[H+]=K☆*(☆☆/☆☆☆)
=4*10<☆☆p>-8*(0.05/0....
Ka = [H+][X-]/[HX] = [H+]×0.05/0.05 = 4×10-8
[H+] = 4×10-8
pH = -log[H+] = -log(4×10-8) = 7.4
27. 某一元弱酸 HX 之 Ka 為 4.0×10-8;今配製一緩衝溶液含有 ..-阿摩線上測驗