【摘要】 <正> 硅胶,由于它的特殊表面性质,具有广泛的用途。在水溶液中它的表面为羟基所复盖,因此研究硅胶表面硅羟基的性质具有重要意义.本文用离子选择电极研究Ca²⁺,Na⁺和K⁺在硅胶表面上的结合方式并求出反应的平衡常数. 在强碱性溶液中,硅胶与Ca²⁺发生两种反应:一是硅胶表面硅羟基与Ca²⁺作用;另一是硅胶溶解生成硅酸根离子,并与Ca²⁺作用生成硅酸钙沉淀.但在中性和弱碱性条件下,硅胶与Ca²⁺只发生表面反应,形成表面配合化合物,而不产生硅酸钙沉淀. 本实验表明:在pH=7～10,[Ca²⁺]<10^-2M条件下,Ca²⁺与硅胶表面反应时H_rel⁺/Ca_c²⁺≈2(Ca_c²⁺为硅胶表面硅羟基上结合的Ca²⁺量,H_rel⁺为Ca²⁺在硅胶表面硅羟基上结合时释放出来的H⁺量).也就是说硅胶表面硅羟基结合一个Ca²⁺就释放出两个H⁺,反应式可表示如下:更多还原

【Abstract】 The reaction of Ca²⁺,Na⁺ and K⁺ with surface silanol groups of silica gel was studied with ion selective electrodes.It has been found that in Ca（NO₃）₂,NaCl and KCl solutions within pH range of 7～10,surface complexes of（≡Si·O）₂Ca,≡Si·ONa and≡Si·OK were formed respectivelv.The formation constants,K,of these complexes were found to be dependent on the concentration of the surface complex,{C},and could be expressed by the equation:pK=pK⁰+α lg （1+b{C}）,wherein α and b being characteristic parameters of each metal ion and K⁰,the formation constant when the concentration effect was eliminated,that is,with{C} equal to naught,then pK=pK⁰.The value of pK⁰ for（≡Si·O）₂Ca was found to be 12;for≡Si·ONa,5.7 and for≡Si·OK,5.4.更多还原