Acid and base strength

Consider the dissolution reaction of a generic acid HA in water:

$$HA(aq) + {\rm H}_2{\rm O}(l)\rightleftharpoons {\rm H}_3{\rm O}^+(aq) + A^- (aq)$$

where A$^-$ is the base conjugate to HA in the Bronsted-Lowry sense.

The equilibrium constant for the reaction is

$$K = {[{\rm H}_3{\rm O}^+][{\rm A}^-] \over [{\rm HA}]}\equiv K_a$$

Again, since the concentration of water is essentially constant, it is not included in the equilibrium constant expression. Here ``a'' stands for acid. $K_a$ is called the acid ionization constant. For strong acids, $K_a$ will be large because the acid nearly completely dissociates and the denominator is small. For very weak acids it will be small. $K_a$ can also range of many orders of magnitude, so it is useful to define a logarithmic measure of acid ionization strength. This is called the pK$_a$ defined by

$${\rm pK}_{\rm a}= -\log_{10}K_a$$

which ranges typically from -11 for very strong acids to 14 for water.

As an example, consider the acid, HCN:

$${\rm HCN}(aq) + {\rm H}_2{\rm O}(l)\rightleftharpoons {\rm H}_3{\rm O}^+(aq) + {\rm CN}^-(aq)$$

The acid ionization constant is

$$K_a = {[{\rm H}_3{\rm O}^+][{\rm CN}^-] \over [{\rm HCN}]} = 6.17\times 10^{-10}$$

and the pK$_a$ is

$${\rm pK}_{\rm a}= -\log_{10}(6.17\times 10^{-10}) = 9.21$$

By analogy, there is a base ionization constant $K_b$. As an example of $K_b$ consider ammonia:

$${\rm H}_2{\rm O}(l) + {\rm N}{\rm H}_3(aq)\rightleftharpoons {\rm N}{\rm H}_4^+(aq) + {\rm OH}^-(aq)$$

$K_b$ is given by

$$K_b = {[{\rm N}{\rm H}_4^+][{\rm OH}^-] \over [{\rm N}{\rm H}_3]}$$

How do we calculate the value of $K_b$? Note that, since $[{\rm H}_3{\rm O}^+][{\rm OH}^-]=K_w$, the above expression becomes:

$$K_b = {[{\rm N}{\rm H}_4^+]K_{\rm w}\over [{\rm N}{\rm H}_3][{\rm H}_3{\rm O}^+]} = {K_{\rm w}\over K_a}$$

where, here $K_a$ is the acid ionization constant for the acid NH$_4^+$:

$$K_a = {[{\rm N}{\rm H}_4^+][{\rm H}_3{\rm O}^+] \over [{\rm N}{\rm H}_3]}$$

Since the values of $K_a$ are tabulated, we do not need to tabulate $K_b$ separately, as they are related by

$$K_a K_b = K_{\rm w}$$

which reflects the fact that when a base association reaction is added to the corresponding acid dissociation reaction, the overall reaction is the water ionization reaction, and the equilibrium constants must be multiplied:

$${\rm NH}_3(aq) + {\rm H}_2{\rm O}(l) \rightleftharpoons {\rm NH}_4^+(aq) + {\rm OH}^-(aq)\;\;\;\;\;\;\;\;\;\;K_b$$

$${\rm NH}_4^+(aq) + {\rm H}_2{\rm O} \rightleftharpoons {\rm H}_3{\rm O}^+(aq) + {\rm NH}_3(aq)\;\;\;\;\;\;\;\;\;\;K_a$$

The overall reaction is

$$2{\rm H}_2{\rm O}(l) \rightleftharpoons {\rm H}_3{\rm O}^+(aq) + {\rm OH}^-(aq)$$

whose equilibrium constant is the product of $K_a$ and $K_b$, but is also, by definition, $K_w$.

Since $K_a$ for NH$_4^+$ = 5.6 $\times 10^{-10}$, the value of $K_b$ is

$$K_b = {1.0\times 10^{-14} \over 5.6\times 10^{-10}} = 1.8 \times 10^{-5}$$

If $K_a$ is large, corresponding to a strong acid, then the conjugate base will be weak, since $K_b$ will be small and vice-versa.

When two bases compete for protons, the stronger base is favored in equilibrium. This can be seen in the following example:

$${\rm HF}(aq) + {\rm CN}^-(aq)\rightleftharpoons {\rm HCN}(aq) + {\rm F}^-(aq)$$

The equilibrium constant $K$ for this reaction is

$$K = {[{\rm HCN}][[{\rm F}^-] \over [{\rm HF}][{\rm CN}^-]}$$

The two bases competing for proton are F$^-$ and CN$^-$. The above reaction is actually the overall reaction produced from the two acid dissociation reactions:

$${\rm HF}(aq) + {\rm H}_2{\rm O}(l) \rightleftharpoons {\rm H}_3{\rm O}^+(aq) + {\rm F}^-(aq)$$

$${\rm HCN}(aq) + {\rm H}_2{\rm O}(l) \rightleftharpoons {\rm H}_3{\rm O}^+(aq) + {\rm CN}^-(aq)$$

by subtraction of the chemical equations, rather than by addition. When two reactions are subtracted, the equilibrium constants must be divided. The two acid ionization constants are:

$$K_a({\rm HF}) = {[{\rm H}_3{\rm O}^+][{\rm F}^-] \over [{\rm... ...m HCN}) = {[{\rm H}_3{\rm O}^+][{\rm CN}^-] \over [{\rm HCN}]}$$

and their ratio produces the equilibrium constant $K$ for the overall reaction:

$$K = {K_a({\rm HF}) \over K_a({\rm HCN})} = {[{\rm H}_3{\rm ... ...}^-}]} = {[{\rm HCN}][{\rm F}^-] \over [{\rm HF}][{\rm CN}^-]}$$

Then $K$ can be computed from the ratio of the acid ionization constants

$$K = {6.6 \times 10^{-4} \over 6.17 \times 10^{-10}} = 1.1\times 10^{6}$$

Since $K$ is large, the products in the overall reaction are favored, meaning that there will be a large concentration of dissociated HF and not much dissociated HCN. Thus, since CN$^-$ is the stronger base, is consumes most of the protons donated by HF producing the acid HCN and a high concentration of the weak base F$^-$.