Weak base

pH, Kb, and Kw

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:

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

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:

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

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

${\displaystyle K_a\times K_b=\frac{[H_3{O}^{+}][N{H}_3]}{[N{H}_4^{+}]}\times \frac{[N{H}_4^{+}][O{H}^{-}]}{[N{H}_3]}=[H_3{O}^{+}][O{H}^{-}]}$

As ${\displaystyle K_w=[H_3{O}^{+}][O{H}^{-}]}$ 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:

${\displaystyle log{K}_a+log{K}_b=log{K}_w}$

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

${\displaystyle p{K}_a+p{K}_b=p{K}_w=14.00}$

With pOH obtained from the pOH formula given above, the pH of the base can then be calculated from ${\displaystyle pH=p{K}_w-pOH}$, 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:

${\displaystyle {\mathrm{K}}_{\mathrm{b}}=\frac{[\mathrm{N}{\mathrm{H}}_4^{+}][\mathrm{O}{\mathrm{H}}^{-}]}{[\mathrm{N}{\mathrm{H}}_3]}}$

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.