

With pOH obtained from the pOH formula given above, the pH of the base can then be calculated from ${\displaystyle pH=pK_{w}-pOH}$, where pK_w = 14.00.

A weak base persists in chemical equilibrium in much the same way as a weak acid does, 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:

${\displaystyle \mathrm {K_{b}={[NH_{4}^{+}][OH^{-}] \over [NH_{3}]}} }$

A base that has a large K_b will ionize more completely and is thus a stronger base. As shown above, the pH of the solution, which depends on the H⁺ concentration, increases with increasing OH⁻ concentration; a greater OH⁻ concentration means a smaller H⁺ concentration, therefore a greater pH. Strong bases have smaller H⁺ concentrations because they are more fully protonated, leaving fewer hydrogen ions in the solution. A smaller H⁺ concentration means a greater OH⁻ concentration and, therefore, a greater K_b and a greater pH.

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.^[1]

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.^[2]

The typical proton transfer equilibrium appears as such:

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

B represents the base.

${\displaystyle Percentage\ protonated={molarity\ of\ HB^{+} \over \ initial\ molarity\ of\ B}\times 100\%={[{HB}^{+}] \over [B]_{initial}}{\times 100\%}}$

In this formula, [B]_initial is the initial molar concentration of the base, assuming that no protonation has occurred.

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.^[3]

First, write the proton transfer equilibrium:

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

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

The equilibrium table, with all concentrations in moles per liter, is

Substitute the equilibrium molarities into the basicity constant	${\displaystyle K_{b}=\mathrm {1.8\times 10^{-9}} ={x\times x \over .20-x}}$
We can assume that x is so small that it will be meaningless by the time we use significant figures.	${\displaystyle \mathrm {1.8\times 10^{-9}} \approx {x^{2} \over .20}}$
Solve for x.	${\displaystyle \mathrm {x} \approx {\sqrt {.20\times (1.8\times 10^{-9})}}=1.9\times 10^{-5}}$
Check the assumption that x << .20	${\displaystyle \mathrm {1} .9\times 10^{-5}\ll .20}$; so the approximation is valid
Find pOH from pOH = -log [OH−] with [OH−]=x	${\displaystyle \mathrm {p} OH\approx -log(1.9\times 10^{-5})=4.7}$
From pH = pKw - pOH,	${\displaystyle \mathrm {p} H\approx 14.00-4.7=9.3}$
From the equation for percentage protonated with [HB+] = x and [B]initial = .20,	${\displaystyle \mathrm {p} ercentage\ protonated={1.9\times 10^{-5} \over .20}\times 100\%=.0095\%}$

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

• Alanine
• Ammonia, NH₃
• Methylamine, CH₃NH₂
• Ammonium hydroxide, NH₄OH

• An example of a weak base is ammonia. It does not contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions.^[4]
• 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.^[5]
• When there is a hydrogen ion gradient between two sides of the biological membrane, the concentration of some weak bases are focused on only one side of the membrane.^[6] Weak bases tend to build up in acidic fluids.^[6] Acid gastric contains a higher concentration of weak base than plasma.^[6] Acid urine, compared to alkaline urine, excretes weak bases at a faster rate.^[6]

• Strong base
• Weak acid