浓度为0.1摩尔每升的磷酸钠溶液的PH 是多少<br>问题描述:<br>Na3PO4水解有三级:<br>PO4 3- + H2O = HPO4 2- + OH- Kh1 = Kw / Ka3 = 10^-14 / 2.2*10^-13 = 0.045<br>HPO4 2- + H2O = H2PO4 - + OH- Kh2 = Kw / Ka2 = 10^-14 / 6.2*10^-8 = 1.61*10^-7<br>因为Kh1 > 1000 Kh2,所以可以忽略第二级及第三级的水解,以第一级计算:<br>PO4 3- + H2O = HPO4 2- + OH-<br>0.1-x x x<br>x2 / (0.1-x) = 0.045 解出x = [OH-] pOH = -lgx<br>pH = 14 - pOH即得.<br>展开全文阅读<br>也许感兴趣的知识<br>© 2021 wesiedu.com | webmaster@wesiedu.com(低俗回答举报邮箱) | 16 q. 0.015 s.