Question #aca6e

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Answer 1 · 2017-01-11T21:01:15

pH = 11.7

We know that

$[{OH}^{-}] = 0.005 mol/L$ $["OH"^"-"] = "0.005 mol/L"$ .

Then

$pOH = - log [{OH}^{-}] = - log (0.005) = 2.3$ $"pOH" = "-"log["OH"^"-"] = "-"log(0.005) = 2.3$

We also know that

$pH + pOH = 14.00$ $"pH + pOH" = 14.00$

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