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Answer 1 · 2017-01-11T21:01:15

pH = 11.7

We know that

#["OH"^"-"] = "0.005 mol/L"#.

Then

#"pOH" = "-"log["OH"^"-"] = "-"log(0.005) = 2.3#

We also know that

#"pH + pOH" = 14.00#

∴ #"pH" = "14.00 - pOH" = "14.00 - 2.3" = 11.7#