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Hey guys please help me |
| Answer» PH = 3.28PH = -LoG {H^+}{H+} = 10 ^-PH{H+} =10^-3.28 = 1.0 x 10 ^-3.28......................(1).{H+} + {OH-} = 1.0 X 10^-14...........(2).So..ConcenTraTioN OF OH-FroM EquaTioN ..(1).{OH-} = 1.0 x 10^-14/{H+}FroM EquaTioN ...(2).{OH-} = 1.0 x 10^-14 / 1.0 x 10^-3.28 = 5.2480746e - 18 WILL Be.=5.3 x 10 ^-10 M | |