Cal. PH and POH for each of the following solutions at 25°c.
i) 1.0×10^-3 OH-
I) 1.0m H+
pH + pOH = 14
pH = - lg [H+]
pOH = - lg [OH-]
1) pOH = - lg [1.0×10-3] = 3
pH = 14 - 3 = 11
2) pH = - lg [1.0] = 0
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