0.005 mol / L = 0.01 OH- ﻋﺪد ﻣﻮﻻت = OH- ﺗﺮﻛﯿﺰ 2= اﻟ اﻟ ﻠﻰ POH = - log [OH-] = - log (0.005) = 2.306 PH = 14 – POH = 14 – 2.306 = 11.694 ١٨٤ 185