On the calculation of pH of phosphoric acid solution
There is a phosphoric acid solution with a concentration of\ (0.250mol/L\), and its\ (pH\) is desired.

Phosphoric acid\ ((H_ {3} PO_ {3}) \) is a medium-strong acid, which is ionized in aqueous solution in steps. The first ionization equation is\ (H_ {3} PO_ {3}\ rightleftharpoons H ^{ + } + H_ {2} PO_ {3 }^{-}\) ， its first ionization constant\ (K_ {a1}\). The second-step ionization equation is\ (H_ {2} PO_ {3} ^ {-}\ rightleftharpoons H ^{ + } + HPO_ {3} ^ {2 - }\) ， its second-order ionization constant\ (K_ {a2}\). Usually, when calculating\ (pH\), the\ (H ^{ + }\) ， produced by the second-order ionization can be ignored.

When the first ionization reaches equilibrium,\ (H ^{ + }\) concentration is\ (xmol/L\), then\ (H_ {2} PO_ {3 }^{-}\) concentration is also\ (xmol/L\),\ (H_ {3} PO_ {3}\) concentration is\ ((0.250 - x) mol/L\). By the ionization constant expression\ (K_ {a1} =\ frac {c (H ^{ + })\ cdot c (H_ {2} PO_ {3 }^{-})}{ c (H_ {3} PO_ {3 })}\) ， i.e.\ (K_ {a1} =\ frac {x\ cdot x} {0.250 - x}\).

Because\ (H_ {3} PO_ {3}\) is a medium strong acid,\ (0.250 - x\ approx0.250\) (when\ (x\) is small relative to\ (0.250\)), then\ (K_ {a1} =\ frac {x ^ {2}} {0.250}\). Find the\ (K_ {a1}\) value of\ (H_ {3} PO_ {3}\), and substitute it into the above formula to get\ (x ^ {2} = K_ {a1}\ times0.250\), and the solution is\ (x =\ sqrt {K_ {a1}\ times0.250}\), this\ (x\) is\ (H ^{ + }\) concentration.

and\ (pH = -\ lg c (H ^{ + })\) ， Substitute the\ (x\) value to calculate the\ (0.250mol/L\) phosphoric acid solution of\ (pH\).