Question #6f85f

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Question #6f85f

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Answer 1 · 2017-07-07T19:58:33

Here's what I get.

Explanation:

a)

$[{OH}^{-}] = 1.83 \times 10^{-7} l mol/L$ $["OH"^"-"] = 1.83 × 10^"-7"color(white)(l) "mol/L"$

$[H_{3} O^{+}] = \frac{1.00 \times 10^{-14}}{1.83 \times 10^{-7}} l mol/L = l 5.46 \times 10^{-8} l mol/L$ $["H"_3"O"^"+"] = (1.00 × 10^"-14")/(1.83 × 10^"-7") color(white)(l)"mol/L" = color(white)(l)5.46 × 10^"-8" color(white)(l)"mol/L"$

b)

$[H_{3} O^{+}] = 8.15 \times 10^{-9} l mol/L$ $["H"_3"O"^"+"] = 8.15 × 10^"-9"color(white)(l) "mol/L"$

$[{OH}^{-}] = \frac{1.00 \times 10^{-14}}{8.15 \times 10^{-9}} l mol/L = l 1.23 \times 10^{-6} l mol/L$ $["OH"^"-"] = (1.00 × 10^"-14")/(8.15 × 10^"-9") color(white)(l)"mol/L" = color(white)(l)1.23 × 10^"-6" color(white)(l)"mol/L"$

c)

$H_{3} O^{+}] = 0.000 665 l mol/L$ $"H"_3"O"^"+"] = "0.000 665" color(white)(l) "mol/L"$

$[{OH}^{-}] = \frac{1.00 \times 10^{-14}}{0.000 665} l mol/L = l 1.50 \times 10^{-11} l mol/L$ $["OH"^"-"] = (1.00 × 10^"-14")/("0.000 665") color(white)(l)"mol/L" = color(white)(l)1.50 × 10^"-11" color(white)(l)"mol/L"$

a) and b) are basic because, in each, $[H_{3} O^{+}] < 1.00 \times 10^{-7} l mol/L$ $["H"_3"O"^"+"] < 1.00 × 10^"-7" color(white)(l)"mol/L"$

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Question #6f85f

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