• Esempio.
  Confrontare la solubilita' di $CaF_2$ in acqua pura e in una soluzione tampone a $pH=1$. Per $CaF_2$ si ha ${K_{PS}}=3.9\TimesTenTo{-11}$, per $HF$ si ha $K_A=6.8\TimesTenTo{-4}$.
  In acqua pura:
  
  $$\begin{array}{lllll} &CaF_2&\stackrel{{K_{PS}}}{=}&{Ca^{2+}}&+2{F^-}\\ t=0&&&0&0\\ t=\infty&&&S&2S \end{array}$$
  
  
  $$\begin{eqnarray*} {K_{PS}}&=&S\Parenthesis{2S}^2\\ S&=&\Parenthesis{\frac{{K_{PS}}}{4}}^{\frac{1}{3}}\\ &=&2.14\TimesTenTo{-4}\;mol/l \end{eqnarray*}$$
  
  
  A $pH=1$ si ha:
  $$\begin{eqnarray*} CaF_2&\stackrel{{K_{PS}}}{=}&{Ca^{2+}}+2{F^-}\\ 2{F^-}+2{H_3O... ...2+2{H_3O^{+}}&\stackrel{{K_{PS}}/K_A^2}{=}&{Ca^{2+}}+2HF+2{H_2O} \end{eqnarray*}$$
  
  
  
  
  $$\begin{array}{lllllll} &CaF_2&+2{H_3O^{+}}&\stackrel{{K_{PS}... ...&&\TenTo{-pH}&&0&0&\\ t=\infty&&\TenTo{-pH}&&S&2S& \end{array}$$
  
  
  $$\begin{eqnarray*} \frac{{K_{PS}}}{K_A^2}&=&\frac{S\Parenthesis{2S}^2}{\Parenthes... ...To{-2pH}}{K_A^2}}^{\frac{1}{3}}\\ &=&5.95\TimesTenTo{-3}\;mol/l \end{eqnarray*}$$